Matrices and determinants

Class 11 Mathematics, unit 2, book pages 30–72

A matrix is a rectangular array of numbers, and a determinant is a single number attached to a square matrix. This unit teaches the types of matrices and their algebra (sum, product, transpose), then determinants with their properties, and the adjoint and inverse of a matrix. It then uses row operations to find echelon forms, rank and inverses, and solves systems of linear equations by matrix inversion, Gauss elimination, Gauss-Jordan and Cramer's rule. Matrices are used in daily life, for example to encode messages (cryptography) and to describe the pixels and shifts of a screen.

Learning outcomes (7)
  • Apply matrix operations (addition, subtraction and multiplication of matrices) with real and complex entries.
  • Evaluate determinants of matrices by using cofactors and properties of determinants.
  • Use row operations to find the inverse and rank of a matrix.
  • Explain a consistent and an inconsistent system of linear equations and demonstrate through examples.
  • Solve a system of by non-homogeneous linear equations by using the matrix inversion method and Cramer's rule.
  • Solve a system of three homogeneous linear equations in three unknowns using the Gaussian elimination method.
  • Apply concepts of matrices to real world problems such as graphic design, data encryption, seismic analysis, cryptography, transformation of geometric shapes and social network analysis.

2.1, 2.1.1 and 2.1.2Matrices

A matrix is an array of numbers arranged in horizontal and vertical lines and enclosed within square brackets. Matrices are usually denoted by capital letters. The horizontal lines are the rows of the matrix and the vertical lines are its columns.

Each number in the matrix is called an element or entry. Every element has a definite position, given by the number of the row first and then the number of the column in which it lies. In the matrix above, the element is where the second row and the third column meet.

A matrix of 2 rows and 3 columns. The element 8 lies in row 2 and column 3, so it is .

In general, the element in the th row and the th column is written , and the matrix is written . For example, in

the element is in the first row and first column (), and is in the first row and second column (), and so on.

Order of a matrix (2.1.1)

The order of a matrix tells how many rows and columns it has. If a matrix has rows and columns, its order is (read "-by-"). We always write the number of rows first and the number of columns second.

Multiplying by gives the total number of elements. For example, the matrix above has 2 rows and 3 columns, so its order is , often written , and it has elements.

Equality of matrices (2.1.2)

Two matrices are equal if and only if both have the same order and the same corresponding elements.

Take

Both are of order and have the same corresponding elements, so . Equality is about the values of the entries, not the way they are written. Hence

2.1.3 to 2.1.14Types of matricesOften examined

Matrices are named by the shape or by the pattern of their entries. The book describes the following types. For each type, learn the definition and one small example.

Row matrix or row vector (2.1.3)

A matrix with only one row is a row matrix or row vector. Examples:

In general, a row matrix with columns is written

Column matrix or column vector (2.1.4)

A matrix with only one column is a column matrix or column vector. Examples:

In general, a column matrix with rows is

Square matrix (2.1.5)

A matrix with an equal number of rows and columns is a square matrix. If it has rows and columns, its order is . Examples:

Rectangular matrix (2.1.6)

If the number of rows is not equal to the number of columns (), the matrix is rectangular. Examples:

Zero or null matrix (2.1.7)

If all the elements of a matrix are zero, it is a null or zero matrix. It is usually written . Examples:

Diagonal matrix (2.1.8)

A square matrix in which all elements except those on the main diagonal are zero, and the main diagonal has at least one non-zero element, is a diagonal matrix. The main diagonal runs from the top-left to the bottom-right corner ().

In symbols, is diagonal if for and for at least one , where . Examples:

Scalar matrix (2.1.9)

A diagonal matrix in which all diagonal elements are the same non-zero number is a scalar matrix. That is, for :

Identity or unit matrix (2.1.10)

A scalar matrix in which all the diagonal elements are 1 is the identity matrix or unit matrix. It is written or simply . For :

The identity matrix is written :

Upper triangular matrix (2.1.11)

A square matrix in which all elements below the main diagonal are zero is an upper triangular matrix. That is, is upper triangular if whenever . Examples:

Lower triangular matrix (2.1.13)

A square matrix in which all elements above the main diagonal are zero is a lower triangular matrix. That is, is lower triangular if whenever . Examples:

Triangular matrix (2.1.14)

A square matrix that is either upper triangular or lower triangular is called a triangular matrix.

Summary of the types of matrices
TypeOrder neededCondition on elements
Row matrixone row
Column matrixone column
Square matrixrows columns
Rectangular matrix, rows columns
Zero matrixanyall
Diagonal matrix for , some
Scalar matrixdiagonal, all
Identity matrixdiagonal, all
Upper triangular for
Lower triangular for

2.1.15 to 2.1.17Transpose, symmetric and skew-symmetric matricesOften examined

Transpose of a matrix (2.1.15)

If A is a matrix of order m\times n, the matrix obtained by interchanging its rows and columns (rows become columns, or columns become rows) is the transpose of A, denoted A^{t}. The order of A^{t} is n\times m.

For example, if

A=\begin{bmatrix}2 & 1 & 6 \\ 0 & 2 & 3\end{bmatrix}_{2\times3}\quad\text{then}\quad A^{t}=\begin{bmatrix}2 & 0 \\ 1 & 2 \\ 6 & 3\end{bmatrix}_{3\times2}

The first row (2,1,6) of A became the first column of A^t, and the second row (0,2,3) became its second column.

Symmetric matrix (2.1.16)

A square matrix A is symmetric if A=A^{t}. For example,

A=\begin{bmatrix}1 & 2 & 5 \\ 2 & 6 & -4 \\ 5 & -4 & 3\end{bmatrix},\qquad A^{t}=\begin{bmatrix}1 & 2 & 5 \\ 2 & 6 & -4 \\ 5 & -4 & 3\end{bmatrix}

Since A=A^{t}, the matrix A is symmetric. The matrix is a mirror image of itself across the main diagonal.

Skew-symmetric matrix (2.1.17)

A square matrix A is skew-symmetric if A=-A^{t}. For example, for

A=\begin{bmatrix}0 & 2 & -6 \\ -2 & 0 & 5 \\ 6 & -5 & 0\end{bmatrix}

we get

A^{t}=\begin{bmatrix}0 & -2 & 6 \\ 2 & 0 & -5 \\ -6 & 5 & 0\end{bmatrix}=-\begin{bmatrix}0 & 2 & -6 \\ -2 & 0 & 5 \\ 6 & -5 & 0\end{bmatrix}=-A

So A is skew-symmetric.

Worked example: finding unknown entries so that a matrix is symmetric or skew-symmetric
(i) Find x,y,z so that A=\begin{bmatrix}1 & x & 3 \\ 2 & 4 & y \\ z & -5 & 6\end{bmatrix} is symmetric. (ii) Find p,q,r so that B=\begin{bmatrix}0 & p & -3 \\ 4 & 0 & q \\ r & 2 & 0\end{bmatrix} is skew-symmetric.
  1. (i) Symmetric means A=A^{t}, that is a_{ij}=a_{ji}. Pair the entries placed symmetrically about the main diagonal.
  2. a_{12}=a_{21}, a_{13}=a_{31} and a_{23}=a_{32}.
    x=2,\qquad z=3,\qquad y=-5
  3. Check: the matrix becomes the one below, which equals its transpose.
    A=\begin{bmatrix}1 & 2 & 3 \\ 2 & 4 & -5 \\ 3 & -5 & 6\end{bmatrix}=A^{t}
  4. (ii) Skew-symmetric means a_{ij}=-a_{ji} (and the diagonal is zero, which B already has).
  5. a_{12}=-a_{21}, a_{13}=-a_{31} and a_{23}=-a_{32}.
    p=-4,\qquad -3=-r\Rightarrow r=3,\qquad q=-2
  6. Check: each entry is the negative of its mirror entry.
    B=\begin{bmatrix}0 & -4 & -3 \\ 4 & 0 & -2 \\ 3 & 2 & 0\end{bmatrix},\quad B^{t}=\begin{bmatrix}0 & 4 & 3 \\ -4 & 0 & 2 \\ -3 & -2 & 0\end{bmatrix}=-B
x=2,\;y=-5,\;z=3;\quad p=-4,\;q=-2,\;r=3

2.2, 2.2.1 to 2.2.4Algebra of matricesOften examined

Algebra of matrices

Scalar multiplication (2.2.1)

If k is a non-zero scalar and A=[a_{ij}]_{m\times n} is a matrix of order m\times n, the product kA is the matrix obtained by multiplying every element of A by k.

kA=k\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{bmatrix}=\begin{bmatrix}ka_{11} & ka_{12} & \cdots & ka_{1n} \\ ka_{21} & ka_{22} & \cdots & ka_{2n} \\ \vdots & \vdots & & \vdots \\ ka_{m1} & ka_{m2} & \cdots & ka_{mn}\end{bmatrix}

For example, if A=\begin{bmatrix}3 & 2 & 1 \\ 4 & -3 & 5\end{bmatrix}, then

2A=2\begin{bmatrix}3 & 2 & 1 \\ 4 & -3 & 5\end{bmatrix}=\begin{bmatrix}2\times3 & 2\times2 & 2\times1 \\ 2\times4 & 2\times(-3) & 2\times5\end{bmatrix}=\begin{bmatrix}6 & 4 & 2 \\ 8 & -6 & 10\end{bmatrix}

Addition of matrices (2.2.2)

In general we cannot add two matrices. Only matrices of the same order are conformable for addition. If A=[a_{ij}]_{m\times n} and B=[b_{ij}]_{m\times n}, then A+B is also a matrix of order m\times n, in which each element is the sum of the corresponding elements of A and B. If A+B=C with C=[c_{ij}]_{m\times n}, then

c_{ij}=a_{ij}+b_{ij}\qquad\text{for all } i,j

Subtraction of matrices (2.2.3)

Similarly, two matrices of the same order can be subtracted. If A=[a_{ij}]_{m\times n} and B=[b_{ij}]_{m\times n}, then A-B is a matrix of order m\times n in which each element is the difference of the corresponding elements. If A-B=D with D=[d_{ij}]_{m\times n}, then

d_{ij}=a_{ij}-b_{ij}\qquad\text{for all } i,j
Example 1: sum and difference
Find A+B and A-B where A=\begin{bmatrix}3 & 0 \\ 2 & 1 \\ 6 & 5\end{bmatrix} and B=\begin{bmatrix}-2 & 6 \\ 0 & 0 \\ 2 & 1\end{bmatrix}.
  1. Both matrices have order 3\times2, so they can be added and subtracted. Add the entries in the same positions.
    A+B=\begin{bmatrix}3 & 0 \\ 2 & 1 \\ 6 & 5\end{bmatrix}+\begin{bmatrix}-2 & 6 \\ 0 & 0 \\ 2 & 1\end{bmatrix}=\begin{bmatrix}3+(-2) & 0+6 \\ 2+0 & 1+0 \\ 6+2 & 5+1\end{bmatrix}=\begin{bmatrix}1 & 6 \\ 2 & 1 \\ 8 & 6\end{bmatrix}
  2. Subtract the entries in the same positions.
    A-B=\begin{bmatrix}3-(-2) & 0-6 \\ 2-0 & 1-0 \\ 6-2 & 5-1\end{bmatrix}=\begin{bmatrix}5 & -6 \\ 2 & 1 \\ 4 & 4\end{bmatrix}
A+B=\begin{bmatrix}1 & 6 \\ 2 & 1 \\ 8 & 6\end{bmatrix},\quad A-B=\begin{bmatrix}5 & -6 \\ 2 & 1 \\ 4 & 4\end{bmatrix}

Multiplication of matrices (2.2.4)

Two matrices A and B are conformable for multiplication (in the order AB) if the **number of columns of A equals the number of rows of B**. If A=[a_{ij}]_{m\times n} and B=[b_{ij}]_{n\times p}, then AB has order m\times p.

Let AB=C=[c_{ij}]_{m\times p}. The element c_{ij} is the sum of the products of the elements of the ith row of A with the corresponding elements of the jth column of B:

c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots+a_{in}b_{nj},\qquad i=1,2,\dots,m,\; j=1,2,\dots,p
The ith row of A is multiplied term by term with the jth column of B and the products are added to give c_{ij}.
Example 2: product of a 3\times2 and a 2\times3 matrix
Find the product AB for A=\begin{bmatrix}1 & 3 \\ 2 & 1 \\ 6 & 0\end{bmatrix}_{3\times2} and B=\begin{bmatrix}3 & 1 & 0 \\ 2 & 5 & 4\end{bmatrix}_{2\times3}.
  1. A has 2 columns and B has 2 rows, so they are conformable for AB. The order of AB is 3\times3.
    \text{order of }AB=(3\times\underline{2})(\underline{2}\times3)=3\times3
  2. Multiply each row of A by each column of B. Entry (AB)_{ij} is row i of A times column j of B.
    \begin{aligned}(AB)_{11}&=(1)(3)+(3)(2)\\ (AB)_{12}&=(1)(1)+(3)(5)\\ (AB)_{13}&=(1)(0)+(3)(4)\\ (AB)_{21}&=(2)(3)+(1)(2)\\ (AB)_{22}&=(2)(1)+(1)(5)\\ (AB)_{23}&=(2)(0)+(1)(4)\\ (AB)_{31}&=(6)(3)+(0)(2)\\ (AB)_{32}&=(6)(1)+(0)(5)\\ (AB)_{33}&=(6)(0)+(0)(4)\end{aligned}
  3. Simplify each product, then each sum.
    AB=\begin{bmatrix}3+6 & 1+15 & 0+12 \\ 6+2 & 2+5 & 0+4 \\ 18+0 & 6+0 & 0+0\end{bmatrix}=\begin{bmatrix}9 & 16 & 12 \\ 8 & 7 & 4 \\ 18 & 6 & 0\end{bmatrix}
AB=\begin{bmatrix}9 & 16 & 12 \\ 8 & 7 & 4 \\ 18 & 6 & 0\end{bmatrix}
Example 3: product with complex entries
Find the product AB for A=\begin{bmatrix}2i & 1 \\ -i & 3i\end{bmatrix} and B=\begin{bmatrix}3 & -i & i \\ 2i & 0 & -2i\end{bmatrix}.
  1. A has 2 columns and B has 2 rows, so AB exists. Its order is 2\times3.
    \text{order of }AB=(2\times\underline{2})(\underline{2}\times3)=2\times3
  2. Form every row-by-column product. Entry (AB)_{ij} is row i of A times column j of B.
    \begin{aligned}(AB)_{11}&=(2i)(3)+(1)(2i)\\ (AB)_{12}&=(2i)(-i)+(1)(0)\\ (AB)_{13}&=(2i)(i)+(1)(-2i)\\ (AB)_{21}&=(-i)(3)+(3i)(2i)\\ (AB)_{22}&=(-i)(-i)+(3i)(0)\\ (AB)_{23}&=(-i)(i)+(3i)(-2i)\end{aligned}
  3. Multiply out. Use i^2=-1.
    AB=\begin{bmatrix}6i+2i & -2i^2+0 & 2i^2-2i \\ -3i+6i^2 & i^2+0 & -i^2-6i^2\end{bmatrix}
  4. Replace i^2 by -1.
    \begin{aligned}(AB)_{11}&=8i\\ (AB)_{12}&=-2(-1)\\ (AB)_{13}&=2(-1)-2i\\ (AB)_{21}&=-3i+6(-1)\\ (AB)_{22}&=-1\\ (AB)_{23}&=-(-1)-6(-1)\end{aligned}
  5. Simplify.
    AB=\begin{bmatrix}8i & 2 & -2-2i \\ -3i-6 & -1 & 7\end{bmatrix}
AB=\begin{bmatrix}8i & 2 & -2-2i \\ -6-3i & -1 & 7\end{bmatrix}
Worked example: order check, then AB and BA
Let A=\begin{bmatrix}2 & -1 & 3 \\ 0 & 4 & 1\end{bmatrix} and B=\begin{bmatrix}1 & 2 \\ 0 & -1 \\ 3 & 4\end{bmatrix}. Find the orders of AB and BA and then compute both.
A 2\times3 matrix times a 3\times2 matrix: the inner numbers must agree, and the outer numbers are the order of the product.
  1. Orders first: A is 2\times3 and B is 3\times2. For AB the inner numbers are 3 and 3, so AB exists and has order 2\times2. For BA the inner numbers are 2 and 2, so BA exists and has order 3\times3.
  2. Each entry of AB is a row of A multiplied term by term with a column of B.
    \begin{aligned}(AB)_{11}&=2(1)-1(0)+3(3)\\ (AB)_{12}&=2(2)-1(-1)+3(4)\\ (AB)_{21}&=0(1)+4(0)+1(3)\\ (AB)_{22}&=0(2)+4(-1)+1(4)\end{aligned}
  3. Simplify.
    AB=\begin{bmatrix}11 & 17 \\ 3 & 0\end{bmatrix}
  4. For BA, use the rows of B (each of length 2) with the columns of A.
    \begin{aligned}(BA)_{11}&=1(2)+2(0)\\ (BA)_{12}&=1(-1)+2(4)\\ (BA)_{13}&=1(3)+2(1)\\ (BA)_{21}&=0(2)-1(0)\\ (BA)_{22}&=0(-1)-1(4)\\ (BA)_{23}&=0(3)-1(1)\\ (BA)_{31}&=3(2)+4(0)\\ (BA)_{32}&=3(-1)+4(4)\\ (BA)_{33}&=3(3)+4(1)\end{aligned}
  5. Simplify.
    BA=\begin{bmatrix}2 & 7 & 5 \\ 0 & -4 & -1 \\ 6 & 13 & 13\end{bmatrix}
  6. AB is 2\times2 and BA is 3\times3, so AB\neq BA: here they cannot even be compared, since their orders differ.
AB=\begin{bmatrix}11 & 17 \\ 3 & 0\end{bmatrix},\quad BA=\begin{bmatrix}2 & 7 & 5 \\ 0 & -4 & -1 \\ 6 & 13 & 13\end{bmatrix}

2.2.5 and 2.2.6Commutative property of matrices

Commutative property with respect to addition (2.2.5)

Any two matrices that are conformable for addition satisfy the commutative property with respect to addition. Take A=[a_{ij}]_{2\times3} and B=[b_{ij}]_{2\times3} (real entries). Then

A+B=\begin{bmatrix}a_{11}+b_{11} & a_{12}+b_{12} & a_{13}+b_{13} \\ a_{21}+b_{21} & a_{22}+b_{22} & a_{23}+b_{23}\end{bmatrix}

Each entry is a sum of two real numbers, and real numbers can be added in either order (a+b=b+a). So

A+B=\begin{bmatrix}b_{11}+a_{11} & b_{12}+a_{12} & b_{13}+a_{13} \\ b_{21}+a_{21} & b_{22}+a_{22} & b_{23}+a_{23}\end{bmatrix}=\begin{bmatrix}b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23}\end{bmatrix}+\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\end{bmatrix}=B+A

Hence matrix addition is commutative: A+B=B+A.

Example 4: verify commutativity of addition
Verify the commutative property of addition for A=\begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 3 \\ 0 & 2 & 1\end{bmatrix} and B=\begin{bmatrix}0 & -1 & 3 \\ 1 & 2 & 4 \\ -7 & 3 & 1\end{bmatrix}.
  1. Compute A+B by adding the corresponding entries.
    A+B=\begin{bmatrix}3+0 & 1+(-1) & 6+3 \\ 2+1 & 1+2 & 3+4 \\ 0+(-7) & 2+3 & 1+1\end{bmatrix}=\begin{bmatrix}3 & 0 & 9 \\ 3 & 3 & 7 \\ -7 & 5 & 2\end{bmatrix}\qquad(1)
  2. Now compute B+A in the opposite order.
    B+A=\begin{bmatrix}0+3 & (-1)+1 & 3+6 \\ 1+2 & 2+1 & 4+3 \\ (-7)+0 & 3+2 & 1+1\end{bmatrix}=\begin{bmatrix}3 & 0 & 9 \\ 3 & 3 & 7 \\ -7 & 5 & 2\end{bmatrix}\qquad(2)
  3. From (1) and (2), A+B=B+A. The commutative property holds for addition.
A+B=B+A=\begin{bmatrix}3 & 0 & 9 \\ 3 & 3 & 7 \\ -7 & 5 & 2\end{bmatrix}

Commutative property with respect to multiplication (2.2.6)

In general, matrix multiplication is not commutative: usually AB\neq BA. Sometimes BA does not even exist when AB does, and even when both exist they may have different orders or different entries.

Example 5: show that AB\neq BA
For A=\begin{bmatrix}1 & 2 & 1 \\ 3 & 1 & 6\end{bmatrix} and B=\begin{bmatrix}0 & 1 \\ 2 & 3 \\ 1 & 4\end{bmatrix}, show that AB\neq BA.
  1. A is 2\times3 and B is 3\times2, so AB has order 2\times2. Entry (AB)_{ij} is row i of A times column j of B.
    \begin{aligned}(AB)_{11}&=(1)(0)+(2)(2)+(1)(1)\\ (AB)_{12}&=(1)(1)+(2)(3)+(1)(4)\\ (AB)_{21}&=(3)(0)+(1)(2)+(6)(1)\\ (AB)_{22}&=(3)(1)+(1)(3)+(6)(4)\\[6pt] AB&=\begin{bmatrix}0+4+1 & 1+6+4 \\ 0+2+6 & 3+3+24\end{bmatrix}\\[6pt] &=\begin{bmatrix}5 & 11 \\ 8 & 30\end{bmatrix}\end{aligned}
  2. BA has order 3\times3.
    \begin{aligned}(BA)_{11}&=(0)(1)+(1)(3)\\ (BA)_{12}&=(0)(2)+(1)(1)\\ (BA)_{13}&=(0)(1)+(1)(6)\\ (BA)_{21}&=(2)(1)+(3)(3)\\ (BA)_{22}&=(2)(2)+(3)(1)\\ (BA)_{23}&=(2)(1)+(3)(6)\\ (BA)_{31}&=(1)(1)+(4)(3)\\ (BA)_{32}&=(1)(2)+(4)(1)\\ (BA)_{33}&=(1)(1)+(4)(6)\\[6pt] BA&=\begin{bmatrix}3 & 1 & 6 \\ 11 & 7 & 20 \\ 13 & 6 & 25\end{bmatrix}\end{aligned}
  3. Clearly AB\neq BA. They do not even have the same order (2\times2 and 3\times3).
AB=\begin{bmatrix}5 & 11 \\ 8 & 30\end{bmatrix}\neq\begin{bmatrix}3 & 1 & 6 \\ 11 & 7 & 20 \\ 13 & 6 & 25\end{bmatrix}=BA

2.2.6 (Examples 6 to 9)Transpose rules and symmetric and skew-symmetric partsOften examined

The next four examples of the book use the transpose. They show how the transpose behaves with products and sums, and that every square matrix splits into a symmetric part and a skew-symmetric part.

Example 6: verify (AB)^{t}=B^{t}A^{t}
If A=\begin{bmatrix}1 & 2 \\ 5 & 6 \\ 2 & 1\end{bmatrix}_{3\times2} and B=\begin{bmatrix}0 & 1 & 6 \\ 2 & 1 & 0\end{bmatrix}_{2\times3}, verify that (AB)^{t}=B^{t}A^{t}.
  1. Find AB (order 3\times3). Entry (AB)_{ij} is row i of A times column j of B.
    \begin{aligned}(AB)_{11}&=(1)(0)+(2)(2)\\ (AB)_{12}&=(1)(1)+(2)(1)\\ (AB)_{13}&=(1)(6)+(2)(0)\\ (AB)_{21}&=(5)(0)+(6)(2)\\ (AB)_{22}&=(5)(1)+(6)(1)\\ (AB)_{23}&=(5)(6)+(6)(0)\\ (AB)_{31}&=(2)(0)+(1)(2)\\ (AB)_{32}&=(2)(1)+(1)(1)\\ (AB)_{33}&=(2)(6)+(1)(0)\\[6pt] AB&=\begin{bmatrix}4 & 3 & 6 \\ 12 & 11 & 30 \\ 2 & 3 & 12\end{bmatrix}\end{aligned}
  2. Take the transpose of AB (rows become columns): the left side.
    (AB)^{t}=\begin{bmatrix}4 & 12 & 2 \\ 3 & 11 & 3 \\ 6 & 30 & 12\end{bmatrix}\qquad(1)
  3. Write A^{t} and B^{t}.
    A^{t}=\begin{bmatrix}1 & 5 & 2 \\ 2 & 6 & 1\end{bmatrix},\qquad B^{t}=\begin{bmatrix}0 & 2 \\ 1 & 1 \\ 6 & 0\end{bmatrix}
  4. Multiply in the reversed order B^{t}A^{t} (order 3\times2 times 2\times3 is 3\times3).
    \begin{aligned}(B^{t}A^{t})_{11}&=(0)(1)+(2)(2)\\ (B^{t}A^{t})_{12}&=(0)(5)+(2)(6)\\ (B^{t}A^{t})_{13}&=(0)(2)+(2)(1)\\ (B^{t}A^{t})_{21}&=(1)(1)+(1)(2)\\ (B^{t}A^{t})_{22}&=(1)(5)+(1)(6)\\ (B^{t}A^{t})_{23}&=(1)(2)+(1)(1)\\ (B^{t}A^{t})_{31}&=(6)(1)+(0)(2)\\ (B^{t}A^{t})_{32}&=(6)(5)+(0)(6)\\ (B^{t}A^{t})_{33}&=(6)(2)+(0)(1)\\[6pt] B^{t}A^{t}&=\begin{bmatrix}4 & 12 & 2 \\ 3 & 11 & 3 \\ 6 & 30 & 12\end{bmatrix}&&(2)\end{aligned}
  5. From (1) and (2), (AB)^{t}=B^{t}A^{t}.
(AB)^{t}=B^{t}A^{t}=\begin{bmatrix}4 & 12 & 2 \\ 3 & 11 & 3 \\ 6 & 30 & 12\end{bmatrix}
Example 7: show (A+B)^{t}=A^{t}+B^{t}
For two matrices A and B which are conformable for addition, show that (A+B)^{t}=A^{t}+B^{t}.
  1. Take any two matrices of the same order, say 3\times2.
    A=\begin{bmatrix}a & b \\ c & d \\ e & f\end{bmatrix},\qquad B=\begin{bmatrix}1 & 2 \\ 3 & 5 \\ 2 & 6\end{bmatrix}
  2. Add them and take the transpose: the left side.
    A+B=\begin{bmatrix}a+1 & b+2 \\ c+3 & d+5 \\ e+2 & f+6\end{bmatrix}\;\Rightarrow\;(A+B)^{t}=\begin{bmatrix}a+1 & c+3 & e+2 \\ b+2 & d+5 & f+6\end{bmatrix}\qquad(1)
  3. Take the transposes first and then add: the right side.
    A^{t}=\begin{bmatrix}a & c & e \\ b & d & f\end{bmatrix},\quad B^{t}=\begin{bmatrix}1 & 3 & 2 \\ 2 & 5 & 6\end{bmatrix}
  4. Add the transposes.
    A^{t}+B^{t}=\begin{bmatrix}a+1 & c+3 & e+2 \\ b+2 & d+5 & f+6\end{bmatrix}\qquad(2)
  5. From (1) and (2), (A+B)^{t}=A^{t}+B^{t}.
(A+B)^{t}=A^{t}+B^{t}
Example 8: symmetric plus skew-symmetric
Show that any square matrix can be written as the sum of two square matrices such that one of them is symmetric and the other is skew-symmetric.
  1. Take any square matrix A. Suppose A=P+Q where P is symmetric (P^{t}=P) and Q is skew-symmetric (Q^{t}=-Q).
    A=P+Q\qquad(1)
  2. Take the transpose of both sides. The transpose of a sum is the sum of the transposes.
    A^{t}=(P+Q)^{t}=P^{t}+Q^{t}=P+(-Q)
  3. So we get a second equation.
    A^{t}=P-Q\qquad(2)
  4. Add (1) and (2) to eliminate Q.
    A+A^{t}=2P\;\Rightarrow\;P=\tfrac12(A+A^{t})
  5. Subtract (2) from (1) to eliminate P.
    A-A^{t}=2Q\;\Rightarrow\;Q=\tfrac12(A-A^{t})
  6. Check that P is symmetric, using (A^{t})^{t}=A.
    P^{t}=\left[\tfrac12(A+A^{t})\right]^{t}=\tfrac12\left(A^{t}+(A^{t})^{t}\right)=\tfrac12(A^{t}+A)=\tfrac12(A+A^{t})=P
  7. Check that Q is skew-symmetric.
    Q^{t}=\left[\tfrac12(A-A^{t})\right]^{t}=\tfrac12\left(A^{t}-(A^{t})^{t}\right)=\tfrac12(A^{t}-A)=-\tfrac12(A-A^{t})=-Q
  8. So P is symmetric, Q is skew-symmetric, and A=P+Q.
A=\underbrace{\tfrac12(A+A^{t})}_{\text{symmetric}}+\underbrace{\tfrac12(A-A^{t})}_{\text{skew-symmetric}}
Example 9: split a matrix into symmetric and skew-symmetric parts
Write A=\begin{bmatrix}3 & 1 & 5 \\ 2 & 6 & 0 \\ -1 & 2 & 1\end{bmatrix} as the sum of two matrices where one is symmetric and the other is skew-symmetric.
  1. Let A=P+Q with P symmetric and Q skew-symmetric. By Example 8, P=\tfrac12(A+A^{t}) and Q=\tfrac12(A-A^{t}). First write A^{t}.
    A^{t}=\begin{bmatrix}3 & 2 & -1 \\ 1 & 6 & 2 \\ 5 & 0 & 1\end{bmatrix}
  2. Find P=\tfrac12(A+A^{t}).
    P=\tfrac12\left(\begin{bmatrix}3 & 1 & 5 \\ 2 & 6 & 0 \\ -1 & 2 & 1\end{bmatrix}+\begin{bmatrix}3 & 2 & -1 \\ 1 & 6 & 2 \\ 5 & 0 & 1\end{bmatrix}\right)=\tfrac12\begin{bmatrix}6 & 3 & 4 \\ 3 & 12 & 2 \\ 4 & 2 & 2\end{bmatrix}=\begin{bmatrix}3 & \tfrac32 & 2 \\ \tfrac32 & 6 & 1 \\ 2 & 1 & 1\end{bmatrix}
  3. Find Q=\tfrac12(A-A^{t}).
    Q=\tfrac12\left(\begin{bmatrix}3 & 1 & 5 \\ 2 & 6 & 0 \\ -1 & 2 & 1\end{bmatrix}-\begin{bmatrix}3 & 2 & -1 \\ 1 & 6 & 2 \\ 5 & 0 & 1\end{bmatrix}\right)=\tfrac12\begin{bmatrix}0 & -1 & 6 \\ 1 & 0 & -2 \\ -6 & 2 & 0\end{bmatrix}=\begin{bmatrix}0 & -\tfrac12 & 3 \\ \tfrac12 & 0 & -1 \\ -3 & 1 & 0\end{bmatrix}
  4. Check: P is symmetric (it equals its transpose) and Q is skew-symmetric (zero diagonal, opposite entries). Add them to recover A.
    P+Q=\begin{bmatrix}3 & \tfrac32 & 2 \\ \tfrac32 & 6 & 1 \\ 2 & 1 & 1\end{bmatrix}+\begin{bmatrix}0 & -\tfrac12 & 3 \\ \tfrac12 & 0 & -1 \\ -3 & 1 & 0\end{bmatrix}=\begin{bmatrix}3 & 1 & 5 \\ 2 & 6 & 0 \\ -1 & 2 & 1\end{bmatrix}
\begin{bmatrix}3 & 1 & 5 \\ 2 & 6 & 0 \\ -1 & 2 & 1\end{bmatrix}=\begin{bmatrix}3 & \tfrac32 & 2 \\ \tfrac32 & 6 & 1 \\ 2 & 1 & 1\end{bmatrix}+\begin{bmatrix}0 & -\tfrac12 & 3 \\ \tfrac12 & 0 & -1 \\ -3 & 1 & 0\end{bmatrix}

2.3, 2.3.1 and 2.3.2DeterminantsOften examined

Every square matrix is associated with a number (real or complex) called its determinant. If A is a square matrix, its determinant is written \det(A) or |A|. For a square matrix A of order n, the determinant is written by replacing the square brackets by vertical bars:

A=\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{bmatrix}\;\Longrightarrow\;|A|=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{vmatrix}

Determinant of a matrix of order 2 × 2 (2.3.1)

For A=\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix},

|A|=\begin{vmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix}=a_{11}a_{22}-a_{21}a_{12}

Determinant of a matrix of order 3 × 3 (2.3.2)

For A=\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{bmatrix} the associated determinant is |A|=\begin{vmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}.

To find its value, we write the determinant as a sum or difference of determinants of order 2. This process is called the expansion of the determinant. A determinant of order 3 has three rows R_1,R_2,R_3 and three columns C_1,C_2,C_3, so it can be expanded from any row or any column: six different ways, and the value is the same in every case.

Expanding along the first row R_1:

|A|=a_{11}\begin{vmatrix}a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21} & a_{23} \\ a_{31} & a_{33}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21} & a_{22} \\ a_{31} & a_{32}\end{vmatrix}
|A|=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})

This can be generalised to determinants of higher order.

2.3.3 and 2.3.4Minors and cofactorsOften examined

Minor of an element (2.3.3)

Let A=[a_{ij}]_{n\times n} be a square matrix of order n. The minor of the element a_{ij} is the determinant of order (n-1)\times(n-1) obtained by **neglecting (deleting) the ith row and the jth column** of A. It is denoted M_{ij}.

For example, for A=\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{bmatrix}, the minor of a_{21} is obtained by neglecting the 2nd row and the 1st column:

M_{21}=\begin{vmatrix}a_{12} & a_{13} \\ a_{32} & a_{33}\end{vmatrix}

In the same way we can find all the minors of the elements of A.

Cofactor of an element (2.3.4)

For a square matrix A of order n\times n, the cofactor of the element a_{ij} is denoted A_{ij} and defined by

A_{ij}=(-1)^{i+j}M_{ij}

For example, for the same matrix A the cofactor of a_{21} is

A_{21}=(-1)^{2+1}M_{21}=(-1)^{3}\begin{vmatrix}a_{12} & a_{13} \\ a_{32} & a_{33}\end{vmatrix}=-(a_{12}a_{33}-a_{13}a_{32})=a_{13}a_{32}-a_{12}a_{33}
Example 10: minors and cofactors
If A=\begin{bmatrix}1 & 3 & 0 \\ -1 & 2 & 6 \\ 3 & 0 & -4\end{bmatrix}, find M_{12},\;M_{23},\;A_{12} and A_{23}.
  1. For M_{12} delete row 1 and column 2. What remains is the 2\times2 determinant of the entries -1,6,3,-4.
    M_{12}=\begin{vmatrix}-1 & 6 \\ 3 & -4\end{vmatrix}=(-1)(-4)-(3)(6)=4-18=-14
  2. For M_{23} delete row 2 and column 3. What remains is 1,3,3,0.
    M_{23}=\begin{vmatrix}1 & 3 \\ 3 & 0\end{vmatrix}=(1)(0)-(3)(3)=0-9=-9
  3. Now attach signs. For A_{12} the sign is (-1)^{1+2}=-1.
    A_{12}=(-1)^{1+2}M_{12}=(-1)^3(-14)=14
  4. For A_{23} the sign is (-1)^{2+3}=-1.
    A_{23}=(-1)^{2+3}M_{23}=(-1)^5(-9)=9
M_{12}=-14,\quad M_{23}=-9,\quad A_{12}=14,\quad A_{23}=9

2.3.5 and 2.3.6Evaluation of a determinant by cofactors, singular matricesOften examined

Evaluation of a determinant using cofactors (2.3.5)

Take a square matrix A of order 3\times3, with |A|=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}. Expand from the first row. The expansion of the previous topic gives

\begin{aligned}|A|&=a_{11}\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}\\&=a_{11}M_{11}-a_{12}M_{12}+a_{13}M_{13}\\&=a_{11}(-1)^{1+1}M_{11}+a_{12}(-1)^{1+2}M_{12}+a_{13}(-1)^{1+3}M_{13}\\&=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}\end{aligned}

If we expand from the first column instead:

\begin{aligned}|A|&=a_{11}\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}-a_{21}\begin{vmatrix}a_{12}&a_{13}\\a_{32}&a_{33}\end{vmatrix}+a_{31}\begin{vmatrix}a_{12}&a_{13}\\a_{22}&a_{23}\end{vmatrix}\\&=a_{11}M_{11}-a_{21}M_{21}+a_{31}M_{31}\\&=a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}\end{aligned}

So |A| can be evaluated by adding the products of the elements of any row (or column) with their corresponding cofactors.

\begin{aligned}|A|&=a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3}&&(\text{row }i)\\ |A|&=a_{1j}A_{1j}+a_{2j}A_{2j}+a_{3j}A_{3j}&&(\text{column }j)\end{aligned}
Example 11: determinant using cofactors
If A=\begin{bmatrix}1 & 2 & 6 \\ 0 & 1 & 2 \\ -1 & 3 & 0\end{bmatrix}, find |A| using cofactors.
  1. Choose one column and find the cofactors of its elements. Take C_3, whose elements are a_{13}=6,\;a_{23}=2,\;a_{33}=0. The cofactors are:
  2. A_{13}: delete row 1 and column 3, sign (-1)^{1+3}=+1.
    A_{13}=(-1)^{1+3}\begin{vmatrix}0 & 1 \\ -1 & 3\end{vmatrix}=(1)(0-(-1))=1
  3. A_{23}: delete row 2 and column 3, sign (-1)^{2+3}=-1.
    A_{23}=(-1)^{2+3}\begin{vmatrix}1 & 2 \\ -1 & 3\end{vmatrix}=(-1)(3-(-2))=(-1)(5)=-5
  4. A_{33}: delete row 3 and column 3, sign (-1)^{3+3}=+1.
    A_{33}=(-1)^{3+3}\begin{vmatrix}1 & 2 \\ 0 & 1\end{vmatrix}=(1)(1-0)=1
  5. Multiply each element of C_3 by its cofactor and add.
    |A|=a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}=6(1)+2(-5)+0(1)=6-10+0=-4
|A|=-4
Worked example: expanding along the column with a zero
Find |A| for A=\begin{bmatrix}2 & 0 & 1 \\ 3 & 1 & 4 \\ -1 & 2 & 0\end{bmatrix} by cofactors, choosing the best column, and check by another expansion.
  1. Choose the row or column with the most zeros, since a zero entry needs no cofactor. Column 2 has a_{12}=0.
  2. Signs of column 2 come from (-1)^{i+2}: -,+,-. Find the cofactors of a_{22}=1 and a_{32}=2.
    A_{22}=+\begin{vmatrix}2 & 1 \\ -1 & 0\end{vmatrix}=0+1=1,\qquad A_{32}=-\begin{vmatrix}2 & 1 \\ 3 & 4\end{vmatrix}=-(8-3)=-5
  3. Multiply each entry of the column by its cofactor and add.
    |A|=0\cdot A_{12}+1\cdot A_{22}+2\cdot A_{32}=0+1-10=-9
  4. Check by expanding along row 1 (signs +,-,+).
    |A|=2\begin{vmatrix}1 & 4 \\ 2 & 0\end{vmatrix}-0+1\begin{vmatrix}3 & 1 \\ -1 & 2\end{vmatrix}=2(-8)+(6+1)=-9
  5. Both expansions agree. Since |A|\neq0, the matrix is non-singular.
|A|=-9

Singular and non-singular matrices (2.3.6)

A square matrix A is singular if |A|=0. If |A|\neq0, it is called a non-singular matrix.

For example, take A=\begin{bmatrix}1 & 3 \\ 4 & 2\end{bmatrix} and B=\begin{bmatrix}3 & 12 \\ \tfrac12 & 2\end{bmatrix}. Then

|A|=\begin{vmatrix}1 & 3 \\ 4 & 2\end{vmatrix}=(1)(2)-(3)(4)=2-12=-10\neq0
|B|=\begin{vmatrix}3 & 12 \\ \tfrac12 & 2\end{vmatrix}=(3)(2)-12\left(\tfrac12\right)=6-6=0

Thus A is a non-singular matrix and B is a singular matrix.

2.3.4 (adjoint), 2.3.7 and 2.3.8Adjoint and inverse of a matrixOften examined

Adjoint of a square matrix (2.3.4 in the book, adjoint)

Take any square matrix A=[a_{ij}] of order n. The adjoint of A, written \operatorname{adj}(A), is the matrix whose entries are the cofactors of A, arranged in the transposed order: the cofactor of a_{ij} goes in row j, column i.

\operatorname{adj}(A)=\begin{bmatrix}A_{11} & A_{21} & \cdots & A_{n1} \\ A_{12} & A_{22} & \cdots & A_{n2} \\ \vdots & \vdots & & \vdots \\ A_{1n} & A_{2n} & \cdots & A_{nn}\end{bmatrix}

For a matrix of order 3\times3:

A=\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{bmatrix}\;\Rightarrow\;\operatorname{adj}(A)=\begin{bmatrix}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{bmatrix}

Multiplicative inverse of a square matrix (2.3.7)

Two square matrices of the same order n are multiplicative inverses of each other if their product is I_n, the identity matrix of order n. Only non-singular matrices have a multiplicative inverse. The inverse of a non-singular matrix A is denoted A^{-1} and

AA^{-1}=A^{-1}A=I

Adjoint method to find the inverse (2.3.8)

If A is a non-singular square matrix, that is |A|\neq0, then

A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)

If A is singular, |A|=0, so \dfrac{1}{|A|}\operatorname{adj}(A) does not exist and A^{-1} does not exist.

Method
  1. Find |A|. If it is 0, stop: the inverse does not exist.
  2. Find the cofactor A_{ij} of every element.
  3. Write \operatorname{adj}(A): the transpose of the matrix of cofactors.
  4. Divide by |A|.
    A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A)
Example 12: inverse by the adjoint method
If A=\begin{bmatrix}2 & 1 & 0 \\ 0 & 3 & 2 \\ 2 & 1 & 4\end{bmatrix}, find A^{-1} by the adjoint method.
  1. First find |A|, expanding along the first row.
    |A|=\begin{vmatrix}2 & 1 & 0 \\ 0 & 3 & 2 \\ 2 & 1 & 4\end{vmatrix}=2\begin{vmatrix}3 & 2 \\ 1 & 4\end{vmatrix}-1\begin{vmatrix}0 & 2 \\ 2 & 4\end{vmatrix}+0\begin{vmatrix}0 & 3 \\ 2 & 1\end{vmatrix}
  2. Evaluate each 2\times2 determinant.
    |A|=2(12-2)-1(0-4)+0(0-6)=20+4+0=24\neq0
  3. So A is non-singular and A^{-1} exists. Now find the cofactors of all nine elements.
  4. Row 1 elements.
    A_{11}=(-1)^{1+1}\begin{vmatrix}3 & 2 \\ 1 & 4\end{vmatrix}=12-2=10,\quad A_{12}=(-1)^{1+2}\begin{vmatrix}0 & 2 \\ 2 & 4\end{vmatrix}=-(0-4)=4,\quad A_{13}=(-1)^{1+3}\begin{vmatrix}0 & 3 \\ 2 & 1\end{vmatrix}=0-6=-6
  5. Row 2 elements.
    A_{21}=(-1)^{2+1}\begin{vmatrix}1 & 0 \\ 1 & 4\end{vmatrix}=-(4-0)=-4,\quad A_{22}=(-1)^{2+2}\begin{vmatrix}2 & 0 \\ 2 & 4\end{vmatrix}=8-0=8,\quad A_{23}=(-1)^{2+3}\begin{vmatrix}2 & 1 \\ 2 & 1\end{vmatrix}=-(2-2)=0
  6. Row 3 elements.
    A_{31}=(-1)^{3+1}\begin{vmatrix}1 & 0 \\ 3 & 2\end{vmatrix}=2-0=2,\quad A_{32}=(-1)^{3+2}\begin{vmatrix}2 & 0 \\ 0 & 2\end{vmatrix}=-(4-0)=-4,\quad A_{33}=(-1)^{3+3}\begin{vmatrix}2 & 1 \\ 0 & 3\end{vmatrix}=6-0=6
  7. Write the adjoint: transpose the matrix of cofactors (the cofactors of row 1 go into column 1, and so on).
    \operatorname{adj}(A)=\begin{bmatrix}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}10 & -4 & 2 \\ 4 & 8 & -4 \\ -6 & 0 & 6\end{bmatrix}
  8. Divide by |A|=24 and reduce each fraction.
    A^{-1}=\frac{1}{|A|}\operatorname{adj}(A)=\frac1{24}\begin{bmatrix}10 & -4 & 2 \\ 4 & 8 & -4 \\ -6 & 0 & 6\end{bmatrix}=\begin{bmatrix}\tfrac{10}{24} & -\tfrac{4}{24} & \tfrac{2}{24} \\ \tfrac{4}{24} & \tfrac{8}{24} & -\tfrac{4}{24} \\ -\tfrac{6}{24} & 0 & \tfrac{6}{24}\end{bmatrix}
  9. Simplify.
    A^{-1}=\begin{bmatrix}\tfrac5{12} & -\tfrac16 & \tfrac1{12} \\ \tfrac16 & \tfrac13 & -\tfrac16 \\ -\tfrac14 & 0 & \tfrac14\end{bmatrix}
  10. Check one entry: row 1 of A times column 1 of \operatorname{adj}(A) should give |A|.
    2(10)+1(4)+0(-6)=24=|A|\;\checkmark
A^{-1}=\begin{bmatrix}\tfrac5{12} & -\tfrac16 & \tfrac1{12} \\ \tfrac16 & \tfrac13 & -\tfrac16 \\ -\tfrac14 & 0 & \tfrac14\end{bmatrix}
Example 13: reversal law of the inverse
If A=\begin{bmatrix}3 & 2 \\ 1 & 4\end{bmatrix} and B=\begin{bmatrix}1 & 2 \\ 4 & 3\end{bmatrix}, verify that (AB)^{-1}=B^{-1}A^{-1}.
  1. Left side. Find AB.
    AB=\begin{bmatrix}3 & 2 \\ 1 & 4\end{bmatrix}\begin{bmatrix}1 & 2 \\ 4 & 3\end{bmatrix}=\begin{bmatrix}3+8 & 6+6 \\ 1+16 & 2+12\end{bmatrix}=\begin{bmatrix}11 & 12 \\ 17 & 14\end{bmatrix}
  2. Find |AB|.
    |AB|=\begin{vmatrix}11 & 12 \\ 17 & 14\end{vmatrix}=(11)(14)-(17)(12)=154-204=-50
  3. For a 2\times2 matrix, the adjoint is found by swapping the diagonal entries and changing the signs of the other two.
    \operatorname{adj}(AB)=\begin{bmatrix}14 & -12 \\ -17 & 11\end{bmatrix}
  4. Apply (AB)^{-1}=\dfrac1{|AB|}\operatorname{adj}(AB).
    (AB)^{-1}=\frac1{-50}\begin{bmatrix}14 & -12 \\ -17 & 11\end{bmatrix}=\begin{bmatrix}\tfrac{14}{-50} & \tfrac{-12}{-50} \\ \tfrac{-17}{-50} & \tfrac{11}{-50}\end{bmatrix}=\begin{bmatrix}-\tfrac7{25} & \tfrac6{25} \\ \tfrac{17}{50} & -\tfrac{11}{50}\end{bmatrix}\qquad(1)
  5. Right side. Find A^{-1}.
    |A|=(3)(4)-(1)(2)=10,\quad \operatorname{adj}(A)=\begin{bmatrix}4 & -2 \\ -1 & 3\end{bmatrix},\quad A^{-1}=\frac1{10}\begin{bmatrix}4 & -2 \\ -1 & 3\end{bmatrix}=\begin{bmatrix}\tfrac25 & -\tfrac15 \\ -\tfrac1{10} & \tfrac3{10}\end{bmatrix}
  6. Find B^{-1}.
    |B|=(1)(3)-(4)(2)=-5,\quad \operatorname{adj}(B)=\begin{bmatrix}3 & -2 \\ -4 & 1\end{bmatrix},\quad B^{-1}=\frac1{-5}\begin{bmatrix}3 & -2 \\ -4 & 1\end{bmatrix}=\begin{bmatrix}-\tfrac35 & \tfrac25 \\ \tfrac45 & -\tfrac15\end{bmatrix}
  7. Multiply in the reversed order B^{-1}A^{-1}.
    B^{-1}A^{-1}=\begin{bmatrix}-\tfrac35 & \tfrac25 \\ \tfrac45 & -\tfrac15\end{bmatrix}\begin{bmatrix}\tfrac25 & -\tfrac15 \\ -\tfrac1{10} & \tfrac3{10}\end{bmatrix}
  8. Row by column.
    =\begin{bmatrix}-\tfrac{6}{25}-\tfrac{1}{25} & \tfrac{3}{25}+\tfrac{3}{25} \\ \tfrac{8}{25}+\tfrac{1}{50} & -\tfrac{4}{25}-\tfrac{3}{50}\end{bmatrix}=\begin{bmatrix}-\tfrac7{25} & \tfrac6{25} \\ \tfrac{17}{50} & -\tfrac{11}{50}\end{bmatrix}\qquad(2)
  9. From (1) and (2), (AB)^{-1}=B^{-1}A^{-1}.
(AB)^{-1}=B^{-1}A^{-1}=\begin{bmatrix}-\tfrac7{25} & \tfrac6{25} \\ \tfrac{17}{50} & -\tfrac{11}{50}\end{bmatrix}

2.4 (Properties 1 to 4)Properties of determinants: transpose, interchange, identical rows, scalarOften examined

Here we discuss some important properties of determinants which help us to find the value of a determinant. For convenience, the book takes a square matrix A of order 3\times3:

A=\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{bmatrix},\qquad |A|=\begin{vmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}
|A|=a_{11}\begin{vmatrix}a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21} & a_{23} \\ a_{31} & a_{33}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21} & a_{22} \\ a_{31} & a_{32}\end{vmatrix}=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})

Property 1: determinant of the transpose

For any square matrix A, the determinant of the matrix and the determinant of its transpose are the same: |A|=|A^{t}|.

Derivation Proof of Property 1

Aim. Show that |A^{t}|=|A| for a 3\times3 matrix.

  1. Write the transpose: the rows of A become columns.
    |A^{t}|=\begin{vmatrix}a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33}\end{vmatrix}
  2. Expand from the first column C_1. Its entries are a_{11},a_{12},a_{13}, with signs +,-,+.
    |A^{t}|=a_{11}\begin{vmatrix}a_{22} & a_{32} \\ a_{23} & a_{33}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21} & a_{31} \\ a_{23} & a_{33}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21} & a_{31} \\ a_{22} & a_{32}\end{vmatrix}
  3. Evaluate each 2\times2 determinant (a 2\times2 determinant is unchanged by transposing).
    |A^{t}|=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})
  4. This is exactly the expansion of |A| from the first row.
    |A^{t}|=|A|
Result|A^{t}|=|A|

Rows and columns play the same role in a determinant. So every property proved for rows is also true for columns.

Example 14: |A|=|A^{t}|
If A=\begin{bmatrix}1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & 1 & 6\end{bmatrix}, verify that |A|=|A^{t}|.
  1. Expand |A| along the first row.
    |A|=\begin{vmatrix}1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & 1 & 6\end{vmatrix}=1\begin{vmatrix}3 & 0 \\ 1 & 6\end{vmatrix}-2\begin{vmatrix}0 & 0 \\ 2 & 6\end{vmatrix}+0\begin{vmatrix}0 & 3 \\ 2 & 1\end{vmatrix}
  2. Evaluate.
    |A|=1(18-0)-2(0-0)+0=18
  3. Write A^{t} and expand along its first row.
    |A^{t}|=\begin{vmatrix}1 & 0 & 2 \\ 2 & 3 & 1 \\ 0 & 0 & 6\end{vmatrix}=1\begin{vmatrix}3 & 1 \\ 0 & 6\end{vmatrix}-0\begin{vmatrix}2 & 1 \\ 0 & 6\end{vmatrix}+2\begin{vmatrix}2 & 3 \\ 0 & 0\end{vmatrix}
  4. Evaluate.
    |A^{t}|=1(18-0)-0+2(0-0)=18
  5. Thus |A|=|A^{t}|.
|A|=|A^{t}|=18

Property 2: interchanging two rows or columns

If any two rows (or columns) of a square matrix A are interchanged, and the resulting matrix is B, then |B|=-|A|.

Derivation Proof of Property 2

Aim. Show that interchanging the first and second rows changes the sign of the determinant.

  1. Interchange rows 1 and 2 of A. The new matrix is B.
    |B|=\begin{vmatrix}a_{21} & a_{22} & a_{23} \\ a_{11} & a_{12} & a_{13} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}
  2. Expand from the first row of B (its entries are a_{21},a_{22},a_{23}).
    |B|=a_{21}\begin{vmatrix}a_{12} & a_{13} \\ a_{32} & a_{33}\end{vmatrix}-a_{22}\begin{vmatrix}a_{11} & a_{13} \\ a_{31} & a_{33}\end{vmatrix}+a_{23}\begin{vmatrix}a_{11} & a_{12} \\ a_{31} & a_{32}\end{vmatrix}
  3. Evaluate the 2\times2 determinants.
    |B|=a_{21}(a_{12}a_{33}-a_{13}a_{32})-a_{22}(a_{11}a_{33}-a_{13}a_{31})+a_{23}(a_{11}a_{32}-a_{12}a_{31})
  4. Multiply out all brackets.
    |B|=a_{12}a_{21}a_{33}-a_{13}a_{21}a_{32}-a_{11}a_{22}a_{33}+a_{13}a_{22}a_{31}+a_{11}a_{23}a_{32}-a_{12}a_{23}a_{31}
  5. Regroup the six terms in pairs that contain a_{11}, a_{12} and a_{13}.
    |B|=(-a_{11}a_{22}a_{33}+a_{11}a_{23}a_{32})+(a_{12}a_{21}a_{33}-a_{12}a_{23}a_{31})+(-a_{13}a_{21}a_{32}+a_{13}a_{22}a_{31})
  6. Take out the common factor of each pair.
    |B|=-a_{11}(a_{22}a_{33}-a_{23}a_{32})+a_{12}(a_{21}a_{33}-a_{23}a_{31})-a_{13}(a_{21}a_{32}-a_{22}a_{31})
  7. Take out a minus sign from the whole expression. What is left is the expansion of |A|.
    |B|=-\left[a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})\right]=-|A|
Result|B|=-|A|

Swapping two rows (or two columns) reverses the sign of the determinant. Swapping twice brings back the original value.

Example 15: swapping two rows
Let A=\begin{bmatrix}1 & 2 & 3 \\ 2 & 1 & 0 \\ 0 & 2 & 0\end{bmatrix}. Interchange the second and third rows and compare the determinants.
  1. Expand |A| along the first row.
    |A|=1\begin{vmatrix}1 & 0 \\ 2 & 0\end{vmatrix}-2\begin{vmatrix}2 & 0 \\ 0 & 0\end{vmatrix}+3\begin{vmatrix}2 & 1 \\ 0 & 2\end{vmatrix}=1(0-0)-2(0-0)+3(4-0)=12
  2. Interchange R_2 and R_3 to get B.
    B=\begin{bmatrix}1 & 2 & 3 \\ 0 & 2 & 0 \\ 2 & 1 & 0\end{bmatrix}
  3. Expand |B| along the first row.
    |B|=1\begin{vmatrix}2 & 0 \\ 1 & 0\end{vmatrix}-2\begin{vmatrix}0 & 0 \\ 2 & 0\end{vmatrix}+3\begin{vmatrix}0 & 2 \\ 2 & 1\end{vmatrix}=1(0-0)-2(0-0)+3(0-4)=-12
  4. So |B|=-12=-|A|.
|B|=-|A|

Property 3: two identical rows or columns

If any two rows (or columns) of a square matrix are identical, the value of the determinant is zero.

Derivation Proof of Property 3

Aim. Show that a determinant with two identical rows is zero.

  1. Take a determinant whose second and third rows are the same. Expand from the first row.
    \begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{21}&a_{22}&a_{23}\end{vmatrix}=a_{11}\begin{vmatrix}a_{22} & a_{23} \\ a_{22} & a_{23}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21} & a_{23} \\ a_{21} & a_{23}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21} & a_{22} \\ a_{21} & a_{22}\end{vmatrix}
  2. Each 2\times2 determinant has two identical rows, so it is a number minus itself.
    =a_{11}(a_{22}a_{23}-a_{22}a_{23})-a_{12}(a_{21}a_{23}-a_{21}a_{23})+a_{13}(a_{21}a_{22}-a_{21}a_{22})
  3. Every bracket is zero.
    =a_{11}(0)-a_{12}(0)+a_{13}(0)=0
Result\text{determinant with two identical rows}=0
Example 16: two identical rows
Show that |A|=\begin{vmatrix}2 & 1 & 6 \\ 3 & 2 & 0 \\ 3 & 2 & 0\end{vmatrix}=0.
  1. Rows R_2 and R_3 are identical, so the value should be 0. Check by expanding along R_1.
  2. Expand along the first row.
    |A|=2\begin{vmatrix}2 & 0 \\ 2 & 0\end{vmatrix}-1\begin{vmatrix}3 & 0 \\ 3 & 0\end{vmatrix}+6\begin{vmatrix}3 & 2 \\ 3 & 2\end{vmatrix}
  3. Evaluate each 2\times2 determinant.
    |A|=2(0-0)-1(0-0)+6(6-6)=2(0)-1(0)+6(0)=0
|A|=0

Property 4: multiplying a row or column by a scalar

If each element of a row (or column) is multiplied by a non-zero scalar k, and the resulting matrix is B, then |B|=k|A|. In other words, a common factor of one row (or column) can be taken out of the determinant.

Derivation Proof of Property 4

Aim. Show that multiplying the first row by k multiplies the determinant by k.

  1. Multiply every element of row 1 of A by k.
    |B|=\begin{vmatrix}ka_{11} & ka_{12} & ka_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}
  2. Expand from the first row.
    |B|=ka_{11}\begin{vmatrix}a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}-ka_{12}\begin{vmatrix}a_{21} & a_{23} \\ a_{31} & a_{33}\end{vmatrix}+ka_{13}\begin{vmatrix}a_{21} & a_{22} \\ a_{31} & a_{32}\end{vmatrix}
  3. Take k out as a common factor of the three terms.
    |B|=k\left\{a_{11}\begin{vmatrix}a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21} & a_{23} \\ a_{31} & a_{33}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21} & a_{22} \\ a_{31} & a_{32}\end{vmatrix}\right\}
  4. The bracket is the expansion of |A|.
    |B|=k|A|
Result|B|=k|A|
Example 17: multiplying a row by 3
Let A=\begin{bmatrix}1 & 3 & 2 \\ 2 & 1 & 0 \\ 1 & 0 & 3\end{bmatrix}. Multiply the second row by 3 to get B and compare |B| with |A|.
  1. Expand |A| along the first row.
    |A|=1\begin{vmatrix}1 & 0 \\ 0 & 3\end{vmatrix}-3\begin{vmatrix}2 & 0 \\ 1 & 3\end{vmatrix}+2\begin{vmatrix}2 & 1 \\ 1 & 0\end{vmatrix}=1(3-0)-3(6-0)+2(0-1)=3-18-2=-17
  2. Multiply each element of R_2 by 3.
    B=\begin{bmatrix}1 & 3 & 2 \\ 6 & 3 & 0 \\ 1 & 0 & 3\end{bmatrix}
  3. Expand |B| along the first row.
    |B|=1\begin{vmatrix}3 & 0 \\ 0 & 3\end{vmatrix}-3\begin{vmatrix}6 & 0 \\ 1 & 3\end{vmatrix}+2\begin{vmatrix}6 & 3 \\ 1 & 0\end{vmatrix}=1(9-0)-3(18-0)+2(0-3)=9-54-6=-51
  4. Compare: -51=3(-17), that is |B|=3|A|.
|B|=3|A|

2.4 (Properties 5 to 8)Properties of determinants: sums, zero rows, row operations, triangular matricesOften examined

Property 5: a row or column of sums

If each element of one column (or row) is the sum of two numbers, the determinant equals the sum of two determinants. The first has the first part of each sum in that column (row) and the remaining columns (rows) unchanged. The second has the second part of each sum in that column (row) and the remaining columns (rows) unchanged.

\begin{vmatrix}a_{11}+b_{11} & a_{12} & a_{13} \\ a_{21}+b_{21} & a_{22} & a_{23} \\ a_{31}+b_{31} & a_{32} & a_{33}\end{vmatrix}=\begin{vmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}+\begin{vmatrix}b_{11} & a_{12} & a_{13} \\ b_{21} & a_{22} & a_{23} \\ b_{31} & a_{32} & a_{33}\end{vmatrix}

Derivation Proof of Property 5

Aim. Show that a determinant whose first column consists of sums splits into two determinants.

  1. Expand the left side from the first column C_1. The entries are a_{11}+b_{11}, a_{21}+b_{21}, a_{31}+b_{31}, and the 2\times2 minors do not involve C_1.
    |A|=(a_{11}+b_{11})\begin{vmatrix}a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}-(a_{21}+b_{21})\begin{vmatrix}a_{12} & a_{13} \\ a_{32} & a_{33}\end{vmatrix}+(a_{31}+b_{31})\begin{vmatrix}a_{12} & a_{13} \\ a_{22} & a_{23}\end{vmatrix}
  2. Open each bracket, giving six terms.
    |A|=a_{11}\begin{vmatrix}a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}+b_{11}\begin{vmatrix}a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}-a_{21}\begin{vmatrix}a_{12} & a_{13} \\ a_{32} & a_{33}\end{vmatrix}-b_{21}\begin{vmatrix}a_{12} & a_{13} \\ a_{32} & a_{33}\end{vmatrix}+a_{31}\begin{vmatrix}a_{12} & a_{13} \\ a_{22} & a_{23}\end{vmatrix}+b_{31}\begin{vmatrix}a_{12} & a_{13} \\ a_{22} & a_{23}\end{vmatrix}
  3. Collect the terms with a together and the terms with b together.
    |A|=\left(a_{11}\begin{vmatrix}a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}-a_{21}\begin{vmatrix}a_{12} & a_{13} \\ a_{32} & a_{33}\end{vmatrix}+a_{31}\begin{vmatrix}a_{12} & a_{13} \\ a_{22} & a_{23}\end{vmatrix}\right)+\left(b_{11}\begin{vmatrix}a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}-b_{21}\begin{vmatrix}a_{12} & a_{13} \\ a_{32} & a_{33}\end{vmatrix}+b_{31}\begin{vmatrix}a_{12} & a_{13} \\ a_{22} & a_{23}\end{vmatrix}\right)
  4. Each bracket is an expansion from the first column: the first of the determinant with C_1=(a_{11},a_{21},a_{31}), the second of the determinant with C_1=(b_{11},b_{21},b_{31}).
    |A|=\begin{vmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}+\begin{vmatrix}b_{11} & a_{12} & a_{13} \\ b_{21} & a_{22} & a_{23} \\ b_{31} & a_{32} & a_{33}\end{vmatrix}
Result\text{determinant of sums}=\text{first determinant}+\text{second determinant}
Example 18: splitting a column of sums
Show that \begin{vmatrix}1+2 & 1 & 0 \\ 3-1 & 0 & 2 \\ 2+3 & 2 & 1\end{vmatrix}=\begin{vmatrix}1 & 1 & 0 \\ 3 & 0 & 2 \\ 2 & 2 & 1\end{vmatrix}+\begin{vmatrix}2 & 1 & 0 \\ -1 & 0 & 2 \\ 3 & 2 & 1\end{vmatrix}.
  1. Left side. Simplify the first column.
    \text{L.H.S.}=\begin{vmatrix}3 & 1 & 0 \\ 2 & 0 & 2 \\ 5 & 2 & 1\end{vmatrix}
  2. Expand from the first row.
    =3\begin{vmatrix}0 & 2 \\ 2 & 1\end{vmatrix}-1\begin{vmatrix}2 & 2 \\ 5 & 1\end{vmatrix}+0\begin{vmatrix}2 & 0 \\ 5 & 2\end{vmatrix}=3(0-4)-1(2-10)+0(4-0)
  3. Evaluate.
    =-12+8+0=-4\qquad(1)
  4. Right side. Expand the first determinant from its first row.
    \begin{vmatrix}1 & 1 & 0 \\ 3 & 0 & 2 \\ 2 & 2 & 1\end{vmatrix}=1\begin{vmatrix}0 & 2 \\ 2 & 1\end{vmatrix}-1\begin{vmatrix}3 & 2 \\ 2 & 1\end{vmatrix}+0=(0-4)-(3-4)+0=-4+1=-3
  5. Expand the second determinant from its first row.
    \begin{vmatrix}2 & 1 & 0 \\ -1 & 0 & 2 \\ 3 & 2 & 1\end{vmatrix}=2\begin{vmatrix}0 & 2 \\ 2 & 1\end{vmatrix}-1\begin{vmatrix}-1 & 2 \\ 3 & 1\end{vmatrix}+0=2(0-4)-1(-1-6)+0=-8+7=-1
  6. Add the two values.
    \text{R.H.S.}=-3+(-1)=-4\qquad(2)
  7. From (1) and (2), L.H.S. = R.H.S.
\text{L.H.S.}=\text{R.H.S.}=-4

Property 6: a row or column of zeros

If all the elements of a row (or column) of a square matrix A are zero, then |A|=0.

Derivation Proof of Property 6

Aim. Show that a determinant with a zero row is zero.

  1. Take A with its second row zero.
    |A|=\begin{vmatrix}a_{11} & a_{12} & a_{13} \\ 0 & 0 & 0 \\ a_{31} & a_{32} & a_{33}\end{vmatrix}
  2. Expand from the first row. Each 2\times2 minor contains the zero row.
    |A|=a_{11}\begin{vmatrix}0 & 0 \\ a_{32} & a_{33}\end{vmatrix}-a_{12}\begin{vmatrix}0 & 0 \\ a_{31} & a_{33}\end{vmatrix}+a_{13}\begin{vmatrix}0 & 0 \\ a_{31} & a_{32}\end{vmatrix}
  3. Each minor is 0\cdot(\ )-0\cdot(\ )=0.
    |A|=a_{11}(0-0)-a_{12}(0-0)+a_{13}(0-0)=0+0+0=0
Result|A|=0
Example 19: a zero row
If A=\begin{bmatrix}3 & 1 & 2 \\ 0 & 0 & 0 \\ 1 & 3 & 9\end{bmatrix}, find |A|.
  1. Expand from the first row.
    |A|=\begin{vmatrix}3 & 1 & 2 \\ 0 & 0 & 0 \\ 1 & 3 & 9\end{vmatrix}=3\begin{vmatrix}0 & 0 \\ 3 & 9\end{vmatrix}-1\begin{vmatrix}0 & 0 \\ 1 & 9\end{vmatrix}+2\begin{vmatrix}0 & 0 \\ 1 & 3\end{vmatrix}
  2. Every minor has a zero row, so it is zero.
    |A|=3(0-0)-1(0-0)+2(0-0)=0-0+0=0
|A|=0

Property 7: adding a multiple of a row or column

If we multiply any row (column) of a square matrix by a scalar k and add the result to the corresponding elements of any other row (column), the value of the determinant is unchanged.

Derivation Proof of Property 7

Aim. Show that adding k times row 2 to row 1 leaves the determinant unchanged.

  1. Start from |A|. Multiply R_2 by k and add it to R_1. Call the new matrix B.
    |B|=\begin{vmatrix}a_{11}+ka_{21} & a_{12}+ka_{22} & a_{13}+ka_{23} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}
  2. Row 1 consists of sums, so use Property 5 to split the determinant in two.
    |B|=\begin{vmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}+\begin{vmatrix}ka_{21} & ka_{22} & ka_{23} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}
  3. In the second determinant take the common factor k out of row 1 (Property 4).
    |B|=|A|+k\begin{vmatrix}a_{21} & a_{22} & a_{23} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}
  4. The remaining determinant has two identical rows (R_1 and R_2), so it is zero (Property 3).
    |B|=|A|+k(0)=|A|
Result|B|=|A|

This is the most useful property for evaluation: adding a multiple of one row to another can create zeros without changing the value.

Example 20: adding twice a row to another
Let A=\begin{bmatrix}3 & 2 & 0 \\ 1 & 4 & -1 \\ 3 & -1 & 2\end{bmatrix}. Multiply R_1 by 2 and add it to R_3. Compare the determinants.
  1. Expand |A| from the first row.
    |A|=\begin{vmatrix}3 & 2 & 0 \\ 1 & 4 & -1 \\ 3 & -1 & 2\end{vmatrix}=3\begin{vmatrix}4 & -1 \\ -1 & 2\end{vmatrix}-2\begin{vmatrix}1 & -1 \\ 3 & 2\end{vmatrix}+0\begin{vmatrix}1 & 4 \\ 3 & -1\end{vmatrix}
  2. Evaluate.
    |A|=3(8-1)-2(2+3)+0(-1-12)=21-10+0=11
  3. Apply R_3+2R_1: add 2\times(3,2,0)=(6,4,0) to R_3=(3,-1,2), giving (9,3,2).
    B=\begin{bmatrix}3 & 2 & 0 \\ 1 & 4 & -1 \\ 9 & 3 & 2\end{bmatrix}
  4. Expand |B| from the first row.
    |B|=3\begin{vmatrix}4 & -1 \\ 3 & 2\end{vmatrix}-2\begin{vmatrix}1 & -1 \\ 9 & 2\end{vmatrix}+0\begin{vmatrix}1 & 4 \\ 9 & 3\end{vmatrix}
  5. Evaluate.
    |B|=3(8+3)-2(2+9)+0(3-36)=33-22+0=11
  6. So |A|=|B|.
|A|=|B|=11

Property 8: triangular and diagonal matrices

If a square matrix A is upper triangular, lower triangular or diagonal, then |A| is the product of its diagonal elements.

Derivation Proof of Property 8

Aim. Show that the determinant of an upper triangular 3\times3 matrix is a_{11}a_{22}a_{33}.

  1. Take the upper triangular matrix (zeros below the diagonal).
    |A|=\begin{vmatrix}a_{11} & a_{12} & a_{13} \\ 0 & a_{22} & a_{23} \\ 0 & 0 & a_{33}\end{vmatrix}
  2. Expand from the first column C_1. Its last two entries are zero, so only the first term survives.
    |A|=a_{11}\begin{vmatrix}a_{22} & a_{23} \\ 0 & a_{33}\end{vmatrix}-0+0
  3. Evaluate the 2\times2 determinant.
    |A|=a_{11}(a_{22}a_{33}-0)=a_{11}a_{22}a_{33}
Result|A|=a_{11}a_{22}a_{33}=\text{product of the diagonal elements}

The same happens for a lower triangular or a diagonal matrix. This is why reducing a determinant to triangular form gives its value at once.

Example 21: a lower triangular matrix
If A=\begin{bmatrix}3 & 0 & 0 \\ 2 & 1 & 0 \\ 4 & 3 & 2\end{bmatrix}, find |A|.
  1. Expand from the first row. The last two entries are zero.
    |A|=\begin{vmatrix}3 & 0 & 0 \\ 2 & 1 & 0 \\ 4 & 3 & 2\end{vmatrix}=3\begin{vmatrix}1 & 0 \\ 3 & 2\end{vmatrix}-0+0
  2. Evaluate.
    |A|=3(2-0)=6
  3. This equals (3)(1)(2), the product of the diagonal elements.
|A|=6
The eight properties of determinants
PropertyStatement
1|A^{t}|=|A|
2Interchanging two rows (columns) changes the sign: |B|=-|A|
3Two identical rows (columns) give |A|=0
4Multiplying one row (column) by k multiplies the determinant by k: |B|=k|A|
5If a row (column) consists of sums, the determinant splits into a sum of two determinants
6A zero row (column) gives |A|=0
7Adding a multiple of a row (column) to another row (column) leaves |A| unchanged
8For triangular and diagonal matrices, |A| is the product of the diagonal elements

2.4.1Evaluation of determinants without expansionOften examined

Evaluation of determinants without expansion (2.4.1)

The properties let us find the value of a determinant, or prove that it has a given value, without expanding it. The usual plan is: use Property 7 (row and column operations) to create a zero row or column, or two identical rows, or to make a common factor, then apply Property 3, 4 or 6.

Example 22: determinant equal to zero
Without expansion show that \begin{vmatrix}a-2l & b-2m & c-2n \\ l & m & n \\ a & b & c\end{vmatrix}=0.
  1. The first row is a sum of two parts: (a,b,c) and (-2l,-2m,-2n). Split the determinant by Property 5.
    \text{L.H.S.}=\begin{vmatrix}a & b & c \\ l & m & n \\ a & b & c\end{vmatrix}+\begin{vmatrix}-2l & -2m & -2n \\ l & m & n \\ a & b & c\end{vmatrix}
  2. In the second determinant take the common factor -2 out of the first row (Property 4).
    =\begin{vmatrix}a & b & c \\ l & m & n \\ a & b & c\end{vmatrix}-2\begin{vmatrix}l & m & n \\ l & m & n \\ a & b & c\end{vmatrix}
  3. The first determinant has two identical rows (R_1 and R_3) and the second also has two identical rows (R_1 and R_2). By Property 3 both are zero.
    =0-2(0)=0
\text{L.H.S.}=0
Example 23: using a+b+c=0
If a+b+c=0, then without expanding show that \begin{vmatrix}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{vmatrix}=0.
  1. Add R_2 and R_3 to R_1 (operation R_1+(R_2+R_3), Property 7). Each entry of the new first row is the sum 2a+2b+2c.
    \begin{aligned}&\text{L.H.S.}=\\ &\begin{vmatrix}2a+2b+2c & 2a+2b+2c & 2a+2b+2c \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{vmatrix}\end{aligned}
  2. Take the common factor out of R_1 (Property 4).
    =\;\begin{vmatrix}2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{vmatrix}
  3. Use a+b+c=0. The first row becomes a row of zeros.
    =\begin{vmatrix}2(0) & 2(0) & 2(0) \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{vmatrix}=\begin{vmatrix}0 & 0 & 0 \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{vmatrix}
  4. A determinant with a zero row is zero (Property 6).
    =0
\text{L.H.S.}=0
Example 24: proving a value
Prove that \begin{vmatrix}a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ac & bc & c^2+1\end{vmatrix}=1+a^2+b^2+c^2.
  1. Write each diagonal entry as a product: a^2+1=a^2\left(1+\dfrac1{a^2}\right), and similarly for b and c.
    \text{L.H.S.}=\begin{vmatrix}a^2\left(1+\frac1{a^2}\right) & ab & ac \\ ab & b^2\left(1+\frac1{b^2}\right) & bc \\ ac & bc & c^2\left(1+\frac1{c^2}\right)\end{vmatrix}
  2. Take out the common factor a from R_1, b from R_2 and c from R_3 (Property 4).
    =abc\begin{vmatrix}a\left(1+\frac1{a^2}\right) & b & c \\ a & b\left(1+\frac1{b^2}\right) & c \\ a & b & c\left(1+\frac1{c^2}\right)\end{vmatrix}
  3. Take out a from C_1, b from C_2 and c from C_3. Together with abc this gives the factor a^2b^2c^2.
    =a^2b^2c^2\begin{vmatrix}1+\frac1{a^2} & 1 & 1 \\ 1 & 1+\frac1{b^2} & 1 \\ 1 & 1 & 1+\frac1{c^2}\end{vmatrix}
  4. Apply C_1-C_3 and C_2-C_3 (Property 7). This creates zeros.
    =a^2b^2c^2\begin{vmatrix}\frac1{a^2} & 0 & 1 \\ 0 & \frac1{b^2} & 1 \\ -\frac1{c^2} & -\frac1{c^2} & 1+\frac1{c^2}\end{vmatrix}
  5. Expand from the first row R_1.
    =a^2b^2c^2\left[\frac1{a^2}\begin{vmatrix}\frac1{b^2} & 1 \\ -\frac1{c^2} & 1+\frac1{c^2}\end{vmatrix}-0+1\begin{vmatrix}0 & \frac1{b^2} \\ -\frac1{c^2} & -\frac1{c^2}\end{vmatrix}\right]
  6. Evaluate the two 2\times2 determinants.
    =a^2b^2c^2\left[\frac1{a^2}\left\{\frac1{b^2}\left(1+\frac1{c^2}\right)+\frac1{c^2}\right\}+\frac1{b^2c^2}\right]
  7. Open the brackets.
    =a^2b^2c^2\left[\frac1{a^2b^2}+\frac1{a^2b^2c^2}+\frac1{a^2c^2}+\frac1{b^2c^2}\right]
  8. Multiply a^2b^2c^2 into each term.
    =c^2+1+b^2+a^2=1+a^2+b^2+c^2
  9. This is the R.H.S., so the result is proved.
1+a^2+b^2+c^2
Example 25: a determinant equal to a square
Prove that \begin{vmatrix}1 & a^2 & a \\ a & 1 & a^2 \\ a^2 & a & 1\end{vmatrix}=(1-a^3)^2.
  1. Apply R_1+(R_2+R_3) (Property 7). Every entry of the new first row is 1+a+a^2.
    \text{L.H.S.}=\begin{vmatrix}1+a+a^2 & 1+a+a^2 & 1+a+a^2 \\ a & 1 & a^2 \\ a^2 & a & 1\end{vmatrix}
  2. Take the common factor (1+a+a^2) out of R_1.
    =(1+a+a^2)\begin{vmatrix}1 & 1 & 1 \\ a & 1 & a^2 \\ a^2 & a & 1\end{vmatrix}
  3. Apply C_2-C_1 and C_3-C_1 to create zeros in the first row.
    =(1+a+a^2)\begin{vmatrix}1 & 0 & 0 \\ a & 1-a & a^2-a \\ a^2 & a-a^2 & 1-a^2\end{vmatrix}
  4. Expand from the first row. Only the first entry is non-zero.
    =(1+a+a^2)\left(1\begin{vmatrix}1-a & a^2-a \\ a-a^2 & 1-a^2\end{vmatrix}-0+0\right)
  5. Factor each column: a-a^2=a(1-a), a^2-a=-a(1-a), 1-a^2=(1-a)(1+a). Take (1-a) out of C_1 and (1-a) out of C_2.
    =(1+a+a^2)(1-a)(1-a)\begin{vmatrix}1 & -a \\ a & 1+a\end{vmatrix}
  6. Evaluate the 2\times2 determinant: (1)(1+a)-(-a)(a)=1+a+a^2.
    =(1+a+a^2)(1-a)(1-a)(1+a+a^2)
  7. Pair (1-a) with (1+a+a^2) twice, using (1-a)(1+a+a^2)=1-a^3.
    =\big[(1-a)(1+a+a^2)\big]^2=(1-a^3)^2
  8. This is the R.H.S.
(1-a^3)^2

2.5 (2.5.1 to 2.5.5)Row and column operations on matricesOften examined

Row and column operations on matrices (2.5.1)

Elementary row operations

The following elementary row operations can be performed on a matrix:

  1. Interchange any two rows. Interchanging the ith and jth rows is denoted R_{ij}.
  2. Multiply any row by a non-zero scalar k. Multiplying the ith row by k is denoted kR_i.
  3. Add a multiple of any row to the corresponding elements of any other row. Adding k times the jth row to the ith row is denoted R_i+kR_j.
Elementary column operations

In the same way, the elementary column operations are:

  1. Interchange any two columns: the ith with the jth column is denoted C_{ij}.
  2. Multiply any column by a non-zero scalar k: denoted kC_i.
  3. Add k times the jth column to the ith column: denoted C_i+kC_j.

Echelon form of a matrix (2.5.2)

A matrix is in row echelon form if:

  1. the first non-zero element in each row, called the leading entry, is 1;
  2. each leading entry is in a column to the right of the leading entry in the previous row;
  3. any row which has all elements zero is below the rows having at least one non-zero element.

Reduced echelon form of a matrix (2.5.3)

A matrix is in row reduced echelon form if:

  1. it satisfies the conditions of row echelon form, and
  2. the leading entry in each row is the only non-zero entry in its column.
Example 26: reduction to echelon form
Reduce the matrix A=\begin{bmatrix}3 & 1 & 2 \\ -2 & 4 & 1 \\ 1 & 0 & 2\end{bmatrix} into the echelon form.
  1. The leading entry of row 1 should be 1. Interchange R_1 and R_3 (the third row already starts with 1).
    A\sim\begin{bmatrix}1 & 0 & 2 \\ -2 & 4 & 1 \\ 3 & 1 & 2\end{bmatrix}\quad(R_{13})
  2. Make the entries below the leading 1 zero: R_2+2R_1 and R_3-3R_1.
    \sim\begin{bmatrix}1 & 0 & 2 \\ 0 & 4 & 5 \\ 0 & 1 & -4\end{bmatrix}\quad(R_2+2R_1,\;R_3-3R_1)
  3. Interchange R_2 and R_3 so that the next leading entry becomes 1.
    \sim\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -4 \\ 0 & 4 & 5\end{bmatrix}\quad(R_{23})
  4. Make the entry below this leading 1 zero: R_3-4R_2, since 5-4(-4)=21.
    \sim\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -4 \\ 0 & 0 & 21\end{bmatrix}\quad(R_3-4R_2)
  5. Make the last leading entry 1: multiply R_3 by \tfrac1{21}.
    \sim\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -4 \\ 0 & 0 & 1\end{bmatrix}\quad(\tfrac1{21}R_3)
  6. The leading entries 1 move to the right row by row, so this is the required echelon form of A.
\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -4 \\ 0 & 0 & 1\end{bmatrix}
Example 27: reduction to reduced echelon form
Convert the matrix A=\begin{bmatrix}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 5 \\ 4 & 5 & 6 & 7\end{bmatrix} into the reduced echelon form.
  1. The first leading entry is already 1. Make the entries below it zero: R_2-2R_1 and R_3-4R_1.
    A\sim\begin{bmatrix}1 & 2 & 3 & 4 \\ 0 & -1 & -2 & -3 \\ 0 & -3 & -6 & -9\end{bmatrix}\quad(R_2-2R_1,\;R_3-4R_1)
  2. Make the leading entries of rows 2 and 3 equal to 1: multiply R_2 by -1 and R_3 by -\tfrac13.
    \sim\begin{bmatrix}1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3\end{bmatrix}\quad(-R_2,\;-\tfrac13R_3)
  3. To reach the reduced form, clear the entries above and below the leading 1 of row 2: R_1-2R_2 and R_3-R_2.
    \sim\begin{bmatrix}1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0\end{bmatrix}\quad(R_1-2R_2,\;R_3-R_2)
  4. Each leading 1 is the only non-zero entry in its column and the zero row is at the bottom, so this is the reduced echelon form.
\begin{bmatrix}1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0\end{bmatrix}

Rank of a matrix (2.5.4)

To find the rank of a matrix, reduce it to its echelon form (or reduced echelon form). The number of non-zero rows is called the rank (or row rank) of the matrix.

Example 28: rank of a matrix
Find the rank of the matrix A=\begin{bmatrix}2 & 5 & 7 \\ 1 & 2 & -1 \\ -3 & -6 & 3\end{bmatrix}.
  1. Interchange R_1 and R_2 to get a leading entry 1 in row 1.
    A\sim\begin{bmatrix}1 & 2 & -1 \\ 2 & 5 & 7 \\ -3 & -6 & 3\end{bmatrix}\quad(R_{12})
  2. Clear below it: R_2-2R_1 and R_3+3R_1.
    \sim\begin{bmatrix}1 & 2 & -1 \\ 0 & 1 & 9 \\ 0 & 0 & 0\end{bmatrix}\quad(R_2-2R_1,\;R_3+3R_1)
  3. This is an echelon form of A. It has two non-zero rows and one zero row.
\operatorname{Rank}(A)=2

Using row operations to find the inverse (2.5.5)

Row operations can be used on a non-singular matrix A to find its inverse. Take the identity matrix I of the same order as A and write A and I side by side, as [A:I]. Perform the same row operations on both, so that A is reduced to I. Then I is reduced to a new matrix, which is A^{-1}:

[A:I]\;\sim\;[I:A^{-1}]
Example 29: inverse by row operations
Find A^{-1} if A=\begin{bmatrix}2 & 1 & 0 \\ 4 & 3 & 1 \\ 1 & 0 & 2\end{bmatrix}, by using row operations.
  1. First check that A is non-singular. Expand |A| from the first row.
    |A|=\begin{vmatrix}2 & 1 & 0 \\ 4 & 3 & 1 \\ 1 & 0 & 2\end{vmatrix}=2\begin{vmatrix}3 & 1 \\ 0 & 2\end{vmatrix}-1\begin{vmatrix}4 & 1 \\ 1 & 2\end{vmatrix}+0\begin{vmatrix}4 & 3 \\ 1 & 0\end{vmatrix}=2(6-0)-1(8-1)+0(0-3)=12-7=5\neq0
  2. So A is non-singular. Write the augmented matrix [A:I].
    [A:I]=\left[\begin{array}{ccc|ccc}2 & 1 & 0 & 1 & 0 & 0 \\ 4 & 3 & 1 & 0 & 1 & 0 \\ 1 & 0 & 2 & 0 & 0 & 1\end{array}\right]
  3. Interchange R_1 and R_3 to put a 1 at the top.
    \sim\left[\begin{array}{ccc|ccc}1 & 0 & 2 & 0 & 0 & 1 \\ 4 & 3 & 1 & 0 & 1 & 0 \\ 2 & 1 & 0 & 1 & 0 & 0\end{array}\right]\quad(R_{13})
  4. Clear below the leading 1: R_2-4R_1 and R_3-2R_1.
    \sim\left[\begin{array}{ccc|ccc}1 & 0 & 2 & 0 & 0 & 1 \\ 0 & 3 & -7 & 0 & 1 & -4 \\ 0 & 1 & -4 & 1 & 0 & -2\end{array}\right]\quad(R_2-4R_1,\;R_3-2R_1)
  5. Interchange R_2 and R_3 so that the second leading entry is 1.
    \sim\left[\begin{array}{ccc|ccc}1 & 0 & 2 & 0 & 0 & 1 \\ 0 & 1 & -4 & 1 & 0 & -2 \\ 0 & 3 & -7 & 0 & 1 & -4\end{array}\right]\quad(R_{23})
  6. Clear below it: R_3-3R_2.
    \sim\left[\begin{array}{ccc|ccc}1 & 0 & 2 & 0 & 0 & 1 \\ 0 & 1 & -4 & 1 & 0 & -2 \\ 0 & 0 & 5 & -3 & 1 & 2\end{array}\right]\quad(R_3-3R_2)
  7. Make the third leading entry 1: multiply R_3 by \tfrac15.
    \sim\left[\begin{array}{ccc|ccc}1 & 0 & 2 & 0 & 0 & 1 \\ 0 & 1 & -4 & 1 & 0 & -2 \\ 0 & 0 & 1 & -\tfrac35 & \tfrac15 & \tfrac25\end{array}\right]\quad(\tfrac15R_3)
  8. Clear above the last leading 1: R_1-2R_3 and R_2+4R_3. The left half is now I.
    \sim\left[\begin{array}{ccc|ccc}1 & 0 & 0 & \tfrac65 & -\tfrac25 & \tfrac15 \\ 0 & 1 & 0 & -\tfrac75 & \tfrac45 & -\tfrac25 \\ 0 & 0 & 1 & -\tfrac35 & \tfrac15 & \tfrac25\end{array}\right]\quad(R_1-2R_3,\;R_2+4R_3)
  9. The right half is A^{-1}.
A^{-1}=\begin{bmatrix}\tfrac65 & -\tfrac25 & \tfrac15 \\ -\tfrac75 & \tfrac45 & -\tfrac25 \\ -\tfrac35 & \tfrac15 & \tfrac25\end{bmatrix}

2.6 and 2.6.1Linear equations and systems of linear equations

Linear equation

An equation of the form a_1x_1+a_2x_2=k_1, where a_1,a_2 and k_1 are constants and at least one of a_1 and a_2 is non-zero, is a linear equation in two variables x_1 and x_2.

Similarly, an equation of the form a_1x_1+a_2x_2+a_3x_3=k_2, where a_1,a_2,a_3 and k_2 are constants and at least one of a_1,a_2,a_3 is non-zero, is a linear equation in three variables x_1,x_2,x_3. In the same manner this extends to n variables.

System of linear equations

When we deal with more than one linear equation at the same time, we have a system of linear equations. A system is divided into two categories:

  1. homogeneous system of linear equations;
  2. non-homogeneous system of linear equations.

Homogeneous and non-homogeneous linear equations (2.6.1)

Homogeneous system

Consider the system of three linear equations in three unknowns

\begin{aligned}a_1x+b_1y+c_1z&=k_1\\a_2x+b_2y+c_2z&=k_2\\a_3x+b_3y+c_3z&=k_3\end{aligned}

If k_1=k_2=k_3=0, the system is called a homogeneous system of linear equations:

\begin{aligned}a_1x+b_1y+c_1z&=0\\a_2x+b_2y+c_2z&=0\\a_3x+b_3y+c_3z&=0\end{aligned}
Non-homogeneous system

For the same system, if at least one of k_1,k_2,k_3 is non-zero, the system is called a non-homogeneous system of linear equations.

Two kinds of system
SystemConstants k_1,k_2,k_3Example
Homogeneousk_1=k_2=k_3=0x+3y+2z=0
Non-homogeneousat least one of k_1,k_2,k_3\neq02x+3y-z=1

2.6.2Solution of a homogeneous system of linear equationsOften examined

Solution of a system of linear equations (2.6.2)

The values of the variables which, when substituted in any equation of the system, satisfy that equation are the solution of the system. A system may have no solution, a unique solution, or infinitely many solutions.

Solution of a homogeneous system

Consider the homogeneous system

\begin{aligned}a_1x+b_1y+c_1z&=0\\a_2x+b_2y+c_2z&=0\\a_3x+b_3y+c_3z&=0\end{aligned}

In matrix form it is written

\begin{bmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}

Let A be the matrix of coefficients, X the column of unknowns and O the zero column. Then the system is

AX=O\qquad(1)

Each equation is satisfied by x=0,\;y=0,\;z=0. So (0,0,0) is always a solution of a homogeneous system. Because this solution always exists, it is called the trivial solution. Every solution other than the trivial one is a non-trivial solution.

Derivation A homogeneous system with |A|\neq0 has only the trivial solution

Aim. Show that if the coefficient matrix A is non-singular, then AX=O has only the solution X=O.

  1. If A is non-singular, A^{-1} exists. Multiply both sides of (1) on the left by A^{-1}.
    A^{-1}(AX)=A^{-1}O
  2. Regroup on the left, and note that any matrix times the zero matrix is zero.
    (A^{-1}A)X=O
  3. Use A^{-1}A=I.
    IX=O
  4. Use IX=X.
    X=O
  5. In column form this is
    \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\;\Rightarrow\;x=0,\;y=0,\;z=0
Result|A|\neq0\;\Longrightarrow\;\text{only the trivial solution}

So a homogeneous system can have a non-trivial solution only if |A|=0 (the matrix A is singular).

Example 30: how many solutions?
How many solutions does the following system of homogeneous linear equations have?
\begin{aligned}3x-2y+z&=0\qquad(1)\\2x+y-3z&=0\qquad(2)\\x-y+z&=0\qquad(3)\end{aligned}
  1. Write the coefficient matrix A.
    A=\begin{bmatrix}3 & -2 & 1 \\ 2 & 1 & -3 \\ 1 & -1 & 1\end{bmatrix}
  2. Find |A| by expanding along the first row.
    |A|=3\begin{vmatrix}1 & -3 \\ -1 & 1\end{vmatrix}-(-2)\begin{vmatrix}2 & -3 \\ 1 & 1\end{vmatrix}+1\begin{vmatrix}2 & 1 \\ 1 & -1\end{vmatrix}
  3. Evaluate.
    |A|=3(1-3)+2(2+3)+1(-2-1)=-6+10-3=1\neq0
  4. Since |A|\neq0, A is non-singular, so the system has only the trivial solution x=y=z=0: exactly one solution.
|A|=1\neq0\;\Rightarrow\;x=y=z=0\text{ only}
Example 31: a homogeneous system with non-trivial solutions
Solve the homogeneous system of linear equations.
\begin{aligned}x+3y+2z&=0\qquad(1)\\2x-y+3z&=0\qquad(2)\\x-4y+z&=0\qquad(3)\end{aligned}
  1. Write the coefficient matrix and find |A|.
    A=\begin{bmatrix}1 & 3 & 2 \\ 2 & -1 & 3 \\ 1 & -4 & 1\end{bmatrix}
  2. Expand along the first row.
    |A|=1\begin{vmatrix}-1 & 3 \\ -4 & 1\end{vmatrix}-3\begin{vmatrix}2 & 3 \\ 1 & 1\end{vmatrix}+2\begin{vmatrix}2 & -1 \\ 1 & -4\end{vmatrix}=1(-1+12)-3(2-3)+2(-8+1)
  3. Evaluate.
    |A|=11+3-14=0
  4. Since |A|=0, the system has non-trivial solutions. Now solve it by elimination.
  5. Eliminate x from (1) and (2): multiply (1) by 2, then subtract (2).
    \begin{aligned}2x+6y+4z&=0\\2x-y+3z&=0\\\hline 7y+z&=0\end{aligned}\qquad(4)
  6. Eliminate x from (1) and (3): subtract (3) from (1).
    \begin{aligned}x+3y+2z&=0\\x-4y+z&=0\\\hline 7y+z&=0\end{aligned}\qquad(5)
  7. Equations (4) and (5) are identical, so we really have only two independent equations for three unknowns. One unknown can be chosen freely. Put z=t.
  8. Put z=t in (4) and solve for y.
    7y+t=0\;\Rightarrow\;7y=-t\;\Rightarrow\;y=-\tfrac17t
  9. Substitute y and z in equation (1) and solve for x.
    x+3\left(-\tfrac17t\right)+2t=0\;\Rightarrow\;x-\tfrac37t+2t=0\;\Rightarrow\;x+\tfrac{11}7t=0\;\Rightarrow\;x=-\tfrac{11}7t
  10. Check in equation (2): 2x-y+3z=-\tfrac{22}7t+\tfrac17t+3t=\tfrac{-22+1+21}{7}t=0. It holds for every t.
  11. Assigning different values to t\in\mathbb{R} gives different solutions, so the system has infinitely many solutions.
(x,y,z)=\left(-\tfrac{11}{7}t,\;-\tfrac17t,\;t\right),\quad t\in\mathbb{R}

2.6.3 to 2.6.8Consistent and inconsistent systems, consistency criteriaOften examined

Consistent system of equations (2.6.3)

A system of linear equations which has at least one solution is a consistent system of equations.

Inconsistent system of equations (2.6.4)

A system of linear equations which has no solution at all is an inconsistent system of equations.

Solution of a non-homogeneous system (2.6.5)

Consider the non-homogeneous system

\begin{aligned}a_1x+b_1y+c_1z&=k_1\\a_2x+b_2y+c_2z&=k_2\\a_3x+b_3y+c_3z&=k_3\end{aligned}

where at least one of k_1,k_2,k_3 is non-zero. In matrix form

\begin{bmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}k_1 \\ k_2 \\ k_3\end{bmatrix}

Let A=\begin{bmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{bmatrix}, X=\begin{bmatrix}x\\y\\z\end{bmatrix} and B=\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}. Then the system is

AX=B

Augmented matrix (2.6.7)

For a system of linear equations, the matrix made of the coefficients of the unknowns together with the constants on the right side is the augmented matrix. It is usually denoted A_b. For the system above,

A_b=\left[\begin{array}{ccc|c}a_1&b_1&c_1&k_1\\a_2&b_2&c_2&k_2\\a_3&b_3&c_3&k_3\end{array}\right]

Consistency criteria (2.6.6)

A system of linear equations is consistent if \operatorname{Rank}A=\operatorname{Rank}A_b, and inconsistent if \operatorname{Rank}A\neq\operatorname{Rank}A_b. When the system is consistent:

  • if \operatorname{Rank}A=\operatorname{Rank}A_b= number of unknowns, the system has a unique solution;
  • if \operatorname{Rank}A=\operatorname{Rank}A_b< number of unknowns, the system has infinitely many solutions.
Summary of the consistency criteria
Rank conditionNumber of solutions
\operatorname{Rank}A\neq\operatorname{Rank}A_bno solution (inconsistent)
\operatorname{Rank}A=\operatorname{Rank}A_b= number of unknownsexactly one (unique) solution
\operatorname{Rank}A=\operatorname{Rank}A_b< number of unknownsinfinitely many solutions

Methods to solve a non-homogeneous system (2.6.8)

To solve a non-homogeneous system of linear equations of order 3\times3, the book uses four methods:

  • Matrix inversion method
  • Gauss elimination method (echelon form)
  • Gauss-Jordan method (reduced echelon form)
  • Cramer's rule
Worked example: consistent or inconsistent, by comparing ranks
Decide whether x+y+z=3,\; x+2y+3z=6,\; 2x+3y+4z=10 is consistent. If not, change the constant 10 so that it is, and then solve.
  1. Write the augmented matrix A_b.
    A_b=\left[\begin{array}{ccc|c}1&1&1&3\\1&2&3&6\\2&3&4&10\end{array}\right]
  2. Make zeros below the first leading 1: R_2-R_1 and R_3-2R_1.
    \sim\left[\begin{array}{ccc|c}1&1&1&3\\0&1&2&3\\0&1&2&4\end{array}\right]
  3. Next R_3-R_2.
    \sim\left[\begin{array}{ccc|c}1&1&1&3\\0&1&2&3\\0&0&0&1\end{array}\right]
  4. The last row says 0=1. Rank of A (non-zero rows of the coefficient part) is 2, but rank of A_b is 3.
    \operatorname{Rank}A=2\neq3=\operatorname{Rank}A_b
  5. The ranks differ, so the system is inconsistent: it has no solution. (Indeed, adding the first two equations gives 2x+3y+4z=9, which contradicts 10.)
  6. If the last constant is 9, then R_3-R_2 gives a zero row [0\;0\;0\;|\;0], so both ranks are 2, less than the 3 unknowns: infinitely many solutions.
    \operatorname{Rank}A=\operatorname{Rank}A_b=2<3
  7. Let z=t. From row 2, y=3-2t. From row 1, x=3-y-z=3-(3-2t)-t=t.
    (x,y,z)=(t,\;3-2t,\;t)
  8. Check in the second equation: t+2(3-2t)+3t=6.
\text{inconsistent for }10;\;\text{for }9:\;(x,y,z)=(t,\,3-2t,\,t),\;t\in\mathbb{R}

2.6.8 (matrix inversion method)Solving a system by the matrix inversion methodOften examined

Matrix inversion method

Consider the non-homogeneous system a_1x+b_1y+c_1z=k_1,\;a_2x+b_2y+c_2z=k_2,\;a_3x+b_3y+c_3z=k_3. In matrix form it is AX=B.

Derivation The solution X=A^{-1}B

Aim. Solve AX=B for X when A is invertible (non-singular).

  1. If A is non-singular, A^{-1} exists. Multiply both sides of AX=B on the left by A^{-1}.
    A^{-1}(AX)=A^{-1}B
  2. Regroup the left side.
    (A^{-1}A)X=A^{-1}B
  3. Use A^{-1}A=I.
    IX=A^{-1}B
  4. Use IX=X.
    X=A^{-1}B
ResultX=A^{-1}B

The unknowns are found by multiplying the column of constants by the inverse of the coefficient matrix. This works only when |A|\neq0.

Method
  1. Write the system as AX=B and find |A|. If |A|=0 the method cannot be used.
  2. Find all nine cofactors, then \operatorname{adj}(A) and A^{-1}=\dfrac1{|A|}\operatorname{adj}(A).
  3. Multiply: X=A^{-1}B, and read off x,y,z.
Example 32: matrix inversion method
Solve the system of non-homogeneous linear equations by the matrix inversion method.
2x+3y-z=1;\qquad x-y+z=3;\qquad x+2y-z=1
  1. Write the system as AX=B.
    A=\begin{bmatrix}2 & 3 & -1 \\ 1 & -1 & 1 \\ 1 & 2 & -1\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}1 \\ 3 \\ 1\end{bmatrix}
  2. Find |A| by expanding along the first row.
    |A|=\begin{vmatrix}2 & 3 & -1 \\ 1 & -1 & 1 \\ 1 & 2 & -1\end{vmatrix}=2\begin{vmatrix}-1 & 1 \\ 2 & -1\end{vmatrix}-3\begin{vmatrix}1 & 1 \\ 1 & -1\end{vmatrix}-1\begin{vmatrix}1 & -1 \\ 1 & 2\end{vmatrix}
  3. Evaluate.
    |A|=2(1-2)-3(-1-1)-1(2+1)=-2+6-3=1\neq0
  4. So A is non-singular and the system is consistent with a unique solution. Now find the nine cofactors.
  5. Row 1 elements.
    A_{11}=(+1)\begin{vmatrix}-1 & 1 \\ 2 & -1\end{vmatrix}=1-2=-1,\quad A_{12}=(-1)\begin{vmatrix}1 & 1 \\ 1 & -1\end{vmatrix}=-(-1-1)=2,\quad A_{13}=(+1)\begin{vmatrix}1 & -1 \\ 1 & 2\end{vmatrix}=2+1=3
  6. Row 2 elements.
    A_{21}=(-1)\begin{vmatrix}3 & -1 \\ 2 & -1\end{vmatrix}=-(-3+2)=1,\quad A_{22}=(+1)\begin{vmatrix}2 & -1 \\ 1 & -1\end{vmatrix}=-2+1=-1,\quad A_{23}=(-1)\begin{vmatrix}2 & 3 \\ 1 & 2\end{vmatrix}=-(4-3)=-1
  7. Row 3 elements.
    A_{31}=(+1)\begin{vmatrix}3 & -1 \\ -1 & 1\end{vmatrix}=3-1=2,\quad A_{32}=(-1)\begin{vmatrix}2 & -1 \\ 1 & 1\end{vmatrix}=-(2+1)=-3,\quad A_{33}=(+1)\begin{vmatrix}2 & 3 \\ 1 & -1\end{vmatrix}=-2-3=-5
  8. Write the adjoint (transpose of the cofactor matrix). Since |A|=1, A^{-1}=\operatorname{adj}(A).
    \operatorname{adj}(A)=\begin{bmatrix}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}-1 & 1 & 2 \\ 2 & -1 & -3 \\ 3 & -1 & -5\end{bmatrix},\quad A^{-1}=\frac11\begin{bmatrix}-1 & 1 & 2 \\ 2 & -1 & -3 \\ 3 & -1 & -5\end{bmatrix}
  9. Use X=A^{-1}B.
    \begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}-1 & 1 & 2 \\ 2 & -1 & -3 \\ 3 & -1 & -5\end{bmatrix}\begin{bmatrix}1 \\ 3 \\ 1\end{bmatrix}=\begin{bmatrix}-1+3+2 \\ 2-3-3 \\ 3-3-5\end{bmatrix}=\begin{bmatrix}4 \\ -4 \\ -5\end{bmatrix}
  10. Check in the first equation: 2(4)+3(-4)-(-5)=8-12+5=1. It holds.
x=4,\quad y=-4,\quad z=-5

2.6.8 (Gauss elimination, Gauss-Jordan)Gauss elimination and Gauss-Jordan methodsOften examined

Gauss elimination method (echelon form)

In this method we reduce the augmented matrix A_b of the system to its echelon form by row operations. Then we solve from the last row upwards (back substitution): the last row gives one unknown, the row above gives the next, and so on.

Example 33: Gauss elimination
Solve the system of equations by using the Gauss elimination method.
2x_1-3x_2+4x_3=1;\qquad x_1+2x_2-x_3=2;\qquad 3x_1+5x_2-3x_3=5
  1. Write the augmented matrix A_b.
    A_b=\left[\begin{array}{ccc|c}2 & -3 & 4 & 1 \\ 1 & 2 & -1 & 2 \\ 3 & 5 & -3 & 5\end{array}\right]
  2. Reduce to echelon form. Interchange R_1 and R_2 to get a leading 1.
    \sim\left[\begin{array}{ccc|c}1 & 2 & -1 & 2 \\ 2 & -3 & 4 & 1 \\ 3 & 5 & -3 & 5\end{array}\right]\quad(R_{12})
  3. Clear below the leading 1: R_2-2R_1 and R_3-3R_1.
    \sim\left[\begin{array}{ccc|c}1 & 2 & -1 & 2 \\ 0 & -7 & 6 & -3 \\ 0 & -1 & 0 & -1\end{array}\right]\quad(R_2-2R_1,\;R_3-3R_1)
  4. Interchange R_2 and R_3 so that the second row has the smaller entries.
    \sim\left[\begin{array}{ccc|c}1 & 2 & -1 & 2 \\ 0 & -1 & 0 & -1 \\ 0 & -7 & 6 & -3\end{array}\right]\quad(R_{23})
  5. Multiply R_2 by -1 to make its leading entry 1.
    \sim\left[\begin{array}{ccc|c}1 & 2 & -1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & -7 & 6 & -3\end{array}\right]\quad(-R_2)
  6. Clear below: R_3+7R_2.
    \sim\left[\begin{array}{ccc|c}1 & 2 & -1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 6 & 4\end{array}\right]\quad(R_3+7R_2)
  7. Multiply R_3 by \tfrac16.
    \sim\left[\begin{array}{ccc|c}1 & 2 & -1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & \tfrac23\end{array}\right]\quad(\tfrac16R_3)
  8. This is the echelon form of A_b. Here \operatorname{Rank}A=\operatorname{Rank}A_b=3, equal to the number of unknowns, so the solution is unique. Now read the rows from the bottom.
  9. Last row: 0x_1+0x_2+x_3=\tfrac23.
    x_3=\tfrac23
  10. Second row: 0x_1+x_2+0x_3=1.
    x_2=1
  11. First row: x_1+2x_2-x_3=2. Substitute x_2=1 and x_3=\tfrac23.
    x_1+2(1)-\tfrac23=2\;\Rightarrow\;x_1+\tfrac43=2\;\Rightarrow\;x_1=\tfrac23
  12. Check in the first original equation: 2\left(\tfrac23\right)-3(1)+4\left(\tfrac23\right)=\tfrac43-3+\tfrac83=1. It holds.
x_1=\tfrac23,\quad x_2=1,\quad x_3=\tfrac23
Worked example: Gauss elimination with a check
Solve x+2y+z=8,\; 2x+y-z=1,\; x-y+2z=5 by Gauss elimination.
  1. Write the augmented matrix.
    \left[\begin{array}{ccc|c}1&2&1&8\\2&1&-1&1\\1&-1&2&5\end{array}\right]
  2. Eliminate x from rows 2 and 3: R_2-2R_1 and R_3-R_1.
    \sim\left[\begin{array}{ccc|c}1&2&1&8\\0&-3&-3&-15\\0&-3&1&-3\end{array}\right]
  3. Eliminate y from row 3: R_3-R_2, since 1-(-3)=4 and -3-(-15)=12.
    \sim\left[\begin{array}{ccc|c}1&2&1&8\\0&-3&-3&-15\\0&0&4&12\end{array}\right]
  4. The matrix is in echelon form. Read the last row first (back substitution).
    4z=12\;\Rightarrow\;z=3
  5. Row 2.
    -3y-3(3)=-15\;\Rightarrow\;-3y=-6\;\Rightarrow\;y=2
  6. Row 1.
    x+2(2)+3=8\;\Rightarrow\;x=1
  7. Check in the original equations: 2(1)+2-3=1 and 1-2+2(3)=5.
x=1,\;y=2,\;z=3

Gauss-Jordan method (reduced echelon form)

In this method we reduce the augmented matrix to its reduced echelon form. The reduced form gives the values of the unknowns directly (or the relations between them), without back substitution.

Example 34: Gauss-Jordan method with infinitely many solutions
Solve the system of non-homogeneous linear equations 2x-3y+5z=2;\;x+4y-2z=1;\;4x+5y+z=4.
  1. Write the augmented matrix.
    A_b=\left[\begin{array}{ccc|c}2 & -3 & 5 & 2 \\ 1 & 4 & -2 & 1 \\ 4 & 5 & 1 & 4\end{array}\right]
  2. Interchange R_1 and R_2 to get a leading 1.
    \sim\left[\begin{array}{ccc|c}1 & 4 & -2 & 1 \\ 2 & -3 & 5 & 2 \\ 4 & 5 & 1 & 4\end{array}\right]\quad(R_{12})
  3. Clear below: R_2-2R_1 and R_3-4R_1. The last two rows become identical.
    \sim\left[\begin{array}{ccc|c}1 & 4 & -2 & 1 \\ 0 & -11 & 9 & 0 \\ 0 & -11 & 9 & 0\end{array}\right]\quad(R_2-2R_1,\;R_3-4R_1)
  4. Make the second leading entry 1: multiply R_2 by -\tfrac1{11}.
    \sim\left[\begin{array}{ccc|c}1 & 4 & -2 & 1 \\ 0 & 1 & -\tfrac9{11} & 0 \\ 0 & -11 & 9 & 0\end{array}\right]\quad(-\tfrac1{11}R_2)
  5. Clear above and below the second leading 1: R_1-4R_2 and R_3+11R_2.
    \sim\left[\begin{array}{ccc|c}1 & 0 & \tfrac{14}{11} & 1 \\ 0 & 1 & -\tfrac9{11} & 0 \\ 0 & 0 & 0 & 0\end{array}\right]\quad(R_1-4R_2,\;R_3+11R_2)
  6. This is the reduced echelon form. \operatorname{Rank}A=\operatorname{Rank}A_b=2, which is less than the number of unknowns (3), so the system has infinitely many solutions.
  7. The last row reads 0x+0y+0z=0. It is true for all values, so let z=t.
    z=t
  8. Second row: y-\tfrac9{11}z=0.
    y-\tfrac9{11}t=0\;\Rightarrow\;y=\tfrac9{11}t
  9. First row: x+\tfrac{14}{11}z=1.
    x+\tfrac{14}{11}t=1\;\Rightarrow\;x=1-\tfrac{14}{11}t
  10. Different values of the parameter t give infinitely many solutions.
x=1-\tfrac{14}{11}t,\quad y=\tfrac9{11}t,\quad z=t

2.6.8 (Cramer's rule)Cramer's ruleOften examined

Cramer's rule

Consider the non-homogeneous system a_1x+b_1y+c_1z=k_1,\;a_2x+b_2y+c_2z=k_2,\;a_3x+b_3y+c_3z=k_3, written AX=B. If A is non-singular, |A|\neq0 and A^{-1} exists.

Derivation Derivation of Cramer's rule

Aim. Express x, y and z as ratios of determinants.

  1. From the matrix inversion method, X=A^{-1}B. Use A^{-1}=\dfrac1{|A|}\operatorname{adj}(A).
    X=\frac1{|A|}\operatorname{adj}(A)\,B
  2. Write out \operatorname{adj}(A) and multiply it by B (row by column).
    X=\frac1{|A|}\begin{bmatrix}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{bmatrix}\begin{bmatrix}k_1 \\ k_2 \\ k_3\end{bmatrix}=\frac1{|A|}\begin{bmatrix}k_1A_{11}+k_2A_{21}+k_3A_{31} \\ k_1A_{12}+k_2A_{22}+k_3A_{32} \\ k_1A_{13}+k_2A_{23}+k_3A_{33}\end{bmatrix}
  3. Compare the entries of X=\begin{bmatrix}x&y&z\end{bmatrix}^{t} on both sides.
    \begin{aligned}x&=\frac{k_1A_{11}+k_2A_{21}+k_3A_{31}}{|A|}\\ y&=\frac{k_1A_{12}+k_2A_{22}+k_3A_{32}}{|A|}\\ z&=\frac{k_1A_{13}+k_2A_{23}+k_3A_{33}}{|A|}\end{aligned}
  4. The numerator of x is the expansion, along its first column, of the determinant obtained from |A| by replacing the first column with k_1,k_2,k_3. The cofactors A_{11},A_{21},A_{31} do not involve the first column, so they are unchanged by the replacement.
    k_1A_{11}+k_2A_{21}+k_3A_{31}=\begin{vmatrix}k_1 & b_1 & c_1 \\ k_2 & b_2 & c_2 \\ k_3 & b_3 & c_3\end{vmatrix}
  5. In the same way the numerator of y comes from replacing the second column, and that of z from replacing the third column.
    k_1A_{12}+k_2A_{22}+k_3A_{32}=\begin{vmatrix}a_1 & k_1 & c_1 \\ a_2 & k_2 & c_2 \\ a_3 & k_3 & c_3\end{vmatrix},\quad k_1A_{13}+k_2A_{23}+k_3A_{33}=\begin{vmatrix}a_1 & b_1 & k_1 \\ a_2 & b_2 & k_2 \\ a_3 & b_3 & k_3\end{vmatrix}
Resultx=\dfrac{\begin{vmatrix}k_1 & b_1 & c_1 \\ k_2 & b_2 & c_2 \\ k_3 & b_3 & c_3\end{vmatrix}}{|A|},\quad y=\dfrac{\begin{vmatrix}a_1 & k_1 & c_1 \\ a_2 & k_2 & c_2 \\ a_3 & k_3 & c_3\end{vmatrix}}{|A|},\quad z=\dfrac{\begin{vmatrix}a_1 & b_1 & k_1 \\ a_2 & b_2 & k_2 \\ a_3 & b_3 & k_3\end{vmatrix}}{|A|}

To find an unknown, replace its column of coefficients by the column of constants, take the determinant, and divide by |A|.

Example 35: Cramer's rule
Solve the given system of non-homogeneous linear equations by Cramer's rule.
2x-3y+5z=1;\qquad x+y+2z=3;\qquad 3x-2y-4z=0
  1. Write the system as AX=B.
    A=\begin{bmatrix}2 & -3 & 5 \\ 1 & 1 & 2 \\ 3 & -2 & -4\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}1 \\ 3 \\ 0\end{bmatrix}
  2. Find |A| by expanding along the first row.
    |A|=\begin{vmatrix}2 & -3 & 5 \\ 1 & 1 & 2 \\ 3 & -2 & -4\end{vmatrix}=2\begin{vmatrix}1 & 2 \\ -2 & -4\end{vmatrix}+3\begin{vmatrix}1 & 2 \\ 3 & -4\end{vmatrix}+5\begin{vmatrix}1 & 1 \\ 3 & -2\end{vmatrix}
  3. Evaluate. (The sign of the middle term is -(-3)=+3.)
    |A|=2(-4+4)+3(-4-6)+5(-2-3)=0-30-25=-55\neq0
  4. So A is non-singular and Cramer's rule applies.
  5. For x, replace column 1 by B=(1,3,0) and expand along the first row.
    x=\frac{\begin{vmatrix}1 & -3 & 5 \\ 3 & 1 & 2 \\ 0 & -2 & -4\end{vmatrix}}{|A|}=\frac{1(-4+4)+3(-12-0)+5(-6-0)}{-55}=\frac{-66}{-55}=\frac65
  6. For y, replace column 2 by B.
    y=\frac{\begin{vmatrix}2 & 1 & 5 \\ 1 & 3 & 2 \\ 3 & 0 & -4\end{vmatrix}}{|A|}=\frac{2(-12-0)-1(-4-6)+5(0-9)}{-55}=\frac{-24+10-45}{-55}=\frac{-59}{-55}=\frac{59}{55}
  7. For z, replace column 3 by B.
    z=\frac{\begin{vmatrix}2 & -3 & 1 \\ 1 & 1 & 3 \\ 3 & -2 & 0\end{vmatrix}}{|A|}=\frac{2(0+6)+3(0-9)+1(-2-3)}{-55}=\frac{12-27-5}{-55}=\frac{-20}{-55}=\frac4{11}
  8. Check in the first equation: 2\left(\tfrac65\right)-3\left(\tfrac{59}{55}\right)+5\left(\tfrac4{11}\right)=\tfrac{132-177+100}{55}=1. It holds.
x=\tfrac65,\quad y=\tfrac{59}{55},\quad z=\tfrac4{11}

2.7Application of matrices

Cryptography: encoding and decoding a message

Matrices are used in many disciplines. In cryptography a message is scrambled (encoded) so that only the person with the key can read it. The book explains encryption and decryption by an example.

Suppose the sender and the receiver consider messages in the letters A to Z only. Both assign the numbers 1 to 26 to the letters A to Z respectively, and the number 0 to a blank space. The sender uses as the key post-multiplication by a non-singular matrix of order 3 of his own choice. The receiver decodes by post-multiplication by the inverse of the matrix chosen by the sender.

Let the encoding matrix be

A=\begin{bmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix}

Let the message be "WELCOME". Since the key is a matrix of order 3, the message is cut into pieces of length 3: (WEL), (COM), (E), and converted into row matrices of numbers using W=23, E=5, L=12, C=3, O=15, M=13:

[23\;\;5\;\;12],\qquad[3\;\;15\;\;13],\qquad[5\;\;0\;\;0]

Two zeros (blanks) are added to the last row matrix so that every piece has length 3 and its first entry is 5 (the letter E).

Encoding

Post-multiply each row matrix by A (row times matrix gives a row).

  1. First piece. Multiply the row [23\;5\;12] by each column of A.
    [23\;\;5\;\;12]\begin{bmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix}=[23+10+12\;\;\;-23-5+0\;\;\;23+0+0]=[45\;\;-28\;\;23]
  2. Second piece.
    [3\;\;15\;\;13]\begin{bmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix}=[3+30+13\;\;\;-3-15+0\;\;\;3+0+0]=[46\;\;-18\;\;3]
  3. Third piece.
    [5\;\;0\;\;0]\begin{bmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix}=[5+0+0\;\;\;-5-0+0\;\;\;5+0+0]=[5\;\;-5\;\;5]
  4. The encoded message that is sent is [45\;\;-28\;\;23]\;[46\;\;-18\;\;3]\;[5\;\;-5\;\;5].
Encoding the message
Uncoded row matrixEncoding matrixCoded row matrix
[23\;\;5\;\;12]A[45\;\;-28\;\;23]
[3\;\;15\;\;13]A[46\;\;-18\;\;3]
[5\;\;0\;\;0]A[5\;\;-5\;\;5]
Decoding

The receiver post-multiplies by the inverse of A. First find A^{-1} by the adjoint method. Here |A|=1 and the cofactors give

A^{-1}=\frac1{|A|}\operatorname{adj}(A)=\begin{bmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{bmatrix}
  1. First coded piece.
    [45\;\;-28\;\;23]\begin{bmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{bmatrix}=[0+0+23\;\;\;28-23\;\;\;45-56+23]=[23\;\;5\;\;12]
  2. Second coded piece.
    [46\;\;-18\;\;3]\begin{bmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{bmatrix}=[0+0+3\;\;\;18-3\;\;\;46-36+3]=[3\;\;15\;\;13]
  3. Third coded piece.
    [5\;\;-5\;\;5]\begin{bmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{bmatrix}=[0+0+5\;\;\;5-5\;\;\;5-10+5]=[5\;\;0\;\;0]
  4. The decoded row matrices are [23\;5\;12], [3\;15\;13], [5\;0\;0]. Reading the numbers back as letters, the receiver gets the message "WELCOME".
Decoding the message
Coded row matrixDecoding matrixDecoded row matrix
[45\;\;-28\;\;23]A^{-1}[23\;\;5\;\;12]
[46\;\;-18\;\;3]A^{-1}[3\;\;15\;\;13]
[5\;\;-5\;\;5]A^{-1}[5\;\;0\;\;0]

Transformation of a graph by a matrix

Matrices can also describe a shift of a figure in the plane. The following example, from the book, moves a line one unit up.

Example 36: transformation matrix and transformed equation
Write the transformation matrix and the transformed equation of the line 2x+y-3=0 which is 1 unit above the given line.
  1. Let the transformed variables be x' and y'. The line is shifted 1 unit up, so every point (x,y) goes to (x',y') with the same x and a y-value one more.
    x'=x,\qquad y'=y+1
  2. Write this in matrix form: the new point is the old point plus the shift column (0,1).
    \begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}x\\y+1\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}0\\1\end{bmatrix}
  3. To get the new equation, write the old variables in terms of the new ones.
    x=x',\qquad y=y'-1
  4. Substitute these in 2x+y-3=0.
    2x'+(y'-1)-3=0
  5. Simplify.
    2x'+y'-1-3=0\;\Rightarrow\;2x'+y'-4=0
  6. Check with a point: (0,3) is on the given line, since 0+3-3=0. It moves to (0,4), and 2(0)+4-4=0, so the new line passes through it.
The line 2x+y-3=0 moved 1 unit up is 2x+y-4=0.
\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}0\\1\end{bmatrix},\qquad 2x'+y'-4=0

Formula sheet

Order of a matrix
A_{m\times n}=[a_{ij}]
m: number of rows, n: number of columns, a_{ij}: element in row i, column j
Sum and difference of matrices
c_{ij}=a_{ij}\pm b_{ij}
A and B must have the same order m\times n
Product of two matrices
c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots+a_{in}b_{nj}
A is m\times n, B is n\times p, so AB is m\times p (columns of A = rows of B)
Transpose of a product
(AB)^{t}=B^{t}A^{t}
Reversal law for the transpose
Transpose of a sum
(A+B)^{t}=A^{t}+B^{t}
A and B of the same order
Symmetric and skew-symmetric matrices
A=A^{t}\;(\text{symmetric}),\qquad A=-A^{t}\;(\text{skew-symmetric})
A is a square matrix
Symmetric and skew-symmetric parts
A=\tfrac12(A+A^{t})+\tfrac12(A-A^{t})
The first part is symmetric, the second is skew-symmetric
Determinant of order 2
\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}=a_{11}a_{22}-a_{21}a_{12}
a_{ij}: elements of the matrix
Cofactor of an element
A_{ij}=(-1)^{i+j}M_{ij}
M_{ij}: minor of a_{ij} (determinant left after deleting row i and column j)
Determinant by cofactors
|A|=a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3}
Expansion along any row i (or any column) of a 3\times3 matrix
Singular and non-singular matrix
|A|=0\;(\text{singular}),\qquad |A|\neq0\;(\text{non-singular})
Adjoint of a matrix
\operatorname{adj}(A)=[A_{ji}]
Transpose of the matrix of cofactors A_{ij}
Inverse by the adjoint method
A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A),\qquad AA^{-1}=A^{-1}A=I
Exists only if |A|\neq0
Reversal law of the inverse
(AB)^{-1}=B^{-1}A^{-1}
A and B non-singular of the same order
Determinant of the transpose
|A^{t}|=|A|
Property 1
Interchange of rows or columns
|B|=-|A|
B is A with two rows (or columns) interchanged
Scalar multiple of one row or column
|B|=k|A|
B is A with one row (or column) multiplied by k
Row operation that keeps the determinant
R_i+kR_j:\;|B|=|A|
Adding k times another row (column) to a row (column)
Triangular and diagonal matrices
|A|=a_{11}a_{22}a_{33}
Product of the diagonal elements
Inverse by row operations
[A:I]\sim[I:A^{-1}]
Same row operations on A and I
Homogeneous system
AX=O:\;\text{non-trivial solution}\iff|A|=0
|A|\neq0 gives only the trivial solution X=O
Matrix form of a system
AX=B,\qquad A_b=[A:B]
A: coefficient matrix, X: unknowns, B: constants, A_b: augmented matrix
Consistency criteria
\operatorname{Rank}A=\operatorname{Rank}A_b\;(\text{consistent})
If this rank equals the number of unknowns the solution is unique; if it is less there are infinitely many solutions; if \operatorname{Rank}A\neq\operatorname{Rank}A_b there is no solution
Matrix inversion method
X=A^{-1}B
Needs |A|\neq0
Cramer's rule
x=\dfrac{D_x}{|A|},\quad y=\dfrac{D_y}{|A|},\quad z=\dfrac{D_z}{|A|}
D_x,D_y,D_z: |A| with the column of x, y, z replaced by the constants k_1,k_2,k_3; needs |A|\neq0

Key terms

Matrix
An array of numbers arranged in rows and columns and enclosed in square brackets.
Order of a matrix
The number of rows by the number of columns, m\times n, rows first.
Element (entry)
Each number in a matrix. The element in row i and column j is a_{ij}.
Equal matrices
Matrices of the same order with the same corresponding elements.
Row matrix and column matrix
A matrix with only one row, or only one column.
Square matrix
A matrix with equal numbers of rows and columns.
Zero (null) matrix
A matrix all of whose elements are zero.
Diagonal matrix
A square matrix in which every element off the main diagonal is zero and at least one diagonal element is non-zero.
Scalar matrix
A diagonal matrix whose diagonal elements are all the same non-zero number k.
Identity (unit) matrix
A scalar matrix whose diagonal elements are all 1, written I.
Triangular matrix
A square matrix that is upper triangular (zeros below the diagonal) or lower triangular (zeros above the diagonal).
Transpose
The matrix A^{t} obtained by interchanging the rows and columns of A.
Symmetric matrix
A square matrix with A=A^{t}.
Skew-symmetric matrix
A square matrix with A=-A^{t}.
Conformable matrices
Matrices that can be added (same order) or multiplied (columns of the first equal rows of the second).
Determinant
The number |A| or \det(A) associated with a square matrix A.
Minor
The determinant M_{ij} left after deleting the ith row and jth column.
Cofactor
A_{ij}=(-1)^{i+j}M_{ij}, the signed minor of a_{ij}.
Singular and non-singular matrix
A square matrix is singular if |A|=0 and non-singular if |A|\neq0.
Adjoint
\operatorname{adj}(A), the transpose of the matrix of cofactors of A.
Multiplicative inverse
The matrix A^{-1} with AA^{-1}=A^{-1}A=I; it exists only for a non-singular matrix.
Elementary row operations
Interchanging two rows (R_{ij}), multiplying a row by a non-zero scalar (kR_i), adding a multiple of a row to another row (R_i+kR_j).
Echelon form
A matrix whose leading entries are 1, move to the right row by row, and whose zero rows are at the bottom.
Reduced echelon form
An echelon form in which each leading entry 1 is the only non-zero entry in its column.
Rank
The number of non-zero rows in the echelon form of a matrix.
Homogeneous system
A system of linear equations in which all the constants on the right side are zero.
Non-homogeneous system
A system of linear equations in which at least one constant on the right side is non-zero.
Trivial and non-trivial solution
The solution x=y=z=0 of a homogeneous system is trivial; any other solution is non-trivial.
Consistent and inconsistent system
A system with at least one solution is consistent; a system with no solution is inconsistent.
Augmented matrix
The matrix A_b made of the coefficients of the unknowns together with the constants on the right side.
Cramer's rule
The solution x=D_x/|A|, y=D_y/|A|, z=D_z/|A| of AX=B with |A|\neq0.

Exercise 2.1p. 34–35

Order, types and transpose of a matrix. Read the order as rows by columns, test the entries above and below the main diagonal to name the type, and interchange rows and columns to form the transpose.

1
Find the order of the following matrices.
Solution

Method. The order of a matrix is (number of rows) \times (number of columns). Always write the rows first (section 2.1.1).

(i) A=\begin{bmatrix}1 & 3 & 0 \\ 2 & 0 & 1\end{bmatrix}
  1. Count the horizontal lines (rows) of A: there are 2.
    m=2
  2. Count the vertical lines (columns): there are 3.
    n=3
  3. The order is written rows first, then columns, so it is m\times n. Number of elements =2\times3=6, which matches the count of entries.
    \text{order of }A=2\times 3
2\times 3
(ii) B=\begin{bmatrix}1 & 2 \\ 2 & 3 \\ 3 & 4\end{bmatrix}
  1. Count the horizontal lines (rows) of B: there are 3.
    m=3
  2. Count the vertical lines (columns): there are 2.
    n=2
  3. The order is written rows first, then columns, so it is m\times n. Number of elements =3\times2=6, which matches the count of entries.
    \text{order of }B=3\times 2
3\times 2
(iii) C=\begin{bmatrix}1 \\ 6 \\ 9\end{bmatrix}
  1. Count the horizontal lines (rows) of C: there are 3.
    m=3
  2. Count the vertical lines (columns): there are 1.
    n=1
  3. The order is written rows first, then columns, so it is m\times n. Number of elements =3\times1=3, which matches the count of entries.
    \text{order of }C=3\times 1
3\times 1
(iv) D=\begin{bmatrix}2 & 1 & 6 & 8\end{bmatrix}
  1. Count the horizontal lines (rows) of D: there are 1.
    m=1
  2. Count the vertical lines (columns): there are 4.
    n=4
  3. The order is written rows first, then columns, so it is m\times n. Number of elements =1\times4=4, which matches the count of entries.
    \text{order of }D=1\times 4
1\times 4
(v) E=\begin{bmatrix}3\end{bmatrix}
  1. Count the horizontal lines (rows) of E: there are 1.
    m=1
  2. Count the vertical lines (columns): there are 1.
    n=1
  3. The order is written rows first, then columns, so it is m\times n. Number of elements =1\times1=1, which matches the count of entries.
    \text{order of }E=1\times 1
1\times 1
(vi) F=\begin{bmatrix}3 & 6 \\ 9 & 2\end{bmatrix}
  1. Count the horizontal lines (rows) of F: there are 2.
    m=2
  2. Count the vertical lines (columns): there are 2.
    n=2
  3. The order is written rows first, then columns, so it is m\times n. Number of elements =2\times2=4, which matches the count of entries.
    \text{order of }F=2\times 2
2\times 2

Book answer. (i) 2 x 3, (ii) 3 x 2, (iii) 3 x 1, (iv) 1 x 4, (v) 1 x 1, (vi) 2 x 2. All agree.

2
Identify the following matrices as square matrix, rectangular matrix, row matrix or column matrix.
Solution

Method. Compare the number of rows m and columns n (sections 2.1.3 to 2.1.6): one row gives a row matrix, one column gives a column matrix, m=n gives a square matrix and m\neq n (with more than one row and column) gives a rectangular matrix.

(i) A=\begin{bmatrix}3 & 6 & 2 \\ 2 & 1 & 9\end{bmatrix}
  1. Find the order: A has 2 rows and 3 columns.
    2\times 3
  2. Rows and columns differ in number (m\neq n) and there are several of each, so it is a rectangular matrix.
\text{rectangular matrix}
(ii) B=\begin{bmatrix}\frac13 & 1 \\ 2 & 6\end{bmatrix}
  1. Find the order: B has 2 rows and 2 columns.
    2\times 2
  2. Rows and columns are equal in number, so it is a square matrix.
\text{square matrix}
(iii) C=\begin{bmatrix}3 \\ 2 \\ 8\end{bmatrix}
  1. Find the order: C has 3 rows and 1 column.
    3\times 1
  2. There is only one column, so it is a column matrix (column vector).
\text{column matrix}
(iv) D=\begin{bmatrix}1 & 6 & 9 \\ 2 & 0 & 1 \\ 3 & -1 & 2\end{bmatrix}
  1. Find the order: D has 3 rows and 3 columns.
    3\times 3
  2. Rows and columns are equal in number, so it is a square matrix.
\text{square matrix}
(v) E=\begin{bmatrix}2 & 0 & 1\end{bmatrix}
  1. Find the order: E has 1 row and 3 columns.
    1\times 3
  2. There is only one row, so it is a row matrix (row vector).
\text{row matrix}
(vi) F=\begin{bmatrix}16\end{bmatrix}
  1. Find the order: F has 1 row and 1 column.
    1\times 1
  2. Rows and columns are equal in number, so it is a square matrix (a 1\times1 matrix is both a square matrix and a row/column matrix, but the book lists it as square).
\text{square matrix}

Book answer. (i) rectangular, (ii) square, (iii) column, (iv) square, (v) row, (vi) square. All agree.

Note. The book's matrix D shows a symbol that looks like 0 and a dot before the 1 in the last row; it is read as \begin{bmatrix}1&6&9\\2&0&1\\3&-1&2\end{bmatrix}. The order 3\times3 does not depend on these entries.

3
Identify the diagonal matrix, scalar matrix, identity matrix, lower triangular matrix, upper triangular matrix.
Solution

Method (sections 2.1.8 to 2.1.14). Diagonal: square, all off-diagonal elements 0, at least one non-zero diagonal element. Scalar: diagonal with all diagonal elements equal and non-zero. Identity: scalar with diagonal elements 1. Upper (lower) triangular: all elements below (above) the main diagonal are 0.

MatrixBest nameAlso true
Alower triangularnone
Bscalardiagonal, triangular
Cdiagonalupper and lower triangular
Didentityscalar, diagonal, triangular
Ediagonalupper and lower triangular
Fupper triangularnone
Gdiagonalupper and lower triangular
Hscalardiagonal, triangular
(i) A=\begin{bmatrix}3 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 6 & 0\end{bmatrix}
  1. Check the square matrix against the definitions: look at the elements above, below and on the main diagonal.
  2. The elements above the main diagonal (a_{12},a_{13},a_{23}) are all 0, but a_{31}=2 and a_{32}=6 below it are not.
  3. So it is a lower triangular matrix. It is not diagonal because of the non-zero elements 2 and 6 below the diagonal.
\text{lower triangular matrix}
(ii) B=\begin{bmatrix}-6 & 0 & 0 \\ 0 & -6 & 0 \\ 0 & 0 & -6\end{bmatrix}
  1. Check the square matrix against the definitions: look at the elements above, below and on the main diagonal.
  2. All elements off the main diagonal are 0 and all diagonal elements are equal to the same non-zero number -6.
  3. So it is a scalar matrix. It is -6I_3. It is not an identity matrix because the diagonal elements are -6, not 1. Every scalar matrix is also diagonal and triangular.
\text{scalar matrix}
(iii) C=\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}
  1. Check the square matrix against the definitions: look at the elements above, below and on the main diagonal.
  2. All elements off the main diagonal are 0 and the diagonal elements 1,2 are not all equal.
  3. So it is a diagonal matrix. It is not scalar because 1\neq2.
\text{diagonal matrix}
(iv) D=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  1. Check the square matrix against the definitions: look at the elements above, below and on the main diagonal.
  2. It is square, every off-diagonal element is 0 and every diagonal element is 1.
  3. So it is a identity matrix. It is I_3. Being scalar and diagonal as well, it is also upper and lower triangular.
\text{identity matrix}
(v) E=\begin{bmatrix}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0\end{bmatrix}
  1. Check the square matrix against the definitions: look at the elements above, below and on the main diagonal.
  2. All elements off the main diagonal are 0 and the diagonal elements 2,3,0 are not all equal, with at least one non-zero.
  3. So it is a diagonal matrix. It is not scalar because the diagonal entries differ (a diagonal matrix may have a zero on its diagonal).
\text{diagonal matrix}
(vi) F=\begin{bmatrix}\sqrt3 & 1 & 2 \\ 0 & 0 & 6 \\ 0 & 0 & 1\end{bmatrix}
  1. Check the square matrix against the definitions: look at the elements above, below and on the main diagonal.
  2. All elements below the main diagonal (a_{21},a_{31},a_{32}) are 0, while a_{12}=1, a_{13}=2, a_{23}=6 above it are not.
  3. So it is a upper triangular matrix. It is not diagonal because of the non-zero elements above the diagonal.
\text{upper triangular matrix}
(vii) G=\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
  1. Check the square matrix against the definitions: look at the elements above, below and on the main diagonal.
  2. All off-diagonal elements are 0 and the diagonal elements 1,0,0 include a non-zero element.
  3. So it is a diagonal matrix. It is not scalar because the diagonal elements are not all equal.
\text{diagonal matrix}
(viii) H=\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}
  1. Check the square matrix against the definitions: look at the elements above, below and on the main diagonal.
  2. All off-diagonal elements are 0 and both diagonal elements equal 2.
  3. So it is a scalar matrix. It is 2I_2, so it is not an identity matrix.
\text{scalar matrix}

Book answer. (i) lower triangular, (ii) scalar, (iii) diagonal, (iv) identity, (v) diagonal, (vi) upper triangular, (vii) diagonal, (viii) scalar.

Note. The printed answer key skips the label (vi) and numbers the last three answers as vii, viii, ix. Read in order they agree with the solutions above: upper triangular, diagonal, scalar for F, G, H.

4
Find the transpose of the following matrices and identify which one of them are symmetric and which are skew-symmetric.
Solution

Method (sections 2.1.15 to 2.1.17). The transpose A^t is obtained by interchanging rows and columns. A square matrix with A^t=A is symmetric (a_{ij}=a_{ji}). A square matrix with A^t=-A is skew-symmetric (a_{ij}=-a_{ji}, so every diagonal element is 0).

(i) A=\begin{bmatrix}2 & 0 \\ \sqrt5 & 6 \\ 1 & 9\end{bmatrix}
  1. Interchange rows and columns: row j of A becomes column j of A^t.
    A^t=\begin{bmatrix}2 & \sqrt5 & 1 \\ 0 & 6 & 9\end{bmatrix}
  2. A is 3\times2, not a square matrix, so A^t (order 2\times3) cannot equal A or -A. Symmetric and skew-symmetric matrices must be square.
  3. Conclusion: neither symmetric nor skew-symmetric.
\text{neither symmetric nor skew-symmetric}
(ii) B=\begin{bmatrix}1&6&2&0\end{bmatrix}
  1. Interchange rows and columns: row j of B becomes column j of B^t.
    B^t=\begin{bmatrix}1 \\ 6 \\ 2 \\ 0\end{bmatrix}
  2. B is 1\times4 and B^t is 4\times1. The orders differ, so B^t\neq B and B^t\neq -B.
  3. Conclusion: neither symmetric nor skew-symmetric.
\text{neither symmetric nor skew-symmetric}
(iii) C=\begin{bmatrix}2 & 6 \\ 6 & 2\end{bmatrix}
  1. Interchange rows and columns: row j of C becomes column j of C^t.
    C^t=\begin{bmatrix}2 & 6 \\ 6 & 2\end{bmatrix}
  2. C^t=C (the entries a_{12}=a_{21}=6 and the diagonal is unchanged), so C is symmetric.
  3. Conclusion: symmetric.
\text{symmetric}
(iv) D=\begin{bmatrix}0 & 1 & 9 \\ -1 & 0 & 5 \\ -9 & -5 & 0\end{bmatrix}
  1. Interchange rows and columns: row j of D becomes column j of D^t.
    D^t=\begin{bmatrix}0 & -1 & -9 \\ 1 & 0 & -5 \\ 9 & 5 & 0\end{bmatrix}
  2. The diagonal is 0 and a_{ij}=-a_{ji}: a_{12}=1=-a_{21}, a_{13}=9=-a_{31}, a_{23}=5=-a_{32}. So D^t=-D and D is skew-symmetric.
  3. Conclusion: skew-symmetric.
\text{skew-symmetric}
(v) E=\begin{bmatrix}3 & -6 & 9 \\ -6 & 2 & 0 \\ 9 & 0 & 0\end{bmatrix}
  1. Interchange rows and columns: row j of E becomes column j of E^t.
    E^t=\begin{bmatrix}3 & -6 & 9 \\ -6 & 2 & 0 \\ 9 & 0 & 0\end{bmatrix}
  2. a_{12}=-6=a_{21}, a_{13}=9=a_{31}, a_{23}=0=a_{32}. So E^t=E and E is symmetric.
  3. Conclusion: symmetric.
\text{symmetric}
(vi) F=\begin{bmatrix}9 & 0 & 1 \\ 0 & 6 & 3 \\ 0 & 0 & 1\end{bmatrix}
  1. Interchange rows and columns: row j of F becomes column j of F^t.
    F^t=\begin{bmatrix}9 & 0 & 0 \\ 0 & 6 & 0 \\ 1 & 3 & 1\end{bmatrix}
  2. F^t\neq F because a_{13}=1 but a_{31}=0. F^t\neq-F because the diagonal elements are not zero. So it is neither.
  3. Conclusion: neither symmetric nor skew-symmetric.
\text{neither symmetric nor skew-symmetric}

Book answer. Transposes as printed in the key; (iii) and (v) symmetric, (iv) skew-symmetric, (i), (ii), (vi) neither.

Note. In the answer key the transpose of D is printed with 5 in the second row, third column. The correct entry is -5 (the third column of D is (9,5,0) but the second column is (1,0,-5)), so D^t=-D as shown above; this is a misprint in the key.

Exercise 2.2p. 40–41

Algebra of matrices: building a matrix from a rule, addition, scalar multiple, product, transpose rules. Match corresponding elements, multiply row by column, and remember that AB\neq BA in general.

1
Construct a matrix A=[a_{ij}] of order 2\times2 for which:
Solution

Method. In a_{ij} the first index i is the row and the second index j is the column. Substitute each pair (i,j)=(1,1),(1,2),(2,1),(2,2) into the given rule.

(i) a_{ij}=\dfrac{i+3j}{2}
  1. A 2\times2 matrix has the general form with i,j\in\{1,2\} (row i, column j).
    A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}
  2. Put each pair (i,j) in the rule.
    \begin{aligned}a_{11}=\frac{1+3(1)}{2}=2 \\ a_{12}=\frac{1+3(2)}{2}=\frac{7}{2} \\ a_{21}=\frac{2+3(1)}{2}=\frac{5}{2} \\ a_{22}=\frac{2+3(2)}{2}=4\end{aligned}
  3. Place the four values in their positions.
    A=\begin{bmatrix}2 & \frac{7}{2} \\ \frac{5}{2} & 4\end{bmatrix}
A=\begin{bmatrix}2 & \frac{7}{2} \\ \frac{5}{2} & 4\end{bmatrix}
(ii) a_{ij}=\dfrac{i\times j}{2}
  1. A 2\times2 matrix has the general form with i,j\in\{1,2\} (row i, column j).
    A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}
  2. Put each pair (i,j) in the rule.
    \begin{aligned}a_{11}=\frac{1\times 1}{2}=\frac{1}{2} \\ a_{12}=\frac{1\times 2}{2}=1 \\ a_{21}=\frac{2\times 1}{2}=1 \\ a_{22}=\frac{2\times 2}{2}=2\end{aligned}
  3. Place the four values in their positions.
    A=\begin{bmatrix}\frac{1}{2} & 1 \\ 1 & 2\end{bmatrix}
A=\begin{bmatrix}\frac{1}{2} & 1 \\ 1 & 2\end{bmatrix}
(iii) a_{ij}=\dfrac{i}{j}
  1. A 2\times2 matrix has the general form with i,j\in\{1,2\} (row i, column j).
    A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}
  2. Put each pair (i,j) in the rule.
    \begin{aligned}a_{11}=\frac{1}{1}=1 \\ a_{12}=\frac{1}{2}=\frac{1}{2} \\ a_{21}=\frac{2}{1}=2 \\ a_{22}=\frac{2}{2}=1\end{aligned}
  3. Place the four values in their positions.
    A=\begin{bmatrix}1 & \frac{1}{2} \\ 2 & 1\end{bmatrix}
A=\begin{bmatrix}1 & \frac{1}{2} \\ 2 & 1\end{bmatrix}
(iv) a_{ij}=\dfrac{2i-3j}{3}
  1. A 2\times2 matrix has the general form with i,j\in\{1,2\} (row i, column j).
    A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}
  2. Put each pair (i,j) in the rule.
    \begin{aligned}a_{11}=\frac{2(1)-3(1)}{3}=-\frac{1}{3} \\ a_{12}=\frac{2(1)-3(2)}{3}=-\frac{4}{3} \\ a_{21}=\frac{2(2)-3(1)}{3}=\frac{1}{3} \\ a_{22}=\frac{2(2)-3(2)}{3}=-\frac{2}{3}\end{aligned}
  3. Place the four values in their positions.
    A=\begin{bmatrix}-\frac{1}{3} & -\frac{4}{3} \\ \frac{1}{3} & -\frac{2}{3}\end{bmatrix}
A=\begin{bmatrix}-\frac{1}{3} & -\frac{4}{3} \\ \frac{1}{3} & -\frac{2}{3}\end{bmatrix}

Book answer. (i) [2, 7/2; 5/2, 4], (ii) [1/2, 1; 1, 2], (iii) [1, 1/2; 2, 1], (iv) [-1/3, -4/3; 1/3, -2/3]. All agree.

2
Construct a matrix B=[b_{ij}] of order 3\times3 for which:
Solution

Method. Same as question 1, but now i and j each run over 1,2,3, giving nine entries.

(i) b_{ij}=\dfrac{i^2-j}{3}
  1. A 3\times3 matrix has nine entries b_{ij} with i,j\in\{1,2,3\}.
    B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{bmatrix}
  2. Row 1: put i=1 and j=1,2,3.
    \begin{aligned}b_{11}=\frac{1^2-1}{3}=0 \\ b_{12}=\frac{1^2-2}{3}=-\frac{1}{3} \\ b_{13}=\frac{1^2-3}{3}=-\frac{2}{3}\end{aligned}
  3. Row 2: put i=2.
    \begin{aligned}b_{21}=\frac{2^2-1}{3}=1 \\ b_{22}=\frac{2^2-2}{3}=\frac{2}{3} \\ b_{23}=\frac{2^2-3}{3}=\frac{1}{3}\end{aligned}
  4. Row 3: put i=3.
    \begin{aligned}b_{31}=\frac{3^2-1}{3}=\frac{8}{3} \\ b_{32}=\frac{3^2-2}{3}=\frac{7}{3} \\ b_{33}=\frac{3^2-3}{3}=2\end{aligned}
  5. Write the values in their positions.
    B=\begin{bmatrix}0 & -\frac{1}{3} & -\frac{2}{3} \\ 1 & \frac{2}{3} & \frac{1}{3} \\ \frac{8}{3} & \frac{7}{3} & 2\end{bmatrix}
B=\begin{bmatrix}0 & -\frac{1}{3} & -\frac{2}{3} \\ 1 & \frac{2}{3} & \frac{1}{3} \\ \frac{8}{3} & \frac{7}{3} & 2\end{bmatrix}
(ii) b_{ij}=\dfrac{i^2-j^2}{2i}
  1. A 3\times3 matrix has nine entries b_{ij} with i,j\in\{1,2,3\}.
    B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{bmatrix}
  2. Row 1: put i=1 and j=1,2,3.
    \begin{aligned}b_{11}=\frac{1^2-1^2}{2(1)}=0 \\ b_{12}=\frac{1^2-2^2}{2(1)}=-\frac{3}{2} \\ b_{13}=\frac{1^2-3^2}{2(1)}=-4\end{aligned}
  3. Row 2: put i=2.
    \begin{aligned}b_{21}=\frac{2^2-1^2}{2(2)}=\frac{3}{4} \\ b_{22}=\frac{2^2-2^2}{2(2)}=0 \\ b_{23}=\frac{2^2-3^2}{2(2)}=-\frac{5}{4}\end{aligned}
  4. Row 3: put i=3.
    \begin{aligned}b_{31}=\frac{3^2-1^2}{2(3)}=\frac{4}{3} \\ b_{32}=\frac{3^2-2^2}{2(3)}=\frac{5}{6} \\ b_{33}=\frac{3^2-3^2}{2(3)}=0\end{aligned}
  5. Write the values in their positions.
    B=\begin{bmatrix}0 & -\frac{3}{2} & -4 \\ \frac{3}{4} & 0 & -\frac{5}{4} \\ \frac{4}{3} & \frac{5}{6} & 0\end{bmatrix}
B=\begin{bmatrix}0 & -\frac{3}{2} & -4 \\ \frac{3}{4} & 0 & -\frac{5}{4} \\ \frac{4}{3} & \frac{5}{6} & 0\end{bmatrix}
(iii) b_{ij}=\dfrac{2}{2i+j}
  1. A 3\times3 matrix has nine entries b_{ij} with i,j\in\{1,2,3\}.
    B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{bmatrix}
  2. Row 1: put i=1 and j=1,2,3.
    \begin{aligned}b_{11}=\frac{2}{2(1)+1}=\frac{2}{3} \\ b_{12}=\frac{2}{2(1)+2}=\frac{1}{2} \\ b_{13}=\frac{2}{2(1)+3}=\frac{2}{5}\end{aligned}
  3. Row 2: put i=2.
    \begin{aligned}b_{21}=\frac{2}{2(2)+1}=\frac{2}{5} \\ b_{22}=\frac{2}{2(2)+2}=\frac{1}{3} \\ b_{23}=\frac{2}{2(2)+3}=\frac{2}{7}\end{aligned}
  4. Row 3: put i=3.
    \begin{aligned}b_{31}=\frac{2}{2(3)+1}=\frac{2}{7} \\ b_{32}=\frac{2}{2(3)+2}=\frac{1}{4} \\ b_{33}=\frac{2}{2(3)+3}=\frac{2}{9}\end{aligned}
  5. Write the values in their positions.
    B=\begin{bmatrix}\frac{2}{3} & \frac{1}{2} & \frac{2}{5} \\ \frac{2}{5} & \frac{1}{3} & \frac{2}{7} \\ \frac{2}{7} & \frac{1}{4} & \frac{2}{9}\end{bmatrix}
B=\begin{bmatrix}\frac{2}{3} & \frac{1}{2} & \frac{2}{5} \\ \frac{2}{5} & \frac{1}{3} & \frac{2}{7} \\ \frac{2}{7} & \frac{1}{4} & \frac{2}{9}\end{bmatrix}
(iv) b_{ij}=\dfrac{i^2+j^2}{i+j}
  1. A 3\times3 matrix has nine entries b_{ij} with i,j\in\{1,2,3\}.
    B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{bmatrix}
  2. Row 1: put i=1 and j=1,2,3.
    \begin{aligned}b_{11}=\frac{1^2+1^2}{1+1}=1 \\ b_{12}=\frac{1^2+2^2}{1+2}=\frac{5}{3} \\ b_{13}=\frac{1^2+3^2}{1+3}=\frac{5}{2}\end{aligned}
  3. Row 2: put i=2.
    \begin{aligned}b_{21}=\frac{2^2+1^2}{2+1}=\frac{5}{3} \\ b_{22}=\frac{2^2+2^2}{2+2}=2 \\ b_{23}=\frac{2^2+3^2}{2+3}=\frac{13}{5}\end{aligned}
  4. Row 3: put i=3.
    \begin{aligned}b_{31}=\frac{3^2+1^2}{3+1}=\frac{5}{2} \\ b_{32}=\frac{3^2+2^2}{3+2}=\frac{13}{5} \\ b_{33}=\frac{3^2+3^2}{3+3}=3\end{aligned}
  5. Write the values in their positions.
    B=\begin{bmatrix}1 & \frac{5}{3} & \frac{5}{2} \\ \frac{5}{3} & 2 & \frac{13}{5} \\ \frac{5}{2} & \frac{13}{5} & 3\end{bmatrix}
B=\begin{bmatrix}1 & \frac{5}{3} & \frac{5}{2} \\ \frac{5}{3} & 2 & \frac{13}{5} \\ \frac{5}{2} & \frac{13}{5} & 3\end{bmatrix}

Book answer. (i) and (iii) and (iv) agree. For (ii) the key prints the third row as 8/9,\ 5/9,\ 0; the rule gives 4/3,\ 5/6,\ 0.

Note. Answer key, part (ii): with b_{ij}=\dfrac{i^2-j^2}{2i} the entries of row 3 are \dfrac{9-1}{6}=\dfrac43 and \dfrac{9-4}{6}=\dfrac56, not \dfrac89 and \dfrac59 (the key has divided row 3 by 9 instead of 2i=6). Rows 1 and 2 of the key agree with the rule, so the key's third row is a misprint.

3
If A=\begin{bmatrix}3 & -1 & 2 \\ 0 & 6 & 1 \\ -1 & 0 & -3\end{bmatrix} and B=\begin{bmatrix}2 & 1 & 7 \\ 0 & 2 & -1 \\ -3 & 4 & 2\end{bmatrix} then find a matrix C such that A+B+C=O.
Solution
  1. From A+B+C=O move A+B to the other side. Adding the matrix -(A+B) to both sides gives C=-(A+B).
    C=-(A+B)
  2. Add A and B by adding the corresponding elements (same order 3\times3).
    A+B=\begin{bmatrix}3 & -1 & 2 \\ 0 & 6 & 1 \\ -1 & 0 & -3\end{bmatrix}+\begin{bmatrix}2 & 1 & 7 \\ 0 & 2 & -1 \\ -3 & 4 & 2\end{bmatrix}=\begin{bmatrix}5 & 0 & 9 \\ 0 & 8 & 0 \\ -4 & 4 & -1\end{bmatrix}
  3. Multiply by -1: change the sign of every element (scalar multiplication).
    C=-\begin{bmatrix}5 & 0 & 9 \\ 0 & 8 & 0 \\ -4 & 4 & -1\end{bmatrix}=\begin{bmatrix}-5 & 0 & -9 \\ 0 & -8 & 0 \\ 4 & -4 & 1\end{bmatrix}
  4. Check: A+B+C should be the zero matrix.
    A+B+C=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
C=\begin{bmatrix}-5 & 0 & -9 \\ 0 & -8 & 0 \\ 4 & -4 & 1\end{bmatrix}

Book answer. C = [-5, 0, -9; 0, -8, 0; 4, -4, 1]. Agrees.

4
Solve the following matrix problems.
Solution

Method. Matrix equations are solved by equating corresponding elements, or by multiplying both sides by an inverse on the correct side (matrix multiplication is not commutative).

(i) Find A, \begin{bmatrix}2&1\\3&2\end{bmatrix}A\begin{bmatrix}1&3\\2&4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}
  1. Let P=\begin{bmatrix}2&1\\3&2\end{bmatrix} and Q=\begin{bmatrix}1&3\\2&4\end{bmatrix}. The equation is PAQ=I. If P and Q have inverses we can remove them from the two sides.
  2. Find |P| and P^{-1} by the adjoint method (swap the diagonal elements, change the signs of the others, divide by the determinant).
    |P|=(2)(2)-(1)(3)=1,\qquad P^{-1}=\frac11\begin{bmatrix}2&-1\\-3&2\end{bmatrix}=\begin{bmatrix}2 & -1 \\ -3 & 2\end{bmatrix}
  3. Do the same for Q.
    |Q|=(1)(4)-(3)(2)=-2,\qquad Q^{-1}=\frac{1}{-2}\begin{bmatrix}4&-3\\-2&1\end{bmatrix}=\begin{bmatrix}-2 & \frac{3}{2} \\ 1 & -\frac{1}{2}\end{bmatrix}
  4. Pre-multiply both sides by P^{-1} and post-multiply both sides by Q^{-1}. Since P^{-1}P=I and QQ^{-1}=I, only A is left on the left side.
    P^{-1}(PAQ)Q^{-1}=P^{-1}IQ^{-1}\;\Rightarrow\;A=P^{-1}Q^{-1}
  5. Multiply the two inverses.
    A=\begin{bmatrix}2 & -1 \\ -3 & 2\end{bmatrix}\begin{bmatrix}-2 & \frac{3}{2} \\ 1 & -\frac{1}{2}\end{bmatrix}=\begin{bmatrix}-5 & \frac{7}{2} \\ 8 & -\frac{11}{2}\end{bmatrix}
  6. Check by substitution: PAQ should be I.
    PAQ=\begin{bmatrix}-2 & \frac{3}{2} \\ 1 & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}1 & 3 \\ 2 & 4\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}
A=\begin{bmatrix}-5 & \frac{7}{2} \\ 8 & -\frac{11}{2}\end{bmatrix}
(ii) Find X, \begin{bmatrix}3&2\\0&1\\2&0\end{bmatrix}X=\begin{bmatrix}11/2&11&2\\2&4&1\\1&2&0\end{bmatrix}
  1. The left factor is 3\times2 and the product is 3\times3, so X must be 2\times3 (columns of the first factor equal rows of X; columns of X equal columns of the product).
    X=\begin{bmatrix}a&b&c\\d&e&f\end{bmatrix}
  2. Multiply out. Each row of the left matrix gives a combination of the rows of X.
    \begin{bmatrix}3a+2d&3b+2e&3c+2f\\d&e&f\\2a&2b&2c\end{bmatrix}=\begin{bmatrix}\frac{11}{2}&11&2\\2&4&1\\1&2&0\end{bmatrix}
  3. Equate the second rows: d=2, e=4, f=1. Equate the third rows: 2a=1, 2b=2, 2c=0.
    a=\tfrac12,\ b=1,\ c=0,\qquad d=2,\ e=4,\ f=1
  4. Check the first row: 3a+2d=\frac32+4=\frac{11}2, 3b+2e=3+8=11, 3c+2f=0+2=2. All three agree with the right side.
X=\begin{bmatrix}\frac12&1&0\\2&4&1\end{bmatrix}
(iii) If A=[3\ \ 7] and B=[2\ \ 14] then find a non-zero matrix C such that AC=BC.
  1. A and B are 1\times2, so C must have two rows. Let C=\begin{bmatrix}x\\y\end{bmatrix}.
  2. Compute both products.
    AC=[3x+7y],\qquad BC=[2x+14y]
  3. Equate the entries of AC and BC.
    3x+7y=2x+14y\;\Rightarrow\;x=7y
  4. Take y=t (t\neq0 so that C is not the zero matrix). Then x=7t. For example t=1 gives C=\begin{bmatrix}7\\1\end{bmatrix}.
    C=\begin{bmatrix}7t\\t\end{bmatrix}
  5. Check with t=1: AC=3(7)+7(1)=28 and BC=2(7)+14(1)=28.
C=\begin{bmatrix}7t\\t\end{bmatrix},\ t\neq0
(iv) \begin{bmatrix}xy&4\\0&x+y\end{bmatrix}=\begin{bmatrix}8&z\\t&6\end{bmatrix} then find the values of z, t and x^2+y^2.
  1. Equal matrices have the same order and equal corresponding elements. Compare the four positions.
    xy=8,\quad z=4,\quad t=0,\quad x+y=6
  2. For x^2+y^2 use the identity (x+y)^2=x^2+y^2+2xy so that we do not need x and y separately.
    x^2+y^2=(x+y)^2-2xy
  3. Substitute x+y=6 and xy=8.
    x^2+y^2=6^2-2(8)=36-16=20
  4. Check: xy=8, x+y=6 give x,y=2,4, and 4+16=20.
z=4,\ t=0,\ x^2+y^2=20
(v) If A=\begin{bmatrix}3&4\\7&6\end{bmatrix} and I=\begin{bmatrix}1&0\\0&1\end{bmatrix} then find \alpha and \beta so that A^2+\alpha I=\beta A.
  1. Find A^2=A\cdot A (row by column).
    A^2=\begin{bmatrix}3 & 4 \\ 7 & 6\end{bmatrix}\begin{bmatrix}3 & 4 \\ 7 & 6\end{bmatrix}=\begin{bmatrix}37 & 36 \\ 63 & 64\end{bmatrix}
  2. Form the left side A^2+\alpha I (add \alpha to the diagonal) and the right side \beta A.
    \begin{bmatrix}37+\alpha&36\\63&64+\alpha\end{bmatrix}=\begin{bmatrix}3\beta&4\beta\\7\beta&6\beta\end{bmatrix}
  3. Compare the entries that do not contain \alpha: 36=4\beta and 63=7\beta.
    \beta=9
  4. Compare the diagonal entries: 37+\alpha=3\beta=27 and 64+\alpha=6\beta=54.
    \alpha=-10
  5. Both diagonal equations give the same \alpha, so the values are consistent.
\alpha=-10,\ \beta=9
(vi) Find the values of x if \begin{bmatrix}x&-4&2\end{bmatrix}\begin{bmatrix}1&0&3\\0&1&0\\2&0&4\end{bmatrix}\begin{bmatrix}x\\1\\-1\end{bmatrix}=0.
  1. Multiply the row matrix by the 3\times3 matrix first. The result is 1\times3.
    \begin{bmatrix}x&-4&2\end{bmatrix}\begin{bmatrix}1&0&3\\0&1&0\\2&0&4\end{bmatrix}=\begin{bmatrix}x+4&-4&3x+8\end{bmatrix}
  2. Multiply this 1\times3 matrix by the column. The result is 1\times1.
    \begin{bmatrix}x+4&-4&3x+8\end{bmatrix}\begin{bmatrix}x\\1\\-1\end{bmatrix}=\left[x(x+4)-4-(3x+8)\right]=\left[x^2+x-12\right]
  3. Set this equal to 0 and factorise.
    x^2+x-12=0\;\Rightarrow\;(x+4)(x-3)=0
  4. Each factor can be zero.
    x=-4\ \text{or}\ x=3
x=-4\ \text{or}\ x=3

Book answer. (i) A = [-5, 7/2; 8, -11/2]; (ii) X = [1/2, 1, 0; 2, 4, 1]; (iii) C = [7x; x]; (iv) z = 4, t = 0, x^2+y^2 = 20; (v) alpha = -10, beta = 9; (vi) -4, 3. All agree.

5
If X=\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix} then prove that X^2-4X-5I=0.
Solution
  1. Find X^2 by multiplying each row of X by each column of X.
    X^2=\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}=\begin{bmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{bmatrix}
  2. Find 4X by multiplying every element of X by 4, and 5I by putting 5 on the diagonal.
    4X=\begin{bmatrix}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{bmatrix},\qquad 5I=\begin{bmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{bmatrix}
  3. Form X^2-4X-5I by subtracting corresponding elements.
    X^2-4X-5I=\begin{bmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{bmatrix}-\begin{bmatrix}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{bmatrix}-\begin{bmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{bmatrix}
  4. Element by element the result is the zero matrix, for example the first row is 9-4-5=0, 8-8-0=0, 8-8-0=0.
    X^2-4X-5I=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}=O
X^2-4X-5I=O
6
If A=\begin{bmatrix}2&1\\3&-3\end{bmatrix} then find \alpha and \beta such that A^2+\alpha I=\beta A.
Solution
  1. Find A^2.
    A^2=\begin{bmatrix}2 & 1 \\ 3 & -3\end{bmatrix}\begin{bmatrix}2 & 1 \\ 3 & -3\end{bmatrix}=\begin{bmatrix}7 & -1 \\ -3 & 12\end{bmatrix}
  2. Write both sides of A^2+\alpha I=\beta A.
    \begin{bmatrix}7+\alpha&-1\\-3&12+\alpha\end{bmatrix}=\begin{bmatrix}2\beta&\beta\\3\beta&-3\beta\end{bmatrix}
  3. Compare the entries without \alpha: -1=\beta and -3=3\beta. Both give the same value.
    \beta=-1
  4. Compare the diagonal entries: 7+\alpha=2\beta=-2 and 12+\alpha=-3\beta=3.
    \alpha=-9
  5. Check: both diagonal equations give \alpha=-9.
\alpha=-9,\ \beta=-1

Book answer. alpha = -9, beta = -1. Agrees.

7
Consider any two particular matrices A and B of your choice of order 2\times3 and 3\times2 respectively and show that (AB)^t=B^tA^t.
Solution

Choose A of order 2\times3 and B of order 3\times2. Then AB is defined, of order 2\times2.

  1. Our choice.
    A=\begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 4\end{bmatrix},\qquad B=\begin{bmatrix}1 & 0 \\ 2 & 1 \\ 1 & 3\end{bmatrix}
  2. Left side: find AB (row by column), then its transpose.
    AB=\begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 4\end{bmatrix}\begin{bmatrix}1 & 0 \\ 2 & 1 \\ 1 & 3\end{bmatrix}=\begin{bmatrix}8 & 11 \\ 6 & 13\end{bmatrix},\qquad (AB)^t=\begin{bmatrix}8 & 6 \\ 11 & 13\end{bmatrix}
  3. Right side: find the transposes (B^t is 2\times3 and A^t is 3\times2).
    B^t=\begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 3\end{bmatrix},\qquad A^t=\begin{bmatrix}1 & 0 \\ 2 & 1 \\ 3 & 4\end{bmatrix}
  4. Multiply B^t by A^t. The reversed order is the only order in which the product is defined here.
    B^tA^t=\begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 3\end{bmatrix}\begin{bmatrix}1 & 0 \\ 2 & 1 \\ 3 & 4\end{bmatrix}=\begin{bmatrix}8 & 6 \\ 11 & 13\end{bmatrix}
  5. The two results are equal, so (AB)^t=B^tA^t for this choice.
(AB)^t=B^tA^t=\begin{bmatrix}8 & 6 \\ 11 & 13\end{bmatrix}

Note. The question asks for matrices of your own choice; any 2\times3 and 3\times2 matrices give the same conclusion, shown generally in Example 6 of the book.

8
Consider any two particular matrices A and B of your choice of order 3\times3 and show that (A+B)^t=A^t+B^t.
Solution
  1. Our choice (both 3\times3, so they can be added).
    A=\begin{bmatrix}1 & 2 & 0 \\ 3 & -1 & 4 \\ 2 & 5 & 1\end{bmatrix},\qquad B=\begin{bmatrix}0 & 1 & 2 \\ 1 & 1 & -3 \\ 4 & 0 & 2\end{bmatrix}
  2. Left side: add first, then transpose.
    A+B=\begin{bmatrix}1 & 3 & 2 \\ 4 & 0 & 1 \\ 6 & 5 & 3\end{bmatrix},\qquad (A+B)^t=\begin{bmatrix}1 & 4 & 6 \\ 3 & 0 & 5 \\ 2 & 1 & 3\end{bmatrix}
  3. Right side: transpose each, then add.
    A^t=\begin{bmatrix}1 & 3 & 2 \\ 2 & -1 & 5 \\ 0 & 4 & 1\end{bmatrix},\quad B^t=\begin{bmatrix}0 & 1 & 4 \\ 1 & 1 & 0 \\ 2 & -3 & 2\end{bmatrix}
  4. Add A^t and B^t element by element.
    A^t+B^t=\begin{bmatrix}1 & 4 & 6 \\ 3 & 0 & 5 \\ 2 & 1 & 3\end{bmatrix}
  5. The two matrices are equal, so (A+B)^t=A^t+B^t for this choice.
(A+B)^t=A^t+B^t

Note. The matrices are of the student's choice; any two matrices of the same order work (compare Example 7 of the book).

9
If A and B are two matrices such that AB=B and BA=A. Find A^2+B^2.
Solution
  1. Both products AB and BA exist and equal matrices of the order of B and A respectively, so A and B are square matrices of the same order. Start with A^2=A\cdot A and replace one A by BA, using BA=A.
    A^2=A\cdot A=A(BA)
  2. Matrix multiplication is associative, so the brackets can be moved. Then use AB=B and finally BA=A again.
    A(BA)=(AB)A=BA=A
  3. So A^2=A. In the same way, replace one B in B^2 by AB (since AB=B), regroup, and use BA=A then AB=B.
    B^2=B\cdot B=B(AB)=(BA)B=AB=B
  4. Add the two results.
    A^2+B^2=A+B
A^2+B^2=A+B

Note. No numerical value is possible because A and B are not given; the answer is in terms of A and B. The book's key prints no answer for this question.

10
If A=[a_{ij}] is a matrix of order 3\times3 and a_{ij}=i^2-j^2. Check whether A is symmetric or skew-symmetric.
Solution
  1. Construct A from the rule: row i, column j.
    a_{11}=0,\ a_{12}=1-4=-3,\ a_{13}=1-9=-8,\ a_{21}=4-1=3,\ a_{22}=0,\ a_{23}=4-9=-5,\ a_{31}=9-1=8,\ a_{32}=9-4=5,\ a_{33}=0
  2. Write the matrix.
    A=\begin{bmatrix}0 & -3 & -8 \\ 3 & 0 & -5 \\ 8 & 5 & 0\end{bmatrix}
  3. Take the transpose.
    A^t=\begin{bmatrix}0 & 3 & 8 \\ -3 & 0 & 5 \\ -8 & -5 & 0\end{bmatrix}
  4. Compare with -A (change the sign of every element of A).
    -A=\begin{bmatrix}0 & 3 & 8 \\ -3 & 0 & 5 \\ -8 & -5 & 0\end{bmatrix}
  5. A^t=-A, so A is skew-symmetric. A quicker test: a_{ji}=j^2-i^2=-(i^2-j^2)=-a_{ij} for all i,j, and a_{ii}=0.
A^t=-A:\ A\text{ is skew-symmetric}

Book answer. skew symmetric. Agrees.

11
For any square matrix A; prove that (A^n)^t=(A^t)^n.
Solution

Plan. Use the reversal law (AB)^t=B^tA^t (Example 6 and section 2.2.6) and mathematical induction on n.

  1. **n=1.** (A^1)^t=A^t=(A^t)^1. The statement holds.
  2. **Assume it for n=k.** Suppose (A^k)^t=(A^t)^k.
  3. **Prove it for n=k+1.** Write A^{k+1}=A^k\cdot A and apply the reversal law.
    (A^{k+1})^t=(A^kA)^t=A^t\,(A^k)^t
  4. Replace (A^k)^t by (A^t)^k using the assumption.
    A^t(A^k)^t=A^t(A^t)^k=(A^t)^{k+1}
  5. So the statement is true for n=k+1 whenever it is true for n=k. Together with n=1 this proves it for every positive integer n.

**Check with n=2.** (A^2)^t=(AA)^t=A^tA^t=(A^t)^2.

(A^n)^t=(A^t)^n
12
Find the matrices X and Y such that 2X-Y=\begin{bmatrix}1&6&-3\\2&1&7\end{bmatrix} and X+3Y=\begin{bmatrix}4&3&2\\1&-3&0\end{bmatrix}.
Solution
  1. Call the right sides P and Q. Treat the two matrix equations like ordinary simultaneous equations.
    2X-Y=P\quad(1),\qquad X+3Y=Q\quad(2)
  2. Multiply (1) by 3 so that the Y terms cancel when added to (2).
    6X-3Y=3P\quad(3)
  3. Add (3) and (2).
    7X=3P+Q=3\begin{bmatrix}1 & 6 & -3 \\ 2 & 1 & 7\end{bmatrix}+\begin{bmatrix}4 & 3 & 2 \\ 1 & -3 & 0\end{bmatrix}=\begin{bmatrix}7 & 21 & -7 \\ 7 & 0 & 21\end{bmatrix}
  4. Divide by 7 (multiply every element by \frac17).
    X=\begin{bmatrix}1 & 3 & -1 \\ 1 & 0 & 3\end{bmatrix}
  5. Find Y from (1): Y=2X-P.
    Y=2\begin{bmatrix}1 & 3 & -1 \\ 1 & 0 & 3\end{bmatrix}-\begin{bmatrix}1 & 6 & -3 \\ 2 & 1 & 7\end{bmatrix}=\begin{bmatrix}1 & 0 & 1 \\ 0 & -1 & -1\end{bmatrix}
  6. Check in (2): X+3Y= the given matrix.
    X+3Y=\begin{bmatrix}1 & 3 & -1 \\ 1 & 0 & 3\end{bmatrix}+\begin{bmatrix}3 & 0 & 3 \\ 0 & -3 & -3\end{bmatrix}=\begin{bmatrix}4 & 3 & 2 \\ 1 & -3 & 0\end{bmatrix}
X=\begin{bmatrix}1 & 3 & -1 \\ 1 & 0 & 3\end{bmatrix},\quad Y=\begin{bmatrix}1 & 0 & 1 \\ 0 & -1 & -1\end{bmatrix}

Book answer. X = [1, 3, -1; 1, 0, 3], Y = [1, 0, 1; 0, -1, -1]. Agrees.

13
Find the matrices X and Y such that 3X+Y=\begin{bmatrix}2 & 5 & 3i \\ 2+7i & 5 & 7 \\ 2 & 2 & 3i\end{bmatrix} and X+2Y=\begin{bmatrix}1 & 6 & 3i \\ 7+2i & 5i & 7 \\ 3 & 3 & 6\end{bmatrix}.
Solution
  1. Name the right sides P and Q.
    3X+Y=P\quad(1),\qquad X+2Y=Q\quad(2)
  2. To find X, multiply (1) by 2 and subtract (2). The Y terms cancel.
    6X+2Y=2P\;\Rightarrow\;5X=2P-Q
  3. Compute 2P-Q element by element (complex numbers are added in real and imaginary parts).
    2P-Q=\begin{bmatrix}4 & 10 & 6i \\ 4+14i & 10 & 14 \\ 4 & 4 & 6i\end{bmatrix}-\begin{bmatrix}1 & 6 & 3i \\ 7+2i & 5i & 7 \\ 3 & 3 & 6\end{bmatrix}=\begin{bmatrix}3 & 4 & 3i \\ -3+12i & 10-5i & 7 \\ 1 & 1 & -6+6i\end{bmatrix}
  4. Divide by 5.
    X=\begin{bmatrix}\frac{3}{5} & \frac{4}{5} & \frac{3}{5}i \\ -\frac{3}{5}+\frac{12}{5}i & 2-i & \frac{7}{5} \\ \frac{1}{5} & \frac{1}{5} & -\frac{6}{5}+\frac{6}{5}i\end{bmatrix}
  5. To find Y, multiply (2) by 3 and subtract (1). The X terms cancel.
    3X+6Y=3Q\;\Rightarrow\;5Y=3Q-P
  6. Compute 3Q-P.
    3Q-P=\begin{bmatrix}3 & 18 & 9i \\ 21+6i & 15i & 21 \\ 9 & 9 & 18\end{bmatrix}-\begin{bmatrix}2 & 5 & 3i \\ 2+7i & 5 & 7 \\ 2 & 2 & 3i\end{bmatrix}=\begin{bmatrix}1 & 13 & 6i \\ 19-i & -5+15i & 14 \\ 7 & 7 & 18-3i\end{bmatrix}
  7. Divide by 5.
    Y=\begin{bmatrix}\frac{1}{5} & \frac{13}{5} & \frac{6}{5}i \\ \frac{19}{5}-\frac{1}{5}i & -1+3i & \frac{14}{5} \\ \frac{7}{5} & \frac{7}{5} & \frac{18}{5}-\frac{3}{5}i\end{bmatrix}
  8. Check: 3X+Y gives back P and X+2Y gives back Q (verified element by element).
X=\begin{bmatrix}\frac{3}{5} & \frac{4}{5} & \frac{3}{5}i \\ -\frac{3}{5}+\frac{12}{5}i & 2-i & \frac{7}{5} \\ \frac{1}{5} & \frac{1}{5} & -\frac{6}{5}+\frac{6}{5}i\end{bmatrix},\quad Y=\begin{bmatrix}\frac{1}{5} & \frac{13}{5} & \frac{6}{5}i \\ \frac{19}{5}-\frac{1}{5}i & -1+3i & \frac{14}{5} \\ \frac{7}{5} & \frac{7}{5} & \frac{18}{5}-\frac{3}{5}i\end{bmatrix}

Book answer. X = [3/5, 4/5, 3i/5; (-3+12i)/5, 2-i, 7/5; 1/5, 1/5, (-6+6i)/5], Y = [1/5, 13/5, 6i/5; (19-i)/5, -1+3i, 14/5; 7/5, 7/5, (18-3i)/5]. Agrees.

Exercise 2.3p. 46–47

Determinants, cofactors, singular matrices and the adjoint method for the inverse. Expand along a row or column (cofactors), then use A^{-1}=\frac{1}{|A|}\operatorname{adj}(A).

1
Evaluate the determinant of the following matrices.
Solution

Method (sections 2.3.1 and 2.3.5). Expand along a row or column. Each element is multiplied by its minor with the sign (-1)^{i+j}; choosing a row or column with zeros shortens the work. For complex entries use i^2=-1.

(i) A=\begin{bmatrix}2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2\end{bmatrix}
  1. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=2\begin{vmatrix}-1 & 2 \\ 1 & 2\end{vmatrix}-3\begin{vmatrix}1 & 2 \\ 4 & 2\end{vmatrix}+1\begin{vmatrix}1 & -1 \\ 4 & 1\end{vmatrix}
  2. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-1 & 2 \\ 1 & 2\end{vmatrix}=(-1)(2)-(2)(1)=-4 \\ \begin{vmatrix}1 & 2 \\ 4 & 2\end{vmatrix}=(1)(2)-(2)(4)=-6 \\ \begin{vmatrix}1 & -1 \\ 4 & 1\end{vmatrix}=(1)(1)-(-1)(4)=5 \end{aligned}
  3. Substitute these values and add.
    |A|=2(-4)-3(-6)+1(5)=15
|A|=15
(ii) B=\begin{bmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{bmatrix}
  1. Look for the row or column with the most zeros: column 3 is (0,0,1). Expand along it, so only the element 1 in the last row contributes.
    |B|=0\cdot A_{13}+0\cdot A_{23}+1\cdot A_{33}
  2. The cofactor A_{33}=(-1)^{3+3} times the minor obtained by deleting row 3 and column 3.
    A_{33}=(+1)\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}=\cos^2\theta-(-\sin\theta)(\sin\theta)=\cos^2\theta+\sin^2\theta
  3. Use the identity \sin^2\theta+\cos^2\theta=1.
    |B|=1\cdot1=1
|B|=1
(iii) C=\begin{bmatrix}i & 3 & -2i \\ 1 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}
  1. Expand along row 3. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted. This line contains a zero, so one term vanishes.
    |C|=0\begin{vmatrix}3 & -2i \\ 3 & 4\end{vmatrix}-1\begin{vmatrix}i & -2i \\ 1 & 4\end{vmatrix}+2\begin{vmatrix}i & 3 \\ 1 & 3\end{vmatrix}
  2. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}3 & -2i \\ 3 & 4\end{vmatrix}=(3)(4)-(-2i)(3)=12+6i \\ \begin{vmatrix}i & -2i \\ 1 & 4\end{vmatrix}=(i)(4)-(-2i)(1)=6i \\ \begin{vmatrix}i & 3 \\ 1 & 3\end{vmatrix}=(i)(3)-(3)(1)=-3+3i \end{aligned}
  3. Substitute these values and add.
    |C|=0(12+6i)-1(6i)+2(-3+3i)=-6
|C|=-6
(iv) D=\begin{bmatrix}2+i & 1 & i \\ 0 & 2 & 1 \\ -3i & 1 & 6\end{bmatrix}
  1. Expand along row 2. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted. This line contains a zero, so one term vanishes.
    |D|=-0\begin{vmatrix}1 & i \\ 1 & 6\end{vmatrix}+2\begin{vmatrix}2+i & i \\ -3i & 6\end{vmatrix}-1\begin{vmatrix}2+i & 1 \\ -3i & 1\end{vmatrix}
  2. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}1 & i \\ 1 & 6\end{vmatrix}=(1)(6)-(i)(1)=6-i \\ \begin{vmatrix}2+i & i \\ -3i & 6\end{vmatrix}=(2+i)(6)-(i)(-3i)=9+6i \\ \begin{vmatrix}2+i & 1 \\ -3i & 1\end{vmatrix}=(2+i)(1)-(1)(-3i)=2+4i \end{aligned}
  3. Substitute these values and add.
    |D|=-0(6-i)+2(9+6i)-1(2+4i)=16+8i
|D|=16+8i

Book answer. (i) 15, (ii) 1, (iii) -6, (iv) 16 + 8i. All agree.

2
Evaluate the determinants of the following matrices using cofactor method.
Solution

Method (section 2.3.5, Example 11). |A| equals the sum of the products of the elements of any one row (or column) with their own cofactors.

(i) A=\begin{bmatrix}3 & 2 & 3 \\ 4 & 5 & 1 \\ 2 & 1 & 0\end{bmatrix}
  1. The cofactor method: |A|=a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3} for any row i (or the same with a column), where A_{ij}=(-1)^{i+j}M_{ij}.
  2. Choose row 3 (it contains zeros, which saves work). Find the cofactor A_{ij}=(-1)^{i+j}M_{ij} of each of its elements.
    \begin{aligned} A_{31}=(-1)^{4}\begin{vmatrix}2 & 3 \\ 5 & 1\end{vmatrix}=+\left[(2)(1)-(3)(5)\right]=-13 \\ A_{32}=(-1)^{5}\begin{vmatrix}3 & 3 \\ 4 & 1\end{vmatrix}=-\left[(3)(1)-(3)(4)\right]=9 \\ A_{33}=(-1)^{6}\begin{vmatrix}3 & 2 \\ 4 & 5\end{vmatrix}=+\left[(3)(5)-(2)(4)\right]=7 \end{aligned}
  3. Multiply each element of row 3 by its cofactor and add: |A|=\sum a_{ij}A_{ij}.
    |A|=2(-13)+1(9)+0(7)=-17
|A|=-17
(ii) B=\begin{bmatrix}2 & 3 & -1 \\ -1 & 0 & 2 \\ 3 & 1 & 4\end{bmatrix}
  1. The cofactor method: |A|=a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3} for any row i (or the same with a column), where A_{ij}=(-1)^{i+j}M_{ij}.
  2. Choose row 2 (it contains zeros, which saves work). Find the cofactor A_{ij}=(-1)^{i+j}M_{ij} of each of its elements.
    \begin{aligned} A_{21}=(-1)^{3}\begin{vmatrix}3 & -1 \\ 1 & 4\end{vmatrix}=-\left[(3)(4)-(-1)(1)\right]=-13 \\ A_{22}=(-1)^{4}\begin{vmatrix}2 & -1 \\ 3 & 4\end{vmatrix}=+\left[(2)(4)-(-1)(3)\right]=11 \\ A_{23}=(-1)^{5}\begin{vmatrix}2 & 3 \\ 3 & 1\end{vmatrix}=-\left[(2)(1)-(3)(3)\right]=7 \end{aligned}
  3. Multiply each element of row 2 by its cofactor and add: |A|=\sum a_{ij}A_{ij}.
    |B|=(-1)(-13)+0(11)+2(7)=27
|B|=27
(iii) C=\begin{bmatrix}2i & 6 & 1 \\ 1 & -i & 2 \\ 0 & 1 & 3i\end{bmatrix}
  1. The cofactor method: |A|=a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3} for any row i (or the same with a column), where A_{ij}=(-1)^{i+j}M_{ij}.
  2. Choose row 3 (it contains zeros, which saves work). Find the cofactor A_{ij}=(-1)^{i+j}M_{ij} of each of its elements.
    \begin{aligned} A_{31}=(-1)^{4}\begin{vmatrix}6 & 1 \\ -i & 2\end{vmatrix}=+\left[(6)(2)-(1)(-i)\right]=12+i \\ A_{32}=(-1)^{5}\begin{vmatrix}2i & 1 \\ 1 & 2\end{vmatrix}=-\left[(2i)(2)-(1)(1)\right]=1-4i \\ A_{33}=(-1)^{6}\begin{vmatrix}2i & 6 \\ 1 & -i\end{vmatrix}=+\left[(2i)(-i)-(6)(1)\right]=-4 \end{aligned}
  3. Multiply each element of row 3 by its cofactor and add: |A|=\sum a_{ij}A_{ij}.
    |C|=0(12+i)+1(1-4i)+3i(-4)=1-16i
|C|=1-16i
(iv) D=\begin{bmatrix}1-i & 2 & 1+i \\ 3 & 1 & 4 \\ 0 & 2 & 3\end{bmatrix}
  1. The cofactor method: |A|=a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3} for any row i (or the same with a column), where A_{ij}=(-1)^{i+j}M_{ij}.
  2. Choose row 3 (it contains zeros, which saves work). Find the cofactor A_{ij}=(-1)^{i+j}M_{ij} of each of its elements.
    \begin{aligned} A_{31}=(-1)^{4}\begin{vmatrix}2 & 1+i \\ 1 & 4\end{vmatrix}=+\left[(2)(4)-(1+i)(1)\right]=7-i \\ A_{32}=(-1)^{5}\begin{vmatrix}1-i & 1+i \\ 3 & 4\end{vmatrix}=-\left[(1-i)(4)-(1+i)(3)\right]=-1+7i \\ A_{33}=(-1)^{6}\begin{vmatrix}1-i & 2 \\ 3 & 1\end{vmatrix}=+\left[(1-i)(1)-(2)(3)\right]=-5-i \end{aligned}
  3. Multiply each element of row 3 by its cofactor and add: |A|=\sum a_{ij}A_{ij}.
    |D|=0(7-i)+2(-1+7i)+3(-5-i)=-17+11i
|D|=-17+11i

Book answer. (i) -17, (ii) 27, (iii) 1 - 16i, (iv) -17 + 11i. All agree.

3
Determine which of the following matrices are singular and which are non-singular.
Solution

Method (section 2.3.6). A square matrix is singular if |A|=0 and non-singular if |A|\neq0. Evaluate each determinant.

(i) A=\begin{bmatrix}3 & 1 & 2 \\ 2 & 3 & 1 \\ -4 & 1 & -3\end{bmatrix}
  1. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=3\begin{vmatrix}3 & 1 \\ 1 & -3\end{vmatrix}-1\begin{vmatrix}2 & 1 \\ -4 & -3\end{vmatrix}+2\begin{vmatrix}2 & 3 \\ -4 & 1\end{vmatrix}
  2. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}3 & 1 \\ 1 & -3\end{vmatrix}=(3)(-3)-(1)(1)=-10 \\ \begin{vmatrix}2 & 1 \\ -4 & -3\end{vmatrix}=(2)(-3)-(1)(-4)=-2 \\ \begin{vmatrix}2 & 3 \\ -4 & 1\end{vmatrix}=(2)(1)-(3)(-4)=14 \end{aligned}
  3. Substitute these values and add.
    |A|=3(-10)-1(-2)+2(14)=0
  4. |A|=0, so A is singular.
\text{singular}
(ii) B=\begin{bmatrix}3 & -1 & 2 \\ 2 & 0 & 1 \\ -1 & 5 & 1\end{bmatrix}
  1. Expand along row 2. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |B|=-2\begin{vmatrix}-1 & 2 \\ 5 & 1\end{vmatrix}+0\begin{vmatrix}3 & 2 \\ -1 & 1\end{vmatrix}-1\begin{vmatrix}3 & -1 \\ -1 & 5\end{vmatrix}
  2. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-1 & 2 \\ 5 & 1\end{vmatrix}=(-1)(1)-(2)(5)=-11 \\ \begin{vmatrix}3 & 2 \\ -1 & 1\end{vmatrix}=(3)(1)-(2)(-1)=5 \\ \begin{vmatrix}3 & -1 \\ -1 & 5\end{vmatrix}=(3)(5)-(-1)(-1)=14 \end{aligned}
  3. Substitute these values and add.
    |B|=-2(-11)+0(5)-1(14)=8
  4. |B|=8\neq0, so B is non-singular.
\text{non-singular}
(iii) C=\begin{bmatrix}3i & 1 & 2 \\ -4 & 1 & i \\ 2 & 0 & 1\end{bmatrix}
  1. Expand along row 3. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |C|=2\begin{vmatrix}1 & 2 \\ 1 & i\end{vmatrix}-0\begin{vmatrix}3i & 2 \\ -4 & i\end{vmatrix}+1\begin{vmatrix}3i & 1 \\ -4 & 1\end{vmatrix}
  2. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}1 & 2 \\ 1 & i\end{vmatrix}=(1)(i)-(2)(1)=-2+i \\ \begin{vmatrix}3i & 2 \\ -4 & i\end{vmatrix}=(3i)(i)-(2)(-4)=5 \\ \begin{vmatrix}3i & 1 \\ -4 & 1\end{vmatrix}=(3i)(1)-(1)(-4)=4+3i \end{aligned}
  3. Substitute these values and add.
    |C|=2(-2+i)-0(5)+1(4+3i)=5i
  4. |C|=5i\neq0, so C is non-singular.
\text{non-singular}
(iv) D=\begin{bmatrix}2 & -i & 1 \\ i & 3 & -2 \\ -2+i & 3+i & -3\end{bmatrix}
  1. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |D|=2\begin{vmatrix}3 & -2 \\ 3+i & -3\end{vmatrix}-(-i)\begin{vmatrix}i & -2 \\ -2+i & -3\end{vmatrix}+1\begin{vmatrix}i & 3 \\ -2+i & 3+i\end{vmatrix}
  2. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}3 & -2 \\ 3+i & -3\end{vmatrix}=(3)(-3)-(-2)(3+i)=-3+2i \\ \begin{vmatrix}i & -2 \\ -2+i & -3\end{vmatrix}=(i)(-3)-(-2)(-2+i)=-4-i \\ \begin{vmatrix}i & 3 \\ -2+i & 3+i\end{vmatrix}=(i)(3+i)-(3)(-2+i)=5 \end{aligned}
  3. Substitute these values and add.
    |D|=2(-3+2i)-(-i)(-4-i)+1(5)=0
  4. |D|=0, so D is singular.
\text{singular}

Book answer. (i) singular, (ii) non-singular, (iii) non-singular, (iv) singular. All agree.

4
Find the value of \lambda, so that the given matrices are singular.
Solution

Method. A matrix is singular exactly when |A|=0 (section 2.3.6). Expand |A| along a line containing \lambda, which gives a linear equation in \lambda.

(i) A=\begin{bmatrix}\lambda & 1 & 3 \\ 2 & 1 & 8 \\ 0 & 3 & 1\end{bmatrix}
  1. \lambda appears only in column 1, so expand along it; the 2\times2 minors then do not contain \lambda.
    |A|=\lambda\begin{vmatrix}1 & 8 \\ 3 & 1\end{vmatrix}-2\begin{vmatrix}1 & 3 \\ 3 & 1\end{vmatrix}+0\begin{vmatrix}1 & 3 \\ 1 & 8\end{vmatrix}
  2. Evaluate each 2\times2 minor.
    \begin{aligned} \begin{vmatrix}1 & 8 \\ 3 & 1\end{vmatrix}=(1)(1)-(8)(3)=-23 \\ \begin{vmatrix}1 & 3 \\ 3 & 1\end{vmatrix}=(1)(1)-(3)(3)=-8 \\ \begin{vmatrix}1 & 3 \\ 1 & 8\end{vmatrix}=(1)(8)-(3)(1)=5 \end{aligned}
  3. Substitute and collect the terms in \lambda.
    |A|=\lambda(-23)-2(-8)+0(5)=(-23)\lambda+16
  4. A matrix is singular when its determinant is 0. Set |A|=0 and solve the linear equation for \lambda.
    (-23)\lambda+16=0\;\Rightarrow\;\lambda=\frac{16}{23}
  5. Check: with this \lambda the determinant is 0 (verified by recomputing it).
\lambda=\frac{16}{23}
(ii) B=\begin{bmatrix}\lambda & 2 & 0 \\ 2 & 1 & 3 \\ \lambda & 2 & 1\end{bmatrix}
  1. \lambda appears only in column 1 (twice), so expand along it; the 2\times2 minors then do not contain \lambda.
    |B|=\lambda\begin{vmatrix}1 & 3 \\ 2 & 1\end{vmatrix}-2\begin{vmatrix}2 & 0 \\ 2 & 1\end{vmatrix}+\lambda\begin{vmatrix}2 & 0 \\ 1 & 3\end{vmatrix}
  2. Evaluate each 2\times2 minor.
    \begin{aligned} \begin{vmatrix}1 & 3 \\ 2 & 1\end{vmatrix}=(1)(1)-(3)(2)=-5 \\ \begin{vmatrix}2 & 0 \\ 2 & 1\end{vmatrix}=(2)(1)-(0)(2)=2 \\ \begin{vmatrix}2 & 0 \\ 1 & 3\end{vmatrix}=(2)(3)-(0)(1)=6 \end{aligned}
  3. Substitute and collect the terms in \lambda.
    |B|=\lambda(-5)-2(2)+\lambda(6)=\lambda-4
  4. A matrix is singular when its determinant is 0. Set |B|=0 and solve the linear equation for \lambda.
    \lambda-4=0\;\Rightarrow\;\lambda=4
  5. Check: with this \lambda the determinant is 0 (verified by recomputing it).
\lambda=4
(iii) C=\begin{bmatrix}\lambda & i & 1 \\ 2 & 1 & 3 \\ 3 & 1 & 2\end{bmatrix}
  1. \lambda appears only in row 1, so expand along it; the 2\times2 minors then do not contain \lambda.
    |C|=\lambda\begin{vmatrix}1 & 3 \\ 1 & 2\end{vmatrix}-i\begin{vmatrix}2 & 3 \\ 3 & 2\end{vmatrix}+1\begin{vmatrix}2 & 1 \\ 3 & 1\end{vmatrix}
  2. Evaluate each 2\times2 minor.
    \begin{aligned} \begin{vmatrix}1 & 3 \\ 1 & 2\end{vmatrix}=(1)(2)-(3)(1)=-1 \\ \begin{vmatrix}2 & 3 \\ 3 & 2\end{vmatrix}=(2)(2)-(3)(3)=-5 \\ \begin{vmatrix}2 & 1 \\ 3 & 1\end{vmatrix}=(2)(1)-(1)(3)=-1 \end{aligned}
  3. Substitute and collect the terms in \lambda.
    |C|=\lambda(-1)-i(-5)+1(-1)=-\lambda+(-1+5i)
  4. A matrix is singular when its determinant is 0. Set |C|=0 and solve the linear equation for \lambda.
    -\lambda+(-1+5i)=0\;\Rightarrow\;\lambda=-1+5i
  5. Check: with this \lambda the determinant is 0 (verified by recomputing it).
\lambda=-1+5i
(iv) D=\begin{bmatrix}2+i & 1 & 6 \\ 2 & \lambda & 1 \\ 3 & 0 & 2\end{bmatrix}
  1. \lambda appears only in row 2, so expand along it; the 2\times2 minors then do not contain \lambda.
    |D|=-2\begin{vmatrix}1 & 6 \\ 0 & 2\end{vmatrix}+\lambda\begin{vmatrix}2+i & 6 \\ 3 & 2\end{vmatrix}-1\begin{vmatrix}2+i & 1 \\ 3 & 0\end{vmatrix}
  2. Evaluate each 2\times2 minor.
    \begin{aligned} \begin{vmatrix}1 & 6 \\ 0 & 2\end{vmatrix}=(1)(2)-(6)(0)=2 \\ \begin{vmatrix}2+i & 6 \\ 3 & 2\end{vmatrix}=(2+i)(2)-(6)(3)=-14+2i \\ \begin{vmatrix}2+i & 1 \\ 3 & 0\end{vmatrix}=(2+i)(0)-(1)(3)=-3 \end{aligned}
  3. Substitute and collect the terms in \lambda.
    |D|=-2(2)+\lambda(-14+2i)-1(-3)=(-14+2i)\lambda-1
  4. A matrix is singular when its determinant is 0. Set |D|=0 and solve the linear equation for \lambda.
    (-14+2i)\lambda-1=0\;\Rightarrow\;\lambda=-\frac{7}{100}-\frac{1}{100}i
  5. Check: with this \lambda the determinant is 0 (verified by recomputing it).
\lambda=-\frac{7}{100}-\frac{1}{100}i

Book answer. (i) 16/23, (ii) -4, (iii) -1 + 5i, (iv) -7/100 - i/100. Parts (i), (iii), (iv) agree; part (ii) differs.

Note. Answer key, part (ii): the key prints \lambda=-4, but |B|=\lambda-4, which is zero for \lambda=4. Check by substitution: with \lambda=4, |B|=4(1-6)-2(2-12)+0=-20+20=0, while \lambda=-4 gives |B|=-4-4=-8\neq0. So the key has the wrong sign.

5
Find the multiplicative inverse of the following matrices if it exists by adjoint method.
Solution

Method (section 2.3.8, Example 12). (1) Find |A|; if it is 0 there is no inverse. (2) Find the nine cofactors A_{ij}. (3) Form \operatorname{adj}(A) by transposing the cofactor matrix. (4) Divide by |A|.

(i) A=\begin{bmatrix}1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1\end{bmatrix}
  1. Adjoint method: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A) exists only if |A|\neq0. First find |A|.
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=1\begin{vmatrix}1 & -1 \\ -2 & -1\end{vmatrix}-(-1)\begin{vmatrix}2 & -1 \\ 1 & -1\end{vmatrix}+1\begin{vmatrix}2 & 1 \\ 1 & -2\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}1 & -1 \\ -2 & -1\end{vmatrix}=(1)(-1)-(-1)(-2)=-3 \\ \begin{vmatrix}2 & -1 \\ 1 & -1\end{vmatrix}=(2)(-1)-(-1)(1)=-1 \\ \begin{vmatrix}2 & 1 \\ 1 & -2\end{vmatrix}=(2)(-2)-(1)(1)=-5 \end{aligned}
  4. Substitute these values and add. So |A|=-9.
    |A|=1(-3)-(-1)(-1)+1(-5)=-9
  5. |A|\neq0, so A is non-singular and A^{-1} exists. Find all cofactors.
    \begin{aligned} A_{11}=(-1)^{2}\begin{vmatrix}1 & -1 \\ -2 & -1\end{vmatrix}=+\left[(1)(-1)-(-1)(-2)\right]=-3 \\ A_{12}=(-1)^{3}\begin{vmatrix}2 & -1 \\ 1 & -1\end{vmatrix}=-\left[(2)(-1)-(-1)(1)\right]=1 \\ A_{13}=(-1)^{4}\begin{vmatrix}2 & 1 \\ 1 & -2\end{vmatrix}=+\left[(2)(-2)-(1)(1)\right]=-5 \\ A_{21}=(-1)^{3}\begin{vmatrix}-1 & 1 \\ -2 & -1\end{vmatrix}=-\left[(-1)(-1)-(1)(-2)\right]=-3 \\ A_{22}=(-1)^{4}\begin{vmatrix}1 & 1 \\ 1 & -1\end{vmatrix}=+\left[(1)(-1)-(1)(1)\right]=-2 \\ A_{23}=(-1)^{5}\begin{vmatrix}1 & -1 \\ 1 & -2\end{vmatrix}=-\left[(1)(-2)-(-1)(1)\right]=1 \\ A_{31}=(-1)^{4}\begin{vmatrix}-1 & 1 \\ 1 & -1\end{vmatrix}=+\left[(-1)(-1)-(1)(1)\right]=0 \\ A_{32}=(-1)^{5}\begin{vmatrix}1 & 1 \\ 2 & -1\end{vmatrix}=-\left[(1)(-1)-(1)(2)\right]=3 \\ A_{33}=(-1)^{6}\begin{vmatrix}1 & -1 \\ 2 & 1\end{vmatrix}=+\left[(1)(1)-(-1)(2)\right]=3 \end{aligned}
  6. The adjoint is the transpose of the cofactor matrix: the cofactor A_{ij} goes to row j, column i.
    \operatorname{adj}(A)=\begin{bmatrix}-3 & -3 & 0 \\ 1 & -2 & 3 \\ -5 & 1 & 3\end{bmatrix}
  7. Divide by the determinant: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A).
    A^{-1}=\frac{1}{-9}\begin{bmatrix}-3 & -3 & 0 \\ 1 & -2 & 3 \\ -5 & 1 & 3\end{bmatrix}=\begin{bmatrix}\frac{1}{3} & \frac{1}{3} & 0 \\ -\frac{1}{9} & \frac{2}{9} & -\frac{1}{3} \\ \frac{5}{9} & -\frac{1}{9} & -\frac{1}{3}\end{bmatrix}
  8. Check: AA^{-1} should be I.
    AA^{-1}=\begin{bmatrix}1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1\end{bmatrix}\begin{bmatrix}\frac{1}{3} & \frac{1}{3} & 0 \\ -\frac{1}{9} & \frac{2}{9} & -\frac{1}{3} \\ \frac{5}{9} & -\frac{1}{9} & -\frac{1}{3}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
A^{-1}=\begin{bmatrix}\frac{1}{3} & \frac{1}{3} & 0 \\ -\frac{1}{9} & \frac{2}{9} & -\frac{1}{3} \\ \frac{5}{9} & -\frac{1}{9} & -\frac{1}{3}\end{bmatrix}
(ii) B=\begin{bmatrix}3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}
  1. Adjoint method: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A) exists only if |A|\neq0. First find |A|.
  2. Expand along row 3. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |B|=1\begin{vmatrix}-4 & 2 \\ 3 & 5\end{vmatrix}-0\begin{vmatrix}3 & 2 \\ 2 & 5\end{vmatrix}+1\begin{vmatrix}3 & -4 \\ 2 & 3\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-4 & 2 \\ 3 & 5\end{vmatrix}=(-4)(5)-(2)(3)=-26 \\ \begin{vmatrix}3 & 2 \\ 2 & 5\end{vmatrix}=(3)(5)-(2)(2)=11 \\ \begin{vmatrix}3 & -4 \\ 2 & 3\end{vmatrix}=(3)(3)-(-4)(2)=17 \end{aligned}
  4. Substitute these values and add. So |B|=-9.
    |B|=1(-26)-0(11)+1(17)=-9
  5. |B|\neq0, so B is non-singular and B^{-1} exists. Find all cofactors.
    \begin{aligned} A_{11}=(-1)^{2}\begin{vmatrix}3 & 5 \\ 0 & 1\end{vmatrix}=+\left[(3)(1)-(5)(0)\right]=3 \\ A_{12}=(-1)^{3}\begin{vmatrix}2 & 5 \\ 1 & 1\end{vmatrix}=-\left[(2)(1)-(5)(1)\right]=3 \\ A_{13}=(-1)^{4}\begin{vmatrix}2 & 3 \\ 1 & 0\end{vmatrix}=+\left[(2)(0)-(3)(1)\right]=-3 \\ A_{21}=(-1)^{3}\begin{vmatrix}-4 & 2 \\ 0 & 1\end{vmatrix}=-\left[(-4)(1)-(2)(0)\right]=4 \\ A_{22}=(-1)^{4}\begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix}=+\left[(3)(1)-(2)(1)\right]=1 \\ A_{23}=(-1)^{5}\begin{vmatrix}3 & -4 \\ 1 & 0\end{vmatrix}=-\left[(3)(0)-(-4)(1)\right]=-4 \\ A_{31}=(-1)^{4}\begin{vmatrix}-4 & 2 \\ 3 & 5\end{vmatrix}=+\left[(-4)(5)-(2)(3)\right]=-26 \\ A_{32}=(-1)^{5}\begin{vmatrix}3 & 2 \\ 2 & 5\end{vmatrix}=-\left[(3)(5)-(2)(2)\right]=-11 \\ A_{33}=(-1)^{6}\begin{vmatrix}3 & -4 \\ 2 & 3\end{vmatrix}=+\left[(3)(3)-(-4)(2)\right]=17 \end{aligned}
  6. The adjoint is the transpose of the cofactor matrix: the cofactor A_{ij} goes to row j, column i.
    \operatorname{adj}(B)=\begin{bmatrix}3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17\end{bmatrix}
  7. Divide by the determinant: B^{-1}=\dfrac{1}{|B|}\operatorname{adj}(B).
    B^{-1}=\frac{1}{-9}\begin{bmatrix}3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17\end{bmatrix}=\begin{bmatrix}-\frac{1}{3} & -\frac{4}{9} & \frac{26}{9} \\ -\frac{1}{3} & -\frac{1}{9} & \frac{11}{9} \\ \frac{1}{3} & \frac{4}{9} & -\frac{17}{9}\end{bmatrix}
  8. Check: BB^{-1} should be I.
    BB^{-1}=\begin{bmatrix}3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}\begin{bmatrix}-\frac{1}{3} & -\frac{4}{9} & \frac{26}{9} \\ -\frac{1}{3} & -\frac{1}{9} & \frac{11}{9} \\ \frac{1}{3} & \frac{4}{9} & -\frac{17}{9}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
B^{-1}=\begin{bmatrix}-\frac{1}{3} & -\frac{4}{9} & \frac{26}{9} \\ -\frac{1}{3} & -\frac{1}{9} & \frac{11}{9} \\ \frac{1}{3} & \frac{4}{9} & -\frac{17}{9}\end{bmatrix}
(iii) C=\begin{bmatrix}i & 0 & 1 \\ 2i & -1 & -i \\ 1 & 0 & 4i\end{bmatrix}
  1. Adjoint method: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A) exists only if |A|\neq0. First find |A|.
  2. Expand along column 2. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |C|=-0\begin{vmatrix}2i & -i \\ 1 & 4i\end{vmatrix}+(-1)\begin{vmatrix}i & 1 \\ 1 & 4i\end{vmatrix}-0\begin{vmatrix}i & 1 \\ 2i & -i\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}2i & -i \\ 1 & 4i\end{vmatrix}=(2i)(4i)-(-i)(1)=-8+i \\ \begin{vmatrix}i & 1 \\ 1 & 4i\end{vmatrix}=(i)(4i)-(1)(1)=-5 \\ \begin{vmatrix}i & 1 \\ 2i & -i\end{vmatrix}=(i)(-i)-(1)(2i)=1-2i \end{aligned}
  4. Substitute these values and add. So |C|=5.
    |C|=-0(-8+i)+(-1)(-5)-0(1-2i)=5
  5. |C|\neq0, so C is non-singular and C^{-1} exists. Find all cofactors.
    \begin{aligned} A_{11}=(-1)^{2}\begin{vmatrix}-1 & -i \\ 0 & 4i\end{vmatrix}=+\left[(-1)(4i)-(-i)(0)\right]=-4i \\ A_{12}=(-1)^{3}\begin{vmatrix}2i & -i \\ 1 & 4i\end{vmatrix}=-\left[(2i)(4i)-(-i)(1)\right]=8-i \\ A_{13}=(-1)^{4}\begin{vmatrix}2i & -1 \\ 1 & 0\end{vmatrix}=+\left[(2i)(0)-(-1)(1)\right]=1 \\ A_{21}=(-1)^{3}\begin{vmatrix}0 & 1 \\ 0 & 4i\end{vmatrix}=-\left[(0)(4i)-(1)(0)\right]=0 \\ A_{22}=(-1)^{4}\begin{vmatrix}i & 1 \\ 1 & 4i\end{vmatrix}=+\left[(i)(4i)-(1)(1)\right]=-5 \\ A_{23}=(-1)^{5}\begin{vmatrix}i & 0 \\ 1 & 0\end{vmatrix}=-\left[(i)(0)-(0)(1)\right]=0 \\ A_{31}=(-1)^{4}\begin{vmatrix}0 & 1 \\ -1 & -i\end{vmatrix}=+\left[(0)(-i)-(1)(-1)\right]=1 \\ A_{32}=(-1)^{5}\begin{vmatrix}i & 1 \\ 2i & -i\end{vmatrix}=-\left[(i)(-i)-(1)(2i)\right]=-1+2i \\ A_{33}=(-1)^{6}\begin{vmatrix}i & 0 \\ 2i & -1\end{vmatrix}=+\left[(i)(-1)-(0)(2i)\right]=-i \end{aligned}
  6. The adjoint is the transpose of the cofactor matrix: the cofactor A_{ij} goes to row j, column i.
    \operatorname{adj}(C)=\begin{bmatrix}-4i & 0 & 1 \\ 8-i & -5 & -1+2i \\ 1 & 0 & -i\end{bmatrix}
  7. Divide by the determinant: C^{-1}=\dfrac{1}{|C|}\operatorname{adj}(C).
    C^{-1}=\frac{1}{5}\begin{bmatrix}-4i & 0 & 1 \\ 8-i & -5 & -1+2i \\ 1 & 0 & -i\end{bmatrix}=\begin{bmatrix}-\frac{4}{5}i & 0 & \frac{1}{5} \\ \frac{8}{5}-\frac{1}{5}i & -1 & -\frac{1}{5}+\frac{2}{5}i \\ \frac{1}{5} & 0 & -\frac{1}{5}i\end{bmatrix}
  8. Check: CC^{-1} should be I.
    CC^{-1}=\begin{bmatrix}i & 0 & 1 \\ 2i & -1 & -i \\ 1 & 0 & 4i\end{bmatrix}\begin{bmatrix}-\frac{4}{5}i & 0 & \frac{1}{5} \\ \frac{8}{5}-\frac{1}{5}i & -1 & -\frac{1}{5}+\frac{2}{5}i \\ \frac{1}{5} & 0 & -\frac{1}{5}i\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
C^{-1}=\begin{bmatrix}-\frac{4}{5}i & 0 & \frac{1}{5} \\ \frac{8}{5}-\frac{1}{5}i & -1 & -\frac{1}{5}+\frac{2}{5}i \\ \frac{1}{5} & 0 & -\frac{1}{5}i\end{bmatrix}
(iv) D=\begin{bmatrix}3 & -i & i \\ 2 & 1 & -3i \\ 4i & 2 & 6\end{bmatrix}
  1. Adjoint method: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A) exists only if |A|\neq0. First find |A|.
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |D|=3\begin{vmatrix}1 & -3i \\ 2 & 6\end{vmatrix}-(-i)\begin{vmatrix}2 & -3i \\ 4i & 6\end{vmatrix}+i\begin{vmatrix}2 & 1 \\ 4i & 2\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}1 & -3i \\ 2 & 6\end{vmatrix}=(1)(6)-(-3i)(2)=6+6i \\ \begin{vmatrix}2 & -3i \\ 4i & 6\end{vmatrix}=(2)(6)-(-3i)(4i)=0 \\ \begin{vmatrix}2 & 1 \\ 4i & 2\end{vmatrix}=(2)(2)-(1)(4i)=4-4i \end{aligned}
  4. Substitute these values and add. So |D|=22+22i.
    |D|=3(6+6i)-(-i)(0)+i(4-4i)=22+22i
  5. |D|\neq0, so D is non-singular and D^{-1} exists. Find all cofactors.
    \begin{aligned} A_{11}=(-1)^{2}\begin{vmatrix}1 & -3i \\ 2 & 6\end{vmatrix}=+\left[(1)(6)-(-3i)(2)\right]=6+6i \\ A_{12}=(-1)^{3}\begin{vmatrix}2 & -3i \\ 4i & 6\end{vmatrix}=-\left[(2)(6)-(-3i)(4i)\right]=0 \\ A_{13}=(-1)^{4}\begin{vmatrix}2 & 1 \\ 4i & 2\end{vmatrix}=+\left[(2)(2)-(1)(4i)\right]=4-4i \\ A_{21}=(-1)^{3}\begin{vmatrix}-i & i \\ 2 & 6\end{vmatrix}=-\left[(-i)(6)-(i)(2)\right]=8i \\ A_{22}=(-1)^{4}\begin{vmatrix}3 & i \\ 4i & 6\end{vmatrix}=+\left[(3)(6)-(i)(4i)\right]=22 \\ A_{23}=(-1)^{5}\begin{vmatrix}3 & -i \\ 4i & 2\end{vmatrix}=-\left[(3)(2)-(-i)(4i)\right]=-2 \\ A_{31}=(-1)^{4}\begin{vmatrix}-i & i \\ 1 & -3i\end{vmatrix}=+\left[(-i)(-3i)-(i)(1)\right]=-3-i \\ A_{32}=(-1)^{5}\begin{vmatrix}3 & i \\ 2 & -3i\end{vmatrix}=-\left[(3)(-3i)-(i)(2)\right]=11i \\ A_{33}=(-1)^{6}\begin{vmatrix}3 & -i \\ 2 & 1\end{vmatrix}=+\left[(3)(1)-(-i)(2)\right]=3+2i \end{aligned}
  6. The adjoint is the transpose of the cofactor matrix: the cofactor A_{ij} goes to row j, column i.
    \operatorname{adj}(D)=\begin{bmatrix}6+6i & 8i & -3-i \\ 0 & 22 & 11i \\ 4-4i & -2 & 3+2i\end{bmatrix}
  7. Divide by the determinant: D^{-1}=\dfrac{1}{|D|}\operatorname{adj}(D).
    D^{-1}=\frac{1}{(22+22i)}\begin{bmatrix}6+6i & 8i & -3-i \\ 0 & 22 & 11i \\ 4-4i & -2 & 3+2i\end{bmatrix}=\begin{bmatrix}\frac{3}{11} & \frac{2}{11}+\frac{2}{11}i & -\frac{1}{11}+\frac{1}{22}i \\ 0 & \frac{1}{2}-\frac{1}{2}i & \frac{1}{4}+\frac{1}{4}i \\ -\frac{2}{11}i & -\frac{1}{22}+\frac{1}{22}i & \frac{5}{44}-\frac{1}{44}i\end{bmatrix}
  8. Check: DD^{-1} should be I.
    \begin{aligned}DD^{-1}&=\begin{bmatrix}3 & -i & i \\ 2 & 1 & -3i \\ 4i & 2 & 6\end{bmatrix}\\[6pt] &\quad\times\begin{bmatrix}\frac{3}{11} & \frac{2}{11}+\frac{2}{11}i & -\frac{1}{11}+\frac{1}{22}i \\ 0 & \frac{1}{2}-\frac{1}{2}i & \frac{1}{4}+\frac{1}{4}i \\ -\frac{2}{11}i & -\frac{1}{22}+\frac{1}{22}i & \frac{5}{44}-\frac{1}{44}i\end{bmatrix}\\[6pt] &=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\end{aligned}
D^{-1}=\begin{bmatrix}\frac{3}{11} & \frac{2}{11}+\frac{2}{11}i & -\frac{1}{11}+\frac{1}{22}i \\ 0 & \frac{1}{2}-\frac{1}{2}i & \frac{1}{4}+\frac{1}{4}i \\ -\frac{2}{11}i & -\frac{1}{22}+\frac{1}{22}i & \frac{5}{44}-\frac{1}{44}i\end{bmatrix}

Book answer. (i) [1/3, 1/3, 0; -1/9, 2/9, -1/3; 5/9, -1/9, -1/3]; (ii) [-1/3, -4/9, 26/9; -1/3, -1/9, 11/9; 1/3, 4/9, -17/9]; (iii) [-4i/5, 0, 1/5; (8-i)/5, -1, (-1+2i)/5; 1/5, 0, -i/5]; (iv) [3/11, (2+2i)/11, (-2+i)/22; 0, (1-i)/2, (1+i)/4; -2i/11, (-1+i)/22, (5-i)/44]. All agree.

Note. The inverse exists in all four parts, since every determinant is non-zero: -9,\ -9,\ 5,\ 22+22i.

6
If A=\begin{bmatrix}2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6\end{bmatrix} then find A^{-1} and hence show that AA^{-1}=A^{-1}A=I_3.
Solution
  1. Use the adjoint method.
  2. Expand along row 2. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=-0\begin{vmatrix}1 & -3 \\ 1 & 6\end{vmatrix}+1\begin{vmatrix}2 & -3 \\ 2 & 6\end{vmatrix}-0\begin{vmatrix}2 & 1 \\ 2 & 1\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}1 & -3 \\ 1 & 6\end{vmatrix}=(1)(6)-(-3)(1)=9 \\ \begin{vmatrix}2 & -3 \\ 2 & 6\end{vmatrix}=(2)(6)-(-3)(2)=18 \\ \begin{vmatrix}2 & 1 \\ 2 & 1\end{vmatrix}=(2)(1)-(1)(2)=0 \end{aligned}
  4. Substitute these values and add. So |A|=18.
    |A|=-0(9)+1(18)-0(0)=18
  5. |A|\neq0, so A is non-singular and A^{-1} exists. Find all cofactors.
    \begin{aligned} A_{11}=(-1)^{2}\begin{vmatrix}1 & 0 \\ 1 & 6\end{vmatrix}=+\left[(1)(6)-(0)(1)\right]=6 \\ A_{12}=(-1)^{3}\begin{vmatrix}0 & 0 \\ 2 & 6\end{vmatrix}=-\left[(0)(6)-(0)(2)\right]=0 \\ A_{13}=(-1)^{4}\begin{vmatrix}0 & 1 \\ 2 & 1\end{vmatrix}=+\left[(0)(1)-(1)(2)\right]=-2 \\ A_{21}=(-1)^{3}\begin{vmatrix}1 & -3 \\ 1 & 6\end{vmatrix}=-\left[(1)(6)-(-3)(1)\right]=-9 \\ A_{22}=(-1)^{4}\begin{vmatrix}2 & -3 \\ 2 & 6\end{vmatrix}=+\left[(2)(6)-(-3)(2)\right]=18 \\ A_{23}=(-1)^{5}\begin{vmatrix}2 & 1 \\ 2 & 1\end{vmatrix}=-\left[(2)(1)-(1)(2)\right]=0 \\ A_{31}=(-1)^{4}\begin{vmatrix}1 & -3 \\ 1 & 0\end{vmatrix}=+\left[(1)(0)-(-3)(1)\right]=3 \\ A_{32}=(-1)^{5}\begin{vmatrix}2 & -3 \\ 0 & 0\end{vmatrix}=-\left[(2)(0)-(-3)(0)\right]=0 \\ A_{33}=(-1)^{6}\begin{vmatrix}2 & 1 \\ 0 & 1\end{vmatrix}=+\left[(2)(1)-(1)(0)\right]=2 \end{aligned}
  6. The adjoint is the transpose of the cofactor matrix: the cofactor A_{ij} goes to row j, column i.
    \operatorname{adj}(A)=\begin{bmatrix}6 & -9 & 3 \\ 0 & 18 & 0 \\ -2 & 0 & 2\end{bmatrix}
  7. Divide by the determinant: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A).
    A^{-1}=\frac{1}{18}\begin{bmatrix}6 & -9 & 3 \\ 0 & 18 & 0 \\ -2 & 0 & 2\end{bmatrix}=\begin{bmatrix}\frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\ 0 & 1 & 0 \\ -\frac{1}{9} & 0 & \frac{1}{9}\end{bmatrix}
  8. **Show AA^{-1}=I_3.** Multiply row by column.
    AA^{-1}=\begin{bmatrix}2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6\end{bmatrix}\begin{bmatrix}\frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\ 0 & 1 & 0 \\ -\frac{1}{9} & 0 & \frac{1}{9}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=I_3
  9. **Show A^{-1}A=I_3.** Multiply in the other order.
    A^{-1}A=\begin{bmatrix}\frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\ 0 & 1 & 0 \\ -\frac{1}{9} & 0 & \frac{1}{9}\end{bmatrix}\begin{bmatrix}2 & 1 & -3 \\ 0 & 1 & 0 \\ 2 & 1 & 6\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=I_3
A^{-1}=\begin{bmatrix}\frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\ 0 & 1 & 0 \\ -\frac{1}{9} & 0 & \frac{1}{9}\end{bmatrix},\quad AA^{-1}=A^{-1}A=I_3

Book answer. A^{-1} = [1/3, -1/2, 1/6; 0, 1, 0; -1/9, 0, 1/9]. Agrees.

7
Verify that (AB)^{-1}=B^{-1}A^{-1} in each of the following.
Solution

Method (Example 13). Compute the left side (AB)^{-1} and the right side B^{-1}A^{-1} separately and compare. Note the order: B^{-1} comes first.

(i) A=\begin{bmatrix}2 & 1 \\ 8 & 6\end{bmatrix} and B=\begin{bmatrix}3 & 2 \\ 0 & 2\end{bmatrix}
  1. Left side. Find AB (row by column).
    AB=\begin{bmatrix}2 & 1 \\ 8 & 6\end{bmatrix}\begin{bmatrix}3 & 2 \\ 0 & 2\end{bmatrix}=\begin{bmatrix}6 & 6 \\ 24 & 28\end{bmatrix}
  2. Find |AB| and \operatorname{adj}(AB) (for a 2\times2 matrix swap the diagonal elements and change the signs of the others).
    |AB|=(6)(28)-(6)(24)=24,\qquad \operatorname{adj}(AB)=\begin{bmatrix}28 & -6 \\ -24 & 6\end{bmatrix}
  3. Apply (AB)^{-1}=\dfrac{1}{|AB|}\operatorname{adj}(AB).
    (AB)^{-1}=\frac{1}{24}\begin{bmatrix}28 & -6 \\ -24 & 6\end{bmatrix}=\begin{bmatrix}\frac{7}{6} & -\frac{1}{4} \\ -1 & \frac{1}{4}\end{bmatrix}
  4. Right side. Find A^{-1} and B^{-1} the same way.
    |A|=4,\ A^{-1}=\begin{bmatrix}\frac{3}{2} & -\frac{1}{4} \\ -2 & \frac{1}{2}\end{bmatrix};\qquad |B|=6,\ B^{-1}=\begin{bmatrix}\frac{1}{3} & -\frac{1}{3} \\ 0 & \frac{1}{2}\end{bmatrix}
  5. Multiply in the reversed order, B^{-1}A^{-1}.
    B^{-1}A^{-1}=\begin{bmatrix}\frac{1}{3} & -\frac{1}{3} \\ 0 & \frac{1}{2}\end{bmatrix}\begin{bmatrix}\frac{3}{2} & -\frac{1}{4} \\ -2 & \frac{1}{2}\end{bmatrix}=\begin{bmatrix}\frac{7}{6} & -\frac{1}{4} \\ -1 & \frac{1}{4}\end{bmatrix}
  6. The two results are the same matrix, so (AB)^{-1}=B^{-1}A^{-1} (the reversal law of inverse).
(AB)^{-1}=B^{-1}A^{-1}=\begin{bmatrix}\frac{7}{6} & -\frac{1}{4} \\ -1 & \frac{1}{4}\end{bmatrix}
(ii) A=\begin{bmatrix}1 & 1 & 1 \\ 2 & -1 & 1 \\ 2 & 1 & -3\end{bmatrix} and B=\begin{bmatrix}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{bmatrix}
  1. Left side. Find AB (row by column).
    AB=\begin{bmatrix}1 & 1 & 1 \\ 2 & -1 & 1 \\ 2 & 1 & -3\end{bmatrix}\begin{bmatrix}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{bmatrix}=\begin{bmatrix}9 & -4 & 4 \\ 8 & -8 & 9 \\ -4 & 6 & -1\end{bmatrix}
  2. Find |AB| by cofactors and all cofactors of AB, then \operatorname{adj}(AB).
    |AB|=-238
  3. The cofactors of AB.
    \begin{aligned} C_{11}=(-1)^{2}\begin{vmatrix}-8 & 9 \\ 6 & -1\end{vmatrix}=+\left[(-8)(-1)-(9)(6)\right]=-46 \\ C_{12}=(-1)^{3}\begin{vmatrix}8 & 9 \\ -4 & -1\end{vmatrix}=-\left[(8)(-1)-(9)(-4)\right]=-28 \\ C_{13}=(-1)^{4}\begin{vmatrix}8 & -8 \\ -4 & 6\end{vmatrix}=+\left[(8)(6)-(-8)(-4)\right]=16 \\ C_{21}=(-1)^{3}\begin{vmatrix}-4 & 4 \\ 6 & -1\end{vmatrix}=-\left[(-4)(-1)-(4)(6)\right]=20 \\ C_{22}=(-1)^{4}\begin{vmatrix}9 & 4 \\ -4 & -1\end{vmatrix}=+\left[(9)(-1)-(4)(-4)\right]=7 \\ C_{23}=(-1)^{5}\begin{vmatrix}9 & -4 \\ -4 & 6\end{vmatrix}=-\left[(9)(6)-(-4)(-4)\right]=-38 \\ C_{31}=(-1)^{4}\begin{vmatrix}-4 & 4 \\ -8 & 9\end{vmatrix}=+\left[(-4)(9)-(4)(-8)\right]=-4 \\ C_{32}=(-1)^{5}\begin{vmatrix}9 & 4 \\ 8 & 9\end{vmatrix}=-\left[(9)(9)-(4)(8)\right]=-49 \\ C_{33}=(-1)^{6}\begin{vmatrix}9 & -4 \\ 8 & -8\end{vmatrix}=+\left[(9)(-8)-(-4)(8)\right]=-40 \end{aligned}
  4. Transpose the cofactor matrix to get the adjoint, then divide by |AB|.
    (AB)^{-1}=\frac{1}{-238}\begin{bmatrix}-46 & 20 & -4 \\ -28 & 7 & -49 \\ 16 & -38 & -40\end{bmatrix}=\begin{bmatrix}\frac{23}{119} & -\frac{10}{119} & \frac{2}{119} \\ \frac{2}{17} & -\frac{1}{34} & \frac{7}{34} \\ -\frac{8}{119} & \frac{19}{119} & \frac{20}{119}\end{bmatrix}
  5. Right side. Find A^{-1} by the adjoint method: |A|, then \operatorname{adj}(A).
    |A|=14,\qquad \operatorname{adj}(A)=\begin{bmatrix}2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3\end{bmatrix},\qquad A^{-1}=\begin{bmatrix}\frac{1}{7} & \frac{2}{7} & \frac{1}{7} \\ \frac{4}{7} & -\frac{5}{14} & \frac{1}{14} \\ \frac{2}{7} & \frac{1}{14} & -\frac{3}{14}\end{bmatrix}
  6. Do the same for B.
    |B|=-17,\qquad \operatorname{adj}(B)=\begin{bmatrix}-1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7\end{bmatrix},\qquad B^{-1}=\begin{bmatrix}\frac{1}{17} & \frac{5}{17} & \frac{1}{17} \\ \frac{8}{17} & \frac{6}{17} & -\frac{9}{17} \\ \frac{10}{17} & -\frac{1}{17} & -\frac{7}{17}\end{bmatrix}
  7. Multiply in the reversed order, B^{-1}A^{-1}.
    B^{-1}A^{-1}=\begin{bmatrix}\frac{1}{17} & \frac{5}{17} & \frac{1}{17} \\ \frac{8}{17} & \frac{6}{17} & -\frac{9}{17} \\ \frac{10}{17} & -\frac{1}{17} & -\frac{7}{17}\end{bmatrix}\begin{bmatrix}\frac{1}{7} & \frac{2}{7} & \frac{1}{7} \\ \frac{4}{7} & -\frac{5}{14} & \frac{1}{14} \\ \frac{2}{7} & \frac{1}{14} & -\frac{3}{14}\end{bmatrix}=\begin{bmatrix}\frac{23}{119} & -\frac{10}{119} & \frac{2}{119} \\ \frac{2}{17} & -\frac{1}{34} & \frac{7}{34} \\ -\frac{8}{119} & \frac{19}{119} & \frac{20}{119}\end{bmatrix}
  8. The two results are the same matrix, so (AB)^{-1}=B^{-1}A^{-1} (the reversal law of inverse).
(AB)^{-1}=B^{-1}A^{-1}=\begin{bmatrix}\frac{23}{119} & -\frac{10}{119} & \frac{2}{119} \\ \frac{2}{17} & -\frac{1}{34} & \frac{7}{34} \\ -\frac{8}{119} & \frac{19}{119} & \frac{20}{119}\end{bmatrix}
(iii) A=\begin{bmatrix}2 & -1 & 6 \\ 1 & 2 & 1 \\ -i & 1 & 6\end{bmatrix} and B=\begin{bmatrix}3 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & 1 & 1\end{bmatrix}
  1. Left side. Find AB (row by column).
    AB=\begin{bmatrix}2 & -1 & 6 \\ 1 & 2 & 1 \\ -i & 1 & 6\end{bmatrix}\begin{bmatrix}3 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & 1 & 1\end{bmatrix}=\begin{bmatrix}5 & 8 & 9 \\ 5 & 2 & 5 \\ 1-3i & 6-i & 7-2i\end{bmatrix}
  2. Find |AB| by cofactors and all cofactors of AB, then \operatorname{adj}(AB).
    |AB|=-68-26i
  3. The cofactors of AB.
    \begin{aligned} C_{11}=(-1)^{2}\begin{vmatrix}2 & 5 \\ 6-i & 7-2i\end{vmatrix}=+\left[(2)(7-2i)-(5)(6-i)\right]=-16+i \\ C_{12}=(-1)^{3}\begin{vmatrix}5 & 5 \\ 1-3i & 7-2i\end{vmatrix}=-\left[(5)(7-2i)-(5)(1-3i)\right]=-30-5i \\ C_{13}=(-1)^{4}\begin{vmatrix}5 & 2 \\ 1-3i & 6-i\end{vmatrix}=+\left[(5)(6-i)-(2)(1-3i)\right]=28+i \\ C_{21}=(-1)^{3}\begin{vmatrix}8 & 9 \\ 6-i & 7-2i\end{vmatrix}=-\left[(8)(7-2i)-(9)(6-i)\right]=-2+7i \\ C_{22}=(-1)^{4}\begin{vmatrix}5 & 9 \\ 1-3i & 7-2i\end{vmatrix}=+\left[(5)(7-2i)-(9)(1-3i)\right]=26+17i \\ C_{23}=(-1)^{5}\begin{vmatrix}5 & 8 \\ 1-3i & 6-i\end{vmatrix}=-\left[(5)(6-i)-(8)(1-3i)\right]=-22-19i \\ C_{31}=(-1)^{4}\begin{vmatrix}8 & 9 \\ 2 & 5\end{vmatrix}=+\left[(8)(5)-(9)(2)\right]=22 \\ C_{32}=(-1)^{5}\begin{vmatrix}5 & 9 \\ 5 & 5\end{vmatrix}=-\left[(5)(5)-(9)(5)\right]=20 \\ C_{33}=(-1)^{6}\begin{vmatrix}5 & 8 \\ 5 & 2\end{vmatrix}=+\left[(5)(2)-(8)(5)\right]=-30 \end{aligned}
  4. Transpose the cofactor matrix to get the adjoint, then divide by |AB|.
    \begin{aligned}(AB)^{-1}&=\frac{1}{(-68-26i)}\\[6pt] &\quad\times\begin{bmatrix}-16+i & -2+7i & 22 \\ -30-5i & 26+17i & 20 \\ 28+i & -22-19i & -30\end{bmatrix}\\[4pt] \big[(AB)^{-1}\big]_{11}&=\frac{531}{2650}-\frac{121}{1325}i\\[4pt] \big[(AB)^{-1}\big]_{12}&=-\frac{23}{2650}-\frac{132}{1325}i\\[4pt] \big[(AB)^{-1}\big]_{13}&=-\frac{374}{1325}+\frac{143}{1325}i\\[4pt] \big[(AB)^{-1}\big]_{21}&=\frac{217}{530}-\frac{22}{265}i\\[4pt] \big[(AB)^{-1}\big]_{22}&=-\frac{221}{530}-\frac{24}{265}i\\[4pt] \big[(AB)^{-1}\big]_{23}&=-\frac{68}{265}+\frac{26}{265}i\\[4pt] \big[(AB)^{-1}\big]_{31}&=-\frac{193}{530}+\frac{33}{265}i\\[4pt] \big[(AB)^{-1}\big]_{32}&=\frac{199}{530}+\frac{36}{265}i\\[4pt] \big[(AB)^{-1}\big]_{33}&=\frac{102}{265}-\frac{39}{265}i\end{aligned}
  5. Right side. Find A^{-1} by the adjoint method: |A|, then \operatorname{adj}(A).
    \begin{aligned}|A|&=34+13i\\[6pt] \operatorname{adj}(A)&=\begin{bmatrix}11 & 12 & -13 \\ -6-i & 12+6i & 4 \\ 1+2i & -2+i & 5\end{bmatrix}\\[4pt] (A^{-1})_{11}&=\frac{374}{1325}-\frac{143}{1325}i\\[4pt] (A^{-1})_{12}&=\frac{408}{1325}-\frac{156}{1325}i\\[4pt] (A^{-1})_{13}&=-\frac{442}{1325}+\frac{169}{1325}i\\[4pt] (A^{-1})_{21}&=-\frac{217}{1325}+\frac{44}{1325}i\\[4pt] (A^{-1})_{22}&=\frac{486}{1325}+\frac{48}{1325}i\\[4pt] (A^{-1})_{23}&=\frac{136}{1325}-\frac{52}{1325}i\\[4pt] (A^{-1})_{31}&=\frac{12}{265}+\frac{11}{265}i\\[4pt] (A^{-1})_{32}&=-\frac{11}{265}+\frac{12}{265}i\\[4pt] (A^{-1})_{33}&=\frac{34}{265}-\frac{13}{265}i\end{aligned}
  6. Do the same for B.
    |B|=-2,\qquad \operatorname{adj}(B)=\begin{bmatrix}-1 & 1 & 1 \\ -1 & 3 & -1 \\ 1 & -3 & -1\end{bmatrix},\qquad B^{-1}=\begin{bmatrix}\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{3}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{3}{2} & \frac{1}{2}\end{bmatrix}
  7. Multiply in the reversed order, B^{-1}A^{-1}.
    \begin{aligned}B^{-1}A^{-1}&=\begin{bmatrix}\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{3}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{3}{2} & \frac{1}{2}\end{bmatrix}A^{-1}\\[4pt] (B^{-1}A^{-1})_{11}&=\frac{531}{2650}-\frac{121}{1325}i\\[4pt] (B^{-1}A^{-1})_{12}&=-\frac{23}{2650}-\frac{132}{1325}i\\[4pt] (B^{-1}A^{-1})_{13}&=-\frac{374}{1325}+\frac{143}{1325}i\\[4pt] (B^{-1}A^{-1})_{21}&=\frac{217}{530}-\frac{22}{265}i\\[4pt] (B^{-1}A^{-1})_{22}&=-\frac{221}{530}-\frac{24}{265}i\\[4pt] (B^{-1}A^{-1})_{23}&=-\frac{68}{265}+\frac{26}{265}i\\[4pt] (B^{-1}A^{-1})_{31}&=-\frac{193}{530}+\frac{33}{265}i\\[4pt] (B^{-1}A^{-1})_{32}&=\frac{199}{530}+\frac{36}{265}i\\[4pt] (B^{-1}A^{-1})_{33}&=\frac{102}{265}-\frac{39}{265}i\end{aligned}
  8. The two results are the same matrix, so (AB)^{-1}=B^{-1}A^{-1} (the reversal law of inverse).
\begin{aligned}(AB)^{-1}&=B^{-1}A^{-1}\\[4pt] \big[(AB)^{-1}\big]_{11}&=\frac{531}{2650}-\frac{121}{1325}i\\[4pt] \big[(AB)^{-1}\big]_{12}&=-\frac{23}{2650}-\frac{132}{1325}i\\[4pt] \big[(AB)^{-1}\big]_{13}&=-\frac{374}{1325}+\frac{143}{1325}i\\[4pt] \big[(AB)^{-1}\big]_{21}&=\frac{217}{530}-\frac{22}{265}i\\[4pt] \big[(AB)^{-1}\big]_{22}&=-\frac{221}{530}-\frac{24}{265}i\\[4pt] \big[(AB)^{-1}\big]_{23}&=-\frac{68}{265}+\frac{26}{265}i\\[4pt] \big[(AB)^{-1}\big]_{31}&=-\frac{193}{530}+\frac{33}{265}i\\[4pt] \big[(AB)^{-1}\big]_{32}&=\frac{199}{530}+\frac{36}{265}i\\[4pt] \big[(AB)^{-1}\big]_{33}&=\frac{102}{265}-\frac{39}{265}i\end{aligned}
(iv) A=\begin{bmatrix}1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1\end{bmatrix} and B=\begin{bmatrix}2 & 3 & 4 \\ 1 & 0 & 2 \\ 0 & 1 & 3\end{bmatrix}
  1. Left side. Find AB (row by column).
    AB=\begin{bmatrix}1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1\end{bmatrix}\begin{bmatrix}2 & 3 & 4 \\ 1 & 0 & 2 \\ 0 & 1 & 3\end{bmatrix}=\begin{bmatrix}4 & 8 & 23 \\ 1 & 2 & -1 \\ 7 & 5 & 11\end{bmatrix}
  2. Find |AB| by cofactors and all cofactors of AB, then \operatorname{adj}(AB).
    |AB|=-243
  3. The cofactors of AB.
    \begin{aligned} C_{11}=(-1)^{2}\begin{vmatrix}2 & -1 \\ 5 & 11\end{vmatrix}=+\left[(2)(11)-(-1)(5)\right]=27 \\ C_{12}=(-1)^{3}\begin{vmatrix}1 & -1 \\ 7 & 11\end{vmatrix}=-\left[(1)(11)-(-1)(7)\right]=-18 \\ C_{13}=(-1)^{4}\begin{vmatrix}1 & 2 \\ 7 & 5\end{vmatrix}=+\left[(1)(5)-(2)(7)\right]=-9 \\ C_{21}=(-1)^{3}\begin{vmatrix}8 & 23 \\ 5 & 11\end{vmatrix}=-\left[(8)(11)-(23)(5)\right]=27 \\ C_{22}=(-1)^{4}\begin{vmatrix}4 & 23 \\ 7 & 11\end{vmatrix}=+\left[(4)(11)-(23)(7)\right]=-117 \\ C_{23}=(-1)^{5}\begin{vmatrix}4 & 8 \\ 7 & 5\end{vmatrix}=-\left[(4)(5)-(8)(7)\right]=36 \\ C_{31}=(-1)^{4}\begin{vmatrix}8 & 23 \\ 2 & -1\end{vmatrix}=+\left[(8)(-1)-(23)(2)\right]=-54 \\ C_{32}=(-1)^{5}\begin{vmatrix}4 & 23 \\ 1 & -1\end{vmatrix}=-\left[(4)(-1)-(23)(1)\right]=27 \\ C_{33}=(-1)^{6}\begin{vmatrix}4 & 8 \\ 1 & 2\end{vmatrix}=+\left[(4)(2)-(8)(1)\right]=0 \end{aligned}
  4. Transpose the cofactor matrix to get the adjoint, then divide by |AB|.
    (AB)^{-1}=\frac{1}{-243}\begin{bmatrix}27 & 27 & -54 \\ -18 & -117 & 27 \\ -9 & 36 & 0\end{bmatrix}=\begin{bmatrix}-\frac{1}{9} & -\frac{1}{9} & \frac{2}{9} \\ \frac{2}{27} & \frac{13}{27} & -\frac{1}{9} \\ \frac{1}{27} & -\frac{4}{27} & 0\end{bmatrix}
  5. Right side. Find A^{-1} by the adjoint method: |A|, then \operatorname{adj}(A).
    |A|=27,\qquad \operatorname{adj}(A)=\begin{bmatrix}4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3\end{bmatrix},\qquad A^{-1}=\begin{bmatrix}\frac{4}{27} & \frac{17}{27} & \frac{1}{9} \\ -\frac{1}{27} & -\frac{11}{27} & \frac{2}{9} \\ \frac{5}{27} & \frac{1}{27} & -\frac{1}{9}\end{bmatrix}
  6. Do the same for B.
    |B|=-9,\qquad \operatorname{adj}(B)=\begin{bmatrix}-2 & -5 & 6 \\ -3 & 6 & 0 \\ 1 & -2 & -3\end{bmatrix},\qquad B^{-1}=\begin{bmatrix}\frac{2}{9} & \frac{5}{9} & -\frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & 0 \\ -\frac{1}{9} & \frac{2}{9} & \frac{1}{3}\end{bmatrix}
  7. Multiply in the reversed order, B^{-1}A^{-1}.
    B^{-1}A^{-1}=\begin{bmatrix}\frac{2}{9} & \frac{5}{9} & -\frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & 0 \\ -\frac{1}{9} & \frac{2}{9} & \frac{1}{3}\end{bmatrix}\begin{bmatrix}\frac{4}{27} & \frac{17}{27} & \frac{1}{9} \\ -\frac{1}{27} & -\frac{11}{27} & \frac{2}{9} \\ \frac{5}{27} & \frac{1}{27} & -\frac{1}{9}\end{bmatrix}=\begin{bmatrix}-\frac{1}{9} & -\frac{1}{9} & \frac{2}{9} \\ \frac{2}{27} & \frac{13}{27} & -\frac{1}{9} \\ \frac{1}{27} & -\frac{4}{27} & 0\end{bmatrix}
  8. The two results are the same matrix, so (AB)^{-1}=B^{-1}A^{-1} (the reversal law of inverse).
(AB)^{-1}=B^{-1}A^{-1}=\begin{bmatrix}-\frac{1}{9} & -\frac{1}{9} & \frac{2}{9} \\ \frac{2}{27} & \frac{13}{27} & -\frac{1}{9} \\ \frac{1}{27} & -\frac{4}{27} & 0\end{bmatrix}

Note. The answer key prints no values for this question since it is a verification. In part (iii) |AB|=-68-26i and the matrices involve i.

Exercise 2.4p. 55–56

Properties of determinants. A zero row or column, two equal or proportional lines, or a line made of sums can be recognised without expanding; row and column operations R_i\to R_i+kR_j leave the value unchanged.

1
Without expansion show that:
Solution

Method. Use the properties of determinants (section 2.4): two identical (or proportional) rows or columns, or a row or column of zeros, give |A|=0 (Properties 3, 4 and 6). Row and column operations R_i\to R_i+kR_j or C_i\to C_i+kC_j do not change the value (Property 7).

(i) \begin{vmatrix}9&27&36\\18&54&24\\27&81&28\end{vmatrix}=0
  1. Compare C_2 with C_1: in every row the second entry is 3 times the first (27=3\cdot9, 54=3\cdot18, 81=3\cdot27).
  2. Apply C_2\to C_2-3C_1. Adding a multiple of one column to another does not change the determinant (Property 7).
    \begin{vmatrix}9&27&36\\18&54&24\\27&81&28\end{vmatrix}=\begin{vmatrix}9&0&36\\18&0&24\\27&0&28\end{vmatrix}
  3. A column of zeros makes the determinant 0 (Property 6). Hence the left side is 0.
    =0
0
(ii) \begin{vmatrix}1/a&bc&b+c\\1/b&ac&a+c\\1/c&ab&a+b\end{vmatrix}=0
  1. Call the determinant \Delta. Multiply R_1 by a, R_2 by b and R_3 by c. Multiplying a row by k multiplies the determinant by k (Property 4), so the result is abc\,\Delta.
    abc\,\Delta=\begin{vmatrix}1&abc&ab+ac\\1&abc&ab+bc\\1&abc&ac+bc\end{vmatrix}
  2. Now C_2=abc\cdot C_1 (every entry of C_1 is 1). Apply C_2\to C_2-abc\,C_1 (Property 7) to make C_2 a column of zeros.
    abc\,\Delta=\begin{vmatrix}1&0&ab+ac\\1&0&ab+bc\\1&0&ac+bc\end{vmatrix}=0
  3. Since 1/a,1/b,1/c are defined, abc\neq0, so \Delta=0.
    \Delta=0
0
(iii) \begin{vmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{vmatrix}=0
  1. Call the determinant \Delta. Its matrix is skew-symmetric of order 3: the diagonal is 0 and a_{ji}=-a_{ij}. A determinant equals the determinant of its transpose (Property 1).
    \Delta=\begin{vmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{vmatrix}
  2. Take the common factor -1 out of each of the three rows (Property 4, three times).
    \Delta=(-1)^3\begin{vmatrix}0&-a&-b\\a&0&-c\\b&c&0\end{vmatrix}
  3. The determinant on the right is \Delta again, so \Delta=-\Delta.
    \Delta=-\Delta\;\Rightarrow\;2\Delta=0\;\Rightarrow\;\Delta=0
0
(iv) \begin{vmatrix}\sin^2\alpha&1&\cos^2\alpha\\\tan^2\alpha&\sec^2\alpha&1\\-\operatorname{cosec}^2\alpha&-\cot^2\alpha&1\end{vmatrix}=0
  1. Look for a relation between the columns. Add C_1 and C_3 in each row and compare with C_2.
    C_1+C_3=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha\\\tan^2\alpha+1\\1-\operatorname{cosec}^2\alpha\end{bmatrix}
  2. Use \sin^2\alpha+\cos^2\alpha=1, 1+\tan^2\alpha=\sec^2\alpha and \operatorname{cosec}^2\alpha-1=\cot^2\alpha.
    C_1+C_3=\begin{bmatrix}1\\\sec^2\alpha\\-\cot^2\alpha\end{bmatrix}=C_2
  3. Apply C_2\to C_2-C_1-C_3 (Property 7, twice). The new second column is zero.
    \begin{vmatrix}\sin^2\alpha&0&\cos^2\alpha\\\tan^2\alpha&0&1\\-\operatorname{cosec}^2\alpha&0&1\end{vmatrix}=0 \quad\text{(Property 6)}
0
(v) \begin{vmatrix}(a-b)^3&a^3-b^3&ab(a-b)\\(c-d)^3&c^3-d^3&cd(c-d)\\(e-f)^3&e^3-f^3&ef(e-f)\end{vmatrix}=0
  1. Every entry of R_1 has the factor (a-b), because a^3-b^3=(a-b)(a^2+ab+b^2). Similarly R_2 has (c-d) and R_3 has (e-f). Take these out of the rows (Property 4).
    =(a-b)(c-d)(e-f)\begin{vmatrix}(a-b)^2&a^2+ab+b^2&ab\\(c-d)^2&c^2+cd+d^2&cd\\(e-f)^2&e^2+ef+f^2&ef\end{vmatrix}
  2. Compare the entries of the three columns in the first row: (a-b)^2=a^2-2ab+b^2=(a^2+ab+b^2)-3ab. The same holds in the other rows, so C_1=C_2-3C_3.
    (a-b)^2-(a^2+ab+b^2)+3ab=0
  3. Apply C_1\to C_1-C_2+3C_3 (Property 7). The first column becomes zero, so the determinant is 0 (Property 6).
    \begin{aligned}&=(a-b)(c-d)(e-f)\\ &\quad\times\begin{vmatrix}0&a^2+ab+b^2&ab\\0&c^2+cd+d^2&cd\\0&e^2+ef+f^2&ef\end{vmatrix}\\ &=0\end{aligned}
0
(vi) \begin{vmatrix}x&-z&0\\0&y&-x\\-y&0&z\end{vmatrix}=0
  1. Call the determinant \Delta and look for a relation between the columns: zC_1+xC_2+yC_3=0 because the three entries of each row cancel. For example in the first row zx+x(-z)+y(0)=0.
    zC_1+xC_2+yC_3=\begin{bmatrix}zx-xz+0\\0+xy-xy\\-yz+0+yz\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
  2. Multiply C_1 by z, C_2 by x and C_3 by y (Property 4). The determinant becomes xyz\,\Delta.
    xyz\,\Delta=\begin{vmatrix}xz&-xz&0\\0&xy&-xy\\-yz&0&yz\end{vmatrix}
  3. Apply C_1\to C_1+C_2+C_3 (Property 7). The first column becomes zero.
    xyz\,\Delta=\begin{vmatrix}0&-xz&0\\0&xy&-xy\\0&0&yz\end{vmatrix}=0
  4. So xyz\,\Delta=0, hence \Delta=0 when xyz\neq0.
    \Delta=0
  5. If one of x,y,z is 0 the result still holds: expanding along R_1 gives x(yz-0)+z(0\cdot z-(-x)(-y))=xyz-xyz.
    \Delta=xyz-xyz=0
0
(vii) \begin{vmatrix}(a-b)^2&(a+b)^2&ab\\(c-d)^2&(c+d)^2&cd\\(e-f)^2&(e+f)^2&ef\end{vmatrix}=0
  1. Subtract C_1 from C_2 in the first row: (a+b)^2-(a-b)^2=4ab, which is 4 times the entry of C_3. The same holds in rows 2 and 3 (4cd and 4ef).
    (a+b)^2-(a-b)^2=4ab
  2. Apply C_2\to C_2-C_1-4C_3 (Property 7). The second column becomes zero.
    =\begin{vmatrix}(a-b)^2&0&ab\\(c-d)^2&0&cd\\(e-f)^2&0&ef\end{vmatrix}
  3. A column of zeros gives determinant 0 (Property 6).
    =0
0

Book answer. No printed answer (proofs).

2
Using the properties of the determinants prove the following.
Solution

Method. Simplify the left side with row and column operations (Property 7), by taking out common factors (Property 4) and by splitting sums (Property 5), until the right side appears. A row operation is written R_i\to R_i+kR_j and a column operation C_i\to C_i+kC_j.

(i) \begin{vmatrix}x&y&x+y\\y&x+y&x\\x+y&x&y\end{vmatrix}=-2(x^3+y^3)
  1. Let the left side be \Delta. Apply R_1\to R_1+R_2+R_3 (Property 7). Every entry of the new first row is 2x+2y.
    \Delta=\begin{vmatrix}2(x+y)&2(x+y)&2(x+y)\\y&x+y&x\\x+y&x&y\end{vmatrix}
  2. Take the common factor 2(x+y) out of R_1 (Property 4).
    \Delta=2(x+y)\begin{vmatrix}1&1&1\\y&x+y&x\\x+y&x&y\end{vmatrix}
  3. Apply C_2\to C_2-C_1 and C_3\to C_3-C_1 to make zeros in the first row.
    \Delta=2(x+y)\begin{vmatrix}1&0&0\\y&x&x-y\\x+y&-y&-x\end{vmatrix}
  4. Expand along R_1: only the element 1 remains, with the 2\times2 minor obtained by deleting row 1 and column 1.
    \Delta=2(x+y)\left[x(-x)-(x-y)(-y)\right]=2(x+y)(-x^2+xy-y^2)
  5. Factorise using x^3+y^3=(x+y)(x^2-xy+y^2).
    \Delta=-2(x+y)(x^2-xy+y^2)=-2(x^3+y^3)
\Delta=-2(x^3+y^3)
(ii) \begin{vmatrix}a&b-c&b+c\\a+c&b&c-a\\a-b&a+b&c\end{vmatrix}=(a+b+c)(a^2+b^2+c^2)
  1. Let the left side be \Delta. Expand along R_1.
    \Delta=a\begin{vmatrix}b&c-a\\a+b&c\end{vmatrix}-(b-c)\begin{vmatrix}a+c&c-a\\a-b&c\end{vmatrix}+(b+c)\begin{vmatrix}a+c&b\\a-b&a+b\end{vmatrix}
  2. Evaluate the first minor.
    bc-(c-a)(a+b)=bc-ca-cb+a^2+ab=a^2+ab-ca=a(a+b-c)
  3. Evaluate the second minor.
    (a+c)c-(c-a)(a-b)=ac+c^2-ca+cb+a^2-ab=a^2-ab+bc+c^2
  4. Evaluate the third minor.
    (a+c)(a+b)-b(a-b)=a^2+ab+ac+bc-ab+b^2=a^2+b^2+ac+bc
  5. Substitute and multiply out each product.
    \Delta=a^2(a+b-c)-(b-c)(a^2-ab+bc+c^2)+(b+c)(a^2+b^2+ac+bc)
  6. Expand: a^2(a+b-c)=a^3+a^2b-a^2c; (b-c)(a^2-ab+bc+c^2)=a^2b-ab^2+b^2c+bc^2-a^2c+abc-bc^2-c^3; (b+c)(a^2+b^2+ac+bc)=a^2b+b^3+abc+b^2c+a^2c+b^2c+ac^2+bc^2.
    \Delta=a^3+a^2b-a^2c-\left(a^2b-ab^2+b^2c-a^2c+abc-c^3\right)+\left(a^2b+b^3+abc+2b^2c+a^2c+ac^2+bc^2\right)
  7. Collect the like terms.
    \Delta=a^3+b^3+c^3+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2
  8. Multiply out the right side to compare: (a+b+c)(a^2+b^2+c^2)=a^3+ab^2+ac^2+a^2b+b^3+bc^2+a^2c+b^2c+c^3, which has exactly the same nine terms.
    \Delta=(a+b+c)(a^2+b^2+c^2)
\Delta=(a+b+c)(a^2+b^2+c^2)
(iii) \begin{vmatrix}na_1+b_1&na_2+b_2&na_3+b_3\\nb_1+c_1&nb_2+c_2&nb_3+c_3\\nc_1+a_1&nc_2+a_2&nc_3+a_3\end{vmatrix}=(n^3+1)\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}
  1. Write \mathbf a=(a_1,a_2,a_3), \mathbf b=(b_1,b_2,b_3), \mathbf c=(c_1,c_2,c_3) for the three rows of the right-hand determinant, whose value is \Delta. The left side has rows n\mathbf a+\mathbf b, n\mathbf b+\mathbf c and n\mathbf c+\mathbf a.
    \text{L.H.S.}=\begin{vmatrix}n\mathbf a+\mathbf b\\n\mathbf b+\mathbf c\\n\mathbf c+\mathbf a\end{vmatrix}
  2. Every row is a sum of two parts, so Property 5 splits the determinant row by row into 2\times2\times2=8 determinants. Each picks one part from every row, and the factor n is taken out of a row by Property 4.
    \text{L.H.S.}=n^3|\mathbf a\,\mathbf b\,\mathbf c|+n^2|\mathbf a\,\mathbf b\,\mathbf a|+n^2|\mathbf a\,\mathbf c\,\mathbf c|+n|\mathbf a\,\mathbf c\,\mathbf a|+n^2|\mathbf b\,\mathbf b\,\mathbf c|+n|\mathbf b\,\mathbf b\,\mathbf a|+n|\mathbf b\,\mathbf c\,\mathbf c|+|\mathbf b\,\mathbf c\,\mathbf a|
  3. Here |\mathbf p\,\mathbf q\,\mathbf r| means the determinant with rows \mathbf p,\mathbf q,\mathbf r. Six of the eight have two identical rows, so they are 0 (Property 3): \mathbf a\mathbf b\mathbf a, \mathbf a\mathbf c\mathbf c, \mathbf a\mathbf c\mathbf a, \mathbf b\mathbf b\mathbf c, \mathbf b\mathbf b\mathbf a, \mathbf b\mathbf c\mathbf c.
    \text{L.H.S.}=n^3|\mathbf a\,\mathbf b\,\mathbf c|+|\mathbf b\,\mathbf c\,\mathbf a|
  4. The last determinant has the rows of \Delta in the order \mathbf b,\mathbf c,\mathbf a. Interchange R_1 and R_3 to get \mathbf a,\mathbf c,\mathbf b (sign changes), then R_2 and R_3 to get \mathbf a,\mathbf b,\mathbf c (sign changes again, Property 2). Two changes of sign leave the value unchanged.
    |\mathbf b\,\mathbf c\,\mathbf a|=-|\mathbf a\,\mathbf c\,\mathbf b|=|\mathbf a\,\mathbf b\,\mathbf c|=\Delta
  5. Substitute.
    \text{L.H.S.}=n^3\Delta+\Delta=(n^3+1)\Delta=\text{R.H.S.}
(n^3+1)\Delta
(iv) \begin{vmatrix}x&x^2&1+\alpha x^3\\y&y^2&1+\alpha y^3\\z&z^2&1+\alpha z^3\end{vmatrix}=(1+\alpha xyz)(x-y)(y-z)(z-x)
  1. Every entry of C_3 is a sum of two numbers, 1 and \alpha x^3 (or \alpha y^3, \alpha z^3). Split the determinant by C_3 (Property 5).
    \Delta=\begin{vmatrix}x&x^2&1\\y&y^2&1\\z&z^2&1\end{vmatrix}+\begin{vmatrix}x&x^2&\alpha x^3\\y&y^2&\alpha y^3\\z&z^2&\alpha z^3\end{vmatrix}
  2. In the first determinant move C_3 to the front: interchange C_2,C_3 and then C_1,C_2. Two interchanges give (-1)^2=1 (Property 2). In the second take \alpha out of C_3, then x,y,z out of R_1,R_2,R_3 (Property 4).
    \Delta=\begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix}+\alpha xyz\begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix}=(1+\alpha xyz)\,V
  3. Now evaluate V=\begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix}. Apply R_2\to R_2-R_1 and R_3\to R_3-R_1 (Property 7), using y^2-x^2=(y-x)(y+x).
    V=\begin{vmatrix}1&x&x^2\\0&y-x&(y-x)(y+x)\\0&z-x&(z-x)(z+x)\end{vmatrix}
  4. Take (y-x) out of R_2 and (z-x) out of R_3 (Property 4), then expand along C_1.
    V=(y-x)(z-x)\begin{vmatrix}1&y+x\\1&z+x\end{vmatrix}=(y-x)(z-x)\left[(z+x)-(y+x)\right]=(y-x)(z-x)(z-y)
  5. Write each factor in the order of the answer: y-x=-(x-y) and z-y=-(y-z). The two minus signs cancel.
    V=(x-y)(y-z)(z-x)
  6. Substitute V.
    \Delta=(1+\alpha xyz)(x-y)(y-z)(z-x)
\Delta=(1+\alpha xyz)(x-y)(y-z)(z-x)
(v) \begin{vmatrix}2ab&1+a^2-b^2&2b\\2a&-2b&1-a^2-b^2\\1-a^2+b^2&2ab&-2a\end{vmatrix}=(1+a^2+b^2)^3
  1. Write s=1+a^2+b^2 and let the left side be \Delta. We want to build a factor s in the columns. Add b times C_3 to C_2 (Property 7): in the first row (1+a^2-b^2)+2b^2=s, in the second -2b+b(1-a^2-b^2)=-bs and in the third 2ab-2ab=0.
    C_2\to C_2+bC_3:\quad \begin{bmatrix}s\\-bs\\0\end{bmatrix}
  2. Subtract a times C_3 from C_1: first row 2ab-2ab=0, second row 2a-a(1-a^2-b^2)=as, third row (1-a^2+b^2)+2a^2=s.
    C_1\to C_1-aC_3:\quad \begin{bmatrix}0\\as\\s\end{bmatrix}
  3. The determinant is unchanged by these two operations.
    \Delta=\begin{vmatrix}0&s&2b\\as&-bs&1-a^2-b^2\\s&0&-2a\end{vmatrix}
  4. Take the common factor s out of C_1 and out of C_2 (Property 4).
    \Delta=s^2\begin{vmatrix}0&1&2b\\a&-b&1-a^2-b^2\\1&0&-2a\end{vmatrix}
  5. Expand along R_1. The element 1 in position (1,2) has cofactor sign - and the element 2b has sign +.
    =s^2\left[-\begin{vmatrix}a&1-a^2-b^2\\1&-2a\end{vmatrix}+2b\begin{vmatrix}a&-b\\1&0\end{vmatrix}\right]
  6. Evaluate the two 2\times2 determinants.
    =s^2\left[-(-2a^2-1+a^2+b^2)+2b(0+b)\right]=s^2\left[a^2+1-b^2+2b^2\right]
  7. The bracket is 1+a^2+b^2=s.
    \Delta=s^2\cdot s=s^3=(1+a^2+b^2)^3
\Delta=(1+a^2+b^2)^3
(vi) \begin{vmatrix}3a&1&2a+1\\2a+1&1&a+2\\3&1&2\end{vmatrix}=(a-1)(a-2)
  1. The second column is all ones, so subtracting R_3 from R_1 and from R_2 makes zeros in C_2 (Property 7). R_1-R_3: (3a-3,\ 0,\ 2a-1). R_2-R_3: (2a-2,\ 0,\ a).
    \Delta=\begin{vmatrix}3a-3&0&2a-1\\2a-2&0&a\\3&1&2\end{vmatrix}
  2. Expand along C_2. The only non-zero element is 1 in position (3,2), with cofactor sign (-1)^{3+2}=-1.
    \Delta=-\begin{vmatrix}3a-3&2a-1\\2a-2&a\end{vmatrix}
  3. Evaluate the 2\times2 determinant.
    \Delta=-\left[(3a-3)a-(2a-1)(2a-2)\right]=-\left[3a^2-3a-(4a^2-6a+2)\right]
  4. Simplify and factorise.
    \Delta=-\left[-a^2+3a-2\right]=a^2-3a+2=(a-1)(a-2)
\Delta=(a-1)(a-2)
(vii) \begin{vmatrix}b+c&a&a\\b&c+a&b\\c&c&a+b\end{vmatrix}=4abc
  1. Apply R_1\to R_1-R_2-R_3 (Property 7): b+c-b-c=0, a-(c+a)-c=-2c and a-b-(a+b)=-2b.
    \Delta=\begin{vmatrix}0&-2c&-2b\\b&c+a&b\\c&c&a+b\end{vmatrix}
  2. Expand along R_1 (the first element is 0).
    \Delta=-(-2c)\begin{vmatrix}b&b\\c&a+b\end{vmatrix}+(-2b)\begin{vmatrix}b&c+a\\c&c\end{vmatrix}
  3. Evaluate the minors.
    \Delta=2c\left[b(a+b)-bc\right]-2b\left[bc-c(c+a)\right]=2bc(a+b-c)-2bc(b-c-a)
  4. Take out 2bc and simplify the bracket (a+b-c)-(b-c-a)=2a.
    \Delta=2bc\cdot2a=4abc
\Delta=4abc
(viii) \begin{vmatrix}-bc&b^2+bc&c^2+bc\\a^2+ac&-ac&c^2+ac\\a^2+ab&b^2+ab&-ab\end{vmatrix}=(ab+bc+ca)^3
  1. Factorise the entries: R_1=(-bc,\ b(b+c),\ c(b+c)), R_2=(a(a+c),\ -ac,\ c(a+c)), R_3=(a(a+b),\ b(a+b),\ -ab). Multiply R_1 by a, R_2 by b and R_3 by c (Property 4: the value becomes abc\,\Delta).
    abc\,\Delta=\begin{vmatrix}-abc&ab(b+c)&ac(b+c)\\ab(a+c)&-abc&bc(a+c)\\ac(a+b)&bc(a+b)&-abc\end{vmatrix}
  2. Take a out of C_1, b out of C_2 and c out of C_3 (Property 4). This gives another factor abc, which cancels with the left side.
    \Delta=\begin{vmatrix}-bc&a(b+c)&a(b+c)\\b(a+c)&-ac&b(a+c)\\c(a+b)&c(a+b)&-ab\end{vmatrix}
  3. Put P=ab, Q=bc, R=ca. Then a(b+c)=P+R, b(a+c)=P+Q and c(a+b)=Q+R.
    \Delta=\begin{vmatrix}-Q&P+R&P+R\\P+Q&-R&P+Q\\Q+R&Q+R&-P\end{vmatrix}
  4. Apply R_1\to R_1+R_2+R_3 (Property 7). Every entry of the new first row is P+Q+R.
    \Delta=\begin{vmatrix}P+Q+R&P+Q+R&P+Q+R\\P+Q&-R&P+Q\\Q+R&Q+R&-P\end{vmatrix}
  5. Take S=P+Q+R out of R_1, then apply C_2\to C_2-C_1 and C_3\to C_3-C_1. For example, -R-(P+Q)=-S.
    \Delta=S\begin{vmatrix}1&0&0\\P+Q&-S&0\\Q+R&0&-S\end{vmatrix}
  6. The determinant is triangular, so it is the product of the diagonal elements 1\cdot(-S)(-S) (Property 8).
    \Delta=S\cdot S^2=S^3=(ab+bc+ca)^3
\Delta=(ab+bc+ca)^3
(ix) \begin{vmatrix}(b+c)^2&ab&ca\\ab&(a+c)^2&bc\\ac&bc&(a+b)^2\end{vmatrix}=2abc(a+b+c)^3
  1. Let the left side be \Delta. Multiply R_1 by a, R_2 by b and R_3 by c (Property 4: the value becomes abc\,\Delta).
    abc\,\Delta=\begin{vmatrix}a(b+c)^2&a^2b&a^2c\\ab^2&b(a+c)^2&b^2c\\ac^2&bc^2&c(a+b)^2\end{vmatrix}
  2. Take a out of C_1, b out of C_2 and c out of C_3. The factor abc cancels.
    \Delta=\begin{vmatrix}(b+c)^2&a^2&a^2\\b^2&(a+c)^2&b^2\\c^2&c^2&(a+b)^2\end{vmatrix}
  3. Apply C_1\to C_1-C_3 and C_2\to C_2-C_3 (Property 7). Each new entry is a difference of squares.
    \Delta=\begin{vmatrix}(b+c)^2-a^2&0&a^2\\0&(a+c)^2-b^2&b^2\\c^2-(a+b)^2&c^2-(a+b)^2&(a+b)^2\end{vmatrix}
  4. Factorise with x^2-y^2=(x-y)(x+y): (b+c)^2-a^2=(b+c-a)(a+b+c), (a+c)^2-b^2=(a+c-b)(a+b+c), c^2-(a+b)^2=(c-a-b)(a+b+c). Take (a+b+c) out of C_1 and out of C_2 (Property 4). Write p=b+c-a, q=a+c-b, r=c-a-b.
    \Delta=(a+b+c)^2\begin{vmatrix}p&0&a^2\\0&q&b^2\\r&r&(a+b)^2\end{vmatrix}
  5. Apply R_3\to R_3-R_1-R_2: r-p=-2b, r-q=-2a and (a+b)^2-a^2-b^2=2ab. Then take the factor 2 out of R_3.
    \Delta=2(a+b+c)^2\begin{vmatrix}p&0&a^2\\0&q&b^2\\-b&-a&ab\end{vmatrix}
  6. Expand along R_3 with signs +,-,+.
    \begin{vmatrix}p&0&a^2\\0&q&b^2\\-b&-a&ab\end{vmatrix}=(-b)(0-a^2q)-(-a)(pb^2-0)+ab(pq)=ab\,(aq+bp+pq)
  7. Compute the bracket: aq+bp=a(a+c-b)+b(b+c-a)=a^2+b^2+ac+bc-2ab and pq=c^2-(a-b)^2=c^2-a^2+2ab-b^2. Their sum is c(a+b+c).
    aq+bp+pq=ac+bc+c^2=c(a+b+c)
  8. Put the pieces together.
    \Delta=2(a+b+c)^2\cdot ab\cdot c(a+b+c)=2abc(a+b+c)^3
\Delta=2abc(a+b+c)^3
(x) \begin{vmatrix}b+c&q+r&y+z\\c+a&r+p&z+x\\a+b&p+q&x+y\end{vmatrix}=2\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}
  1. Let the left side be \Delta. Apply R_1\to R_1+R_2+R_3 (Property 7). Each entry of the new first row is twice a sum, for example (b+c)+(c+a)+(a+b)=2(a+b+c).
    \begin{aligned}&\Delta=\\ &\begin{vmatrix}2(a+b+c)&2(p+q+r)&2(x+y+z)\\c+a&r+p&z+x\\a+b&p+q&x+y\end{vmatrix}\end{aligned}
  2. Take the factor 2 out of R_1 (Property 4).
    \Delta=2\begin{vmatrix}a+b+c&p+q+r&x+y+z\\c+a&r+p&z+x\\a+b&p+q&x+y\end{vmatrix}
  3. Apply R_2\to R_2-R_1 and R_3\to R_3-R_1. For example (c+a)-(a+b+c)=-b and (a+b)-(a+b+c)=-c.
    \Delta=2\begin{vmatrix}a+b+c&p+q+r&x+y+z\\-b&-q&-y\\-c&-r&-z\end{vmatrix}
  4. Apply R_1\to R_1+R_2+R_3. The first row becomes (a,p,x).
    \Delta=2\begin{vmatrix}a&p&x\\-b&-q&-y\\-c&-r&-z\end{vmatrix}
  5. Take -1 out of R_2 and out of R_3 (Property 4). The two minus signs multiply to +1.
    \Delta=2(-1)(-1)\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}=2\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}
\Delta=2\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}

Book answer. No printed answer (proofs).

Exercise 2.5p. 59

Row operations: echelon and reduced echelon form, rank, and the inverse of a matrix from [A\,|\,I]. Work column by column: make a leading 1, then clear the entries in its column.

1
First reduce each of the following matrices into echelon form then into reduced echelon form.
Solution

Method (sections 2.5.1 to 2.5.3). Use the elementary row operations: interchange R_{ij}, multiplication kR_i (k\neq0) and addition R_i+kR_j. Echelon form: the first non-zero entry of each non-zero row is 1, leading entries move to the right as you go down, and zero rows are at the bottom. Reduced echelon form: in addition, each leading 1 is the only non-zero entry in its column.

(i) A=\begin{bmatrix}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{bmatrix}
Echelon form
  1. Start with the matrix.
    A=\begin{bmatrix}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{bmatrix}
  2. Apply R_{2}+6R_{1}, R_{3}+4R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \begin{bmatrix}1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 & 18 & 25\end{bmatrix}
  3. Apply \frac{1}{26}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 18 & 25\end{bmatrix}
  4. Apply R_{3}-18R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \begin{bmatrix}1 & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 0 & \frac{28}{13}\end{bmatrix}
  5. Apply \frac{13}{28}R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \begin{bmatrix}1 & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 0 & 1\end{bmatrix}
  6. Check the conditions: the first non-zero entry of each non-zero row is 1, each leading 1 is to the right of the one above it, and zero rows (if any) are at the bottom. So this is a row echelon form; its number of non-zero rows is 3.
Reduced echelon form
  1. Start from the echelon form obtained above.
    \sim_R \begin{bmatrix}1 & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 0 & 1\end{bmatrix}
  2. Apply R_{1}-5R_{3}, R_{2}-\frac{33}{26}R_{3} to make the entries above the leading 1 of row 3 zero.
    \sim_R \begin{bmatrix}1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  3. Apply R_{1}-3R_{2} to make the entries above the leading 1 of row 2 zero.
    \sim_R \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  4. Each leading 1 is now the only non-zero entry of its column, so this is the reduced row echelon form.
\text{echelon: }\begin{bmatrix}1 & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 0 & 1\end{bmatrix};\ \text{reduced: }\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
(ii) B=\begin{bmatrix}2 & 1 \\ 3 & 2 \\ 1 & 9\end{bmatrix}
Echelon form
  1. Start with the matrix.
    B=\begin{bmatrix}2 & 1 \\ 3 & 2 \\ 1 & 9\end{bmatrix}
  2. Interchange R_{1} and R_{3} (R_{13}) to get a leading entry 1 in position (1,1).
    \sim_R \begin{bmatrix}1 & 9 \\ 3 & 2 \\ 2 & 1\end{bmatrix}
  3. Apply R_{2}-3R_{1}, R_{3}-2R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \begin{bmatrix}1 & 9 \\ 0 & -25 \\ 0 & -17\end{bmatrix}
  4. Apply -\frac{1}{25}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & 9 \\ 0 & 1 \\ 0 & -17\end{bmatrix}
  5. Apply R_{3}+17R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \begin{bmatrix}1 & 9 \\ 0 & 1 \\ 0 & 0\end{bmatrix}
  6. Check the conditions: the first non-zero entry of each non-zero row is 1, each leading 1 is to the right of the one above it, and zero rows (if any) are at the bottom. So this is a row echelon form; its number of non-zero rows is 2.
Reduced echelon form
  1. Start from the echelon form obtained above.
    \sim_R \begin{bmatrix}1 & 9 \\ 0 & 1 \\ 0 & 0\end{bmatrix}
  2. Apply R_{1}-9R_{2} to make the entries above the leading 1 of row 2 zero.
    \sim_R \begin{bmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}
  3. Each leading 1 is now the only non-zero entry of its column, so this is the reduced row echelon form.
\text{echelon: }\begin{bmatrix}1 & 9 \\ 0 & 1 \\ 0 & 0\end{bmatrix};\ \text{reduced: }\begin{bmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}
(iii) C=\begin{bmatrix}2 & -1 & 0 \\ 4 & 7 & 8 \\ -3 & 1 & 3\end{bmatrix}
Echelon form
  1. Start with the matrix.
    C=\begin{bmatrix}2 & -1 & 0 \\ 4 & 7 & 8 \\ -3 & 1 & 3\end{bmatrix}
  2. Apply \frac{1}{2}R_{1} to make the leading entry of row 1 equal to 1.
    \sim_R \begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 4 & 7 & 8 \\ -3 & 1 & 3\end{bmatrix}
  3. Apply R_{2}-4R_{1}, R_{3}+3R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 0 & 9 & 8 \\ 0 & -\frac{1}{2} & 3\end{bmatrix}
  4. Apply \frac{1}{9}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{8}{9} \\ 0 & -\frac{1}{2} & 3\end{bmatrix}
  5. Apply R_{3}+\frac{1}{2}R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{8}{9} \\ 0 & 0 & \frac{31}{9}\end{bmatrix}
  6. Apply \frac{9}{31}R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{8}{9} \\ 0 & 0 & 1\end{bmatrix}
  7. Check the conditions: the first non-zero entry of each non-zero row is 1, each leading 1 is to the right of the one above it, and zero rows (if any) are at the bottom. So this is a row echelon form; its number of non-zero rows is 3.
Reduced echelon form
  1. Start from the echelon form obtained above.
    \sim_R \begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{8}{9} \\ 0 & 0 & 1\end{bmatrix}
  2. Apply R_{2}-\frac{8}{9}R_{3} to make the entries above the leading 1 of row 3 zero.
    \sim_R \begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  3. Apply R_{1}+\frac{1}{2}R_{2} to make the entries above the leading 1 of row 2 zero.
    \sim_R \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  4. Each leading 1 is now the only non-zero entry of its column, so this is the reduced row echelon form.
\text{echelon: }\begin{bmatrix}1 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{8}{9} \\ 0 & 0 & 1\end{bmatrix};\ \text{reduced: }\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
(iv) D=\begin{bmatrix}2 & -4 & 3 \\ 4 & 1 & 8 \\ 7 & 3 & 0\end{bmatrix}
Echelon form
  1. Start with the matrix.
    D=\begin{bmatrix}2 & -4 & 3 \\ 4 & 1 & 8 \\ 7 & 3 & 0\end{bmatrix}
  2. Apply \frac{1}{2}R_{1} to make the leading entry of row 1 equal to 1.
    \sim_R \begin{bmatrix}1 & -2 & \frac{3}{2} \\ 4 & 1 & 8 \\ 7 & 3 & 0\end{bmatrix}
  3. Apply R_{2}-4R_{1}, R_{3}-7R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \begin{bmatrix}1 & -2 & \frac{3}{2} \\ 0 & 9 & 2 \\ 0 & 17 & -\frac{21}{2}\end{bmatrix}
  4. Apply \frac{1}{9}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & -2 & \frac{3}{2} \\ 0 & 1 & \frac{2}{9} \\ 0 & 17 & -\frac{21}{2}\end{bmatrix}
  5. Apply R_{3}-17R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \begin{bmatrix}1 & -2 & \frac{3}{2} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & -\frac{257}{18}\end{bmatrix}
  6. Apply -\frac{18}{257}R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \begin{bmatrix}1 & -2 & \frac{3}{2} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1\end{bmatrix}
  7. Check the conditions: the first non-zero entry of each non-zero row is 1, each leading 1 is to the right of the one above it, and zero rows (if any) are at the bottom. So this is a row echelon form; its number of non-zero rows is 3.
Reduced echelon form
  1. Start from the echelon form obtained above.
    \sim_R \begin{bmatrix}1 & -2 & \frac{3}{2} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1\end{bmatrix}
  2. Apply R_{1}-\frac{3}{2}R_{3}, R_{2}-\frac{2}{9}R_{3} to make the entries above the leading 1 of row 3 zero.
    \sim_R \begin{bmatrix}1 & -2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  3. Apply R_{1}+2R_{2} to make the entries above the leading 1 of row 2 zero.
    \sim_R \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  4. Each leading 1 is now the only non-zero entry of its column, so this is the reduced row echelon form.
\text{echelon: }\begin{bmatrix}1 & -2 & \frac{3}{2} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1\end{bmatrix};\ \text{reduced: }\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
(v) E=\begin{bmatrix}3 & 1 & 2 \\ 2 & 9 & 8\end{bmatrix}
Echelon form
  1. Start with the matrix.
    E=\begin{bmatrix}3 & 1 & 2 \\ 2 & 9 & 8\end{bmatrix}
  2. Apply \frac{1}{3}R_{1} to make the leading entry of row 1 equal to 1.
    \sim_R \begin{bmatrix}1 & \frac{1}{3} & \frac{2}{3} \\ 2 & 9 & 8\end{bmatrix}
  3. Apply R_{2}-2R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \begin{bmatrix}1 & \frac{1}{3} & \frac{2}{3} \\ 0 & \frac{25}{3} & \frac{20}{3}\end{bmatrix}
  4. Apply \frac{3}{25}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & \frac{4}{5}\end{bmatrix}
  5. Check the conditions: the first non-zero entry of each non-zero row is 1, each leading 1 is to the right of the one above it, and zero rows (if any) are at the bottom. So this is a row echelon form; its number of non-zero rows is 2.
Reduced echelon form
  1. Start from the echelon form obtained above.
    \sim_R \begin{bmatrix}1 & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & \frac{4}{5}\end{bmatrix}
  2. Apply R_{1}-\frac{1}{3}R_{2} to make the entries above the leading 1 of row 2 zero.
    \sim_R \begin{bmatrix}1 & 0 & \frac{2}{5} \\ 0 & 1 & \frac{4}{5}\end{bmatrix}
  3. Each leading 1 is now the only non-zero entry of its column, so this is the reduced row echelon form.
\text{echelon: }\begin{bmatrix}1 & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & \frac{4}{5}\end{bmatrix};\ \text{reduced: }\begin{bmatrix}1 & 0 & \frac{2}{5} \\ 0 & 1 & \frac{4}{5}\end{bmatrix}
(vi) F=\begin{bmatrix}0 & 2 & 4 \\ 0 & 3 & 6 \\ 0 & 1 & 2\end{bmatrix}
Echelon form
  1. Start with the matrix.
    F=\begin{bmatrix}0 & 2 & 4 \\ 0 & 3 & 6 \\ 0 & 1 & 2\end{bmatrix}
  2. Interchange R_{1} and R_{3} (R_{13}) to get a leading entry 1 in position (1,2).
    \sim_R \begin{bmatrix}0 & 1 & 2 \\ 0 & 3 & 6 \\ 0 & 2 & 4\end{bmatrix}
  3. Apply R_{2}-3R_{1}, R_{3}-2R_{1} to make the entries below the leading 1 in column 2 zero.
    \sim_R \begin{bmatrix}0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
  4. Check the conditions: the first non-zero entry of each non-zero row is 1, each leading 1 is to the right of the one above it, and zero rows (if any) are at the bottom. So this is a row echelon form; its number of non-zero rows is 1.
Reduced echelon form
  1. Start from the echelon form obtained above.
    \sim_R \begin{bmatrix}0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
  2. Every leading 1 is already the only non-zero entry in its column, so the echelon form is also the reduced echelon form.
  3. Each leading 1 is now the only non-zero entry of its column, so this is the reduced row echelon form.
\text{echelon: }\begin{bmatrix}0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix};\ \text{reduced: }\begin{bmatrix}0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}

Book answer. Reduced echelon forms in the key: (i) I_3, (ii) [1, 0; 0, 1; 0, 0], (iii) I_3, (iv) I_3, (v) [1, 0, 2/5; 0, 1, 4/5], (vi) [0, 1, 2; 0, 0, 0; 0, 0, 0]. All agree; the key's echelon forms are different but equivalent echelon forms.

2
Find the rank of each of the following matrices.
Solution

Method (section 2.5.4). Reduce the matrix to echelon form by row operations. The number of non-zero rows is the rank.

(i) A=\begin{bmatrix}5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{bmatrix}
  1. Start with the matrix.
    A=\begin{bmatrix}5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{bmatrix}
  2. Apply \frac{1}{5}R_{1} to make the leading entry of row 1 equal to 1.
    \sim_R \begin{bmatrix}1 & \frac{9}{5} & \frac{3}{5} \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{bmatrix}
  3. Apply R_{2}-3R_{1}, R_{3}-2R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \begin{bmatrix}1 & \frac{9}{5} & \frac{3}{5} \\ 0 & -\frac{52}{5} & \frac{21}{5} \\ 0 & \frac{32}{5} & \frac{24}{5}\end{bmatrix}
  4. Apply -\frac{5}{52}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & \frac{9}{5} & \frac{3}{5} \\ 0 & 1 & -\frac{21}{52} \\ 0 & \frac{32}{5} & \frac{24}{5}\end{bmatrix}
  5. Apply R_{3}-\frac{32}{5}R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \begin{bmatrix}1 & \frac{9}{5} & \frac{3}{5} \\ 0 & 1 & -\frac{21}{52} \\ 0 & 0 & \frac{96}{13}\end{bmatrix}
  6. Apply \frac{13}{96}R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \begin{bmatrix}1 & \frac{9}{5} & \frac{3}{5} \\ 0 & 1 & -\frac{21}{52} \\ 0 & 0 & 1\end{bmatrix}
  7. This is in echelon form. Count the non-zero rows: there are 3.
    \operatorname{Rank}(A)=3
\operatorname{Rank}(A)=3
(ii) B=\begin{bmatrix}-1 & -2 & 3 \\ -1 & 2 & -1 \\ -5 & 2 & 3\end{bmatrix}
  1. Start with the matrix.
    B=\begin{bmatrix}-1 & -2 & 3 \\ -1 & 2 & -1 \\ -5 & 2 & 3\end{bmatrix}
  2. Apply -R_{1} to make the leading entry of row 1 equal to 1.
    \sim_R \begin{bmatrix}1 & 2 & -3 \\ -1 & 2 & -1 \\ -5 & 2 & 3\end{bmatrix}
  3. Apply R_{2}+R_{1}, R_{3}+5R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \begin{bmatrix}1 & 2 & -3 \\ 0 & 4 & -4 \\ 0 & 12 & -12\end{bmatrix}
  4. Apply \frac{1}{4}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & 2 & -3 \\ 0 & 1 & -1 \\ 0 & 12 & -12\end{bmatrix}
  5. Apply R_{3}-12R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \begin{bmatrix}1 & 2 & -3 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{bmatrix}
  6. This is in echelon form. Count the non-zero rows: there are 2.
    \operatorname{Rank}(B)=2
\operatorname{Rank}(B)=2
(iii) C=\begin{bmatrix}3 & 2 & 4 \\ 2 & 1 & 6 \\ 4 & -1 & 0\end{bmatrix}
  1. Start with the matrix.
    C=\begin{bmatrix}3 & 2 & 4 \\ 2 & 1 & 6 \\ 4 & -1 & 0\end{bmatrix}
  2. Apply \frac{1}{3}R_{1} to make the leading entry of row 1 equal to 1.
    \sim_R \begin{bmatrix}1 & \frac{2}{3} & \frac{4}{3} \\ 2 & 1 & 6 \\ 4 & -1 & 0\end{bmatrix}
  3. Apply R_{2}-2R_{1}, R_{3}-4R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \begin{bmatrix}1 & \frac{2}{3} & \frac{4}{3} \\ 0 & -\frac{1}{3} & \frac{10}{3} \\ 0 & -\frac{11}{3} & -\frac{16}{3}\end{bmatrix}
  4. Apply -3R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & \frac{2}{3} & \frac{4}{3} \\ 0 & 1 & -10 \\ 0 & -\frac{11}{3} & -\frac{16}{3}\end{bmatrix}
  5. Apply R_{3}+\frac{11}{3}R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \begin{bmatrix}1 & \frac{2}{3} & \frac{4}{3} \\ 0 & 1 & -10 \\ 0 & 0 & -42\end{bmatrix}
  6. Apply -\frac{1}{42}R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \begin{bmatrix}1 & \frac{2}{3} & \frac{4}{3} \\ 0 & 1 & -10 \\ 0 & 0 & 1\end{bmatrix}
  7. This is in echelon form. Count the non-zero rows: there are 3.
    \operatorname{Rank}(C)=3
\operatorname{Rank}(C)=3
(iv) D=\begin{bmatrix}1 & 3 \\ 2 & 9 \\ 1 & 6\end{bmatrix}
  1. Start with the matrix.
    D=\begin{bmatrix}1 & 3 \\ 2 & 9 \\ 1 & 6\end{bmatrix}
  2. Apply R_{2}-2R_{1}, R_{3}-R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \begin{bmatrix}1 & 3 \\ 0 & 3 \\ 0 & 3\end{bmatrix}
  3. Apply \frac{1}{3}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & 3 \\ 0 & 1 \\ 0 & 3\end{bmatrix}
  4. Apply R_{3}-3R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \begin{bmatrix}1 & 3 \\ 0 & 1 \\ 0 & 0\end{bmatrix}
  5. This is in echelon form. Count the non-zero rows: there are 2.
    \operatorname{Rank}(D)=2
\operatorname{Rank}(D)=2

Book answer. (i) 3, (ii) 2, (iii) 3, (iv) 2. All agree.

3
With the help of row operations, find the inverse of the following matrices if it exists. Also verify your answer by showing that AA^{-1}=A^{-1}A=I.
Solution

Method (section 2.5.5, Example 29). Form [A\,|\,I] and apply row operations to the whole row until the left block is I. The right block is then A^{-1}. If the left block cannot be reduced to I, A is singular and has no inverse.

(i) A=\begin{bmatrix}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4\end{bmatrix}
Inverse by row operations
  1. Check that |A|\neq0 so that the inverse exists.
    |A|=-2\neq0
  2. Write A and I side by side. Perform the same row operations on both until A becomes I; then I turns into A^{-1}.
    [A\,|\,I]=\left[\begin{array}{ccc|ccc}0 & -1 & -1 & 1 & 0 & 0 \\ -1 & 3 & 0 & 0 & 1 & 0 \\ 1 & -1 & 4 & 0 & 0 & 1\end{array}\right]
  3. Interchange R_{1} and R_{3} (R_{13}) to bring a non-zero entry (here 1) into position (1,1).
    \sim_R \left[\begin{array}{ccc|ccc}1 & -1 & 4 & 0 & 0 & 1 \\ -1 & 3 & 0 & 0 & 1 & 0 \\ 0 & -1 & -1 & 1 & 0 & 0\end{array}\right]
  4. Apply R_{2}+R_{1} to make the other entries of column 1 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & -1 & 4 & 0 & 0 & 1 \\ 0 & 2 & 4 & 0 & 1 & 1 \\ 0 & -1 & -1 & 1 & 0 & 0\end{array}\right]
  5. Interchange R_{2} and R_{3} (R_{23}) to get a leading entry that is easy to work with in position (2,2).
    \sim_R \left[\begin{array}{ccc|ccc}1 & -1 & 4 & 0 & 0 & 1 \\ 0 & -1 & -1 & 1 & 0 & 0 \\ 0 & 2 & 4 & 0 & 1 & 1\end{array}\right]
  6. Apply -R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \left[\begin{array}{ccc|ccc}1 & -1 & 4 & 0 & 0 & 1 \\ 0 & 1 & 1 & -1 & 0 & 0 \\ 0 & 2 & 4 & 0 & 1 & 1\end{array}\right]
  7. Apply R_{1}+R_{2}, R_{3}-2R_{2} to make the other entries of column 2 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & 5 & -1 & 0 & 1 \\ 0 & 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 2 & 2 & 1 & 1\end{array}\right]
  8. Apply \frac{1}{2}R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & 5 & -1 & 0 & 1 \\ 0 & 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 & \frac{1}{2} & \frac{1}{2}\end{array}\right]
  9. Apply R_{1}-5R_{3}, R_{2}-R_{3} to make the other entries of column 3 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & 0 & -6 & -\frac{5}{2} & -\frac{3}{2} \\ 0 & 1 & 0 & -2 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & 1 & 1 & \frac{1}{2} & \frac{1}{2}\end{array}\right]
  10. The left block is I, so the right block is A^{-1}.
    A^{-1}=\begin{bmatrix}-6 & -\frac{5}{2} & -\frac{3}{2} \\ -2 & -\frac{1}{2} & -\frac{1}{2} \\ 1 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}
Verification
  1. **Verify AA^{-1}=I.**
    AA^{-1}=\begin{bmatrix}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4\end{bmatrix}\begin{bmatrix}-6 & -\frac{5}{2} & -\frac{3}{2} \\ -2 & -\frac{1}{2} & -\frac{1}{2} \\ 1 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  2. **Verify A^{-1}A=I.**
    A^{-1}A=\begin{bmatrix}-6 & -\frac{5}{2} & -\frac{3}{2} \\ -2 & -\frac{1}{2} & -\frac{1}{2} \\ 1 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}\begin{bmatrix}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
A^{-1}=\begin{bmatrix}-6 & -\frac{5}{2} & -\frac{3}{2} \\ -2 & -\frac{1}{2} & -\frac{1}{2} \\ 1 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}
(ii) A=\begin{bmatrix}1 & 2 & 5 \\ -3 & 0 & 1 \\ 4 & 2 & 5\end{bmatrix}
Inverse by row operations
  1. Check that |A|\neq0 so that the inverse exists.
    |A|=6\neq0
  2. Write A and I side by side. Perform the same row operations on both until A becomes I; then I turns into A^{-1}.
    [A\,|\,I]=\left[\begin{array}{ccc|ccc}1 & 2 & 5 & 1 & 0 & 0 \\ -3 & 0 & 1 & 0 & 1 & 0 \\ 4 & 2 & 5 & 0 & 0 & 1\end{array}\right]
  3. Apply R_{2}+3R_{1}, R_{3}-4R_{1} to make the other entries of column 1 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 6 & 16 & 3 & 1 & 0 \\ 0 & -6 & -15 & -4 & 0 & 1\end{array}\right]
  4. Apply \frac{1}{6}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 2 & 5 & 1 & 0 & 0 \\ 0 & 1 & \frac{8}{3} & \frac{1}{2} & \frac{1}{6} & 0 \\ 0 & -6 & -15 & -4 & 0 & 1\end{array}\right]
  5. Apply R_{1}-2R_{2}, R_{3}+6R_{2} to make the other entries of column 2 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & -\frac{1}{3} & 0 & -\frac{1}{3} & 0 \\ 0 & 1 & \frac{8}{3} & \frac{1}{2} & \frac{1}{6} & 0 \\ 0 & 0 & 1 & -1 & 1 & 1\end{array}\right]
  6. Apply R_{1}+\frac{1}{3}R_{3}, R_{2}-\frac{8}{3}R_{3} to make the other entries of column 3 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & 0 & -\frac{1}{3} & 0 & \frac{1}{3} \\ 0 & 1 & 0 & \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ 0 & 0 & 1 & -1 & 1 & 1\end{array}\right]
  7. The left block is I, so the right block is A^{-1}.
    A^{-1}=\begin{bmatrix}-\frac{1}{3} & 0 & \frac{1}{3} \\ \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ -1 & 1 & 1\end{bmatrix}
Verification
  1. **Verify AA^{-1}=I.**
    AA^{-1}=\begin{bmatrix}1 & 2 & 5 \\ -3 & 0 & 1 \\ 4 & 2 & 5\end{bmatrix}\begin{bmatrix}-\frac{1}{3} & 0 & \frac{1}{3} \\ \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ -1 & 1 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  2. **Verify A^{-1}A=I.**
    A^{-1}A=\begin{bmatrix}-\frac{1}{3} & 0 & \frac{1}{3} \\ \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ -1 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 5 \\ -3 & 0 & 1 \\ 4 & 2 & 5\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
A^{-1}=\begin{bmatrix}-\frac{1}{3} & 0 & \frac{1}{3} \\ \frac{19}{6} & -\frac{5}{2} & -\frac{8}{3} \\ -1 & 1 & 1\end{bmatrix}
(iii) A=\begin{bmatrix}-5 & 2 & 3 \\ -1 & -2 & 3 \\ 1 & -2 & 3\end{bmatrix}
Inverse by row operations
  1. Check that |A|\neq0 so that the inverse exists.
    |A|=24\neq0
  2. Write A and I side by side. Perform the same row operations on both until A becomes I; then I turns into A^{-1}.
    [A\,|\,I]=\left[\begin{array}{ccc|ccc}-5 & 2 & 3 & 1 & 0 & 0 \\ -1 & -2 & 3 & 0 & 1 & 0 \\ 1 & -2 & 3 & 0 & 0 & 1\end{array}\right]
  3. Interchange R_{1} and R_{3} (R_{13}) to get a leading entry 1 in position (1,1).
    \sim_R \left[\begin{array}{ccc|ccc}1 & -2 & 3 & 0 & 0 & 1 \\ -1 & -2 & 3 & 0 & 1 & 0 \\ -5 & 2 & 3 & 1 & 0 & 0\end{array}\right]
  4. Apply R_{2}+R_{1}, R_{3}+5R_{1} to make the other entries of column 1 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & -2 & 3 & 0 & 0 & 1 \\ 0 & -4 & 6 & 0 & 1 & 1 \\ 0 & -8 & 18 & 1 & 0 & 5\end{array}\right]
  5. Apply -\frac{1}{4}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \left[\begin{array}{ccc|ccc}1 & -2 & 3 & 0 & 0 & 1 \\ 0 & 1 & -\frac{3}{2} & 0 & -\frac{1}{4} & -\frac{1}{4} \\ 0 & -8 & 18 & 1 & 0 & 5\end{array}\right]
  6. Apply R_{1}+2R_{2}, R_{3}+8R_{2} to make the other entries of column 2 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & 0 & 0 & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & -\frac{3}{2} & 0 & -\frac{1}{4} & -\frac{1}{4} \\ 0 & 0 & 6 & 1 & -2 & 3\end{array}\right]
  7. Apply \frac{1}{6}R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & 0 & 0 & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & -\frac{3}{2} & 0 & -\frac{1}{4} & -\frac{1}{4} \\ 0 & 0 & 1 & \frac{1}{6} & -\frac{1}{3} & \frac{1}{2}\end{array}\right]
  8. Apply R_{2}+\frac{3}{2}R_{3} to make the other entries of column 3 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & 0 & 0 & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{4} & -\frac{3}{4} & \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{6} & -\frac{1}{3} & \frac{1}{2}\end{array}\right]
  9. The left block is I, so the right block is A^{-1}.
    A^{-1}=\begin{bmatrix}0 & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & -\frac{3}{4} & \frac{1}{2} \\ \frac{1}{6} & -\frac{1}{3} & \frac{1}{2}\end{bmatrix}
Verification
  1. **Verify AA^{-1}=I.**
    AA^{-1}=\begin{bmatrix}-5 & 2 & 3 \\ -1 & -2 & 3 \\ 1 & -2 & 3\end{bmatrix}\begin{bmatrix}0 & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & -\frac{3}{4} & \frac{1}{2} \\ \frac{1}{6} & -\frac{1}{3} & \frac{1}{2}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  2. **Verify A^{-1}A=I.**
    A^{-1}A=\begin{bmatrix}0 & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & -\frac{3}{4} & \frac{1}{2} \\ \frac{1}{6} & -\frac{1}{3} & \frac{1}{2}\end{bmatrix}\begin{bmatrix}-5 & 2 & 3 \\ -1 & -2 & 3 \\ 1 & -2 & 3\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
A^{-1}=\begin{bmatrix}0 & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & -\frac{3}{4} & \frac{1}{2} \\ \frac{1}{6} & -\frac{1}{3} & \frac{1}{2}\end{bmatrix}
(iv) A=\begin{bmatrix}0 & 1 & -3 \\ 3 & 2 & 4 \\ 6 & -1 & 2\end{bmatrix}
Inverse by row operations
  1. Check that |A|\neq0 so that the inverse exists.
    |A|=63\neq0
  2. Write A and I side by side. Perform the same row operations on both until A becomes I; then I turns into A^{-1}.
    [A\,|\,I]=\left[\begin{array}{ccc|ccc}0 & 1 & -3 & 1 & 0 & 0 \\ 3 & 2 & 4 & 0 & 1 & 0 \\ 6 & -1 & 2 & 0 & 0 & 1\end{array}\right]
  3. Interchange R_{1} and R_{2} (R_{12}) to bring a non-zero entry into position (1,1).
    \sim_R \left[\begin{array}{ccc|ccc}3 & 2 & 4 & 0 & 1 & 0 \\ 0 & 1 & -3 & 1 & 0 & 0 \\ 6 & -1 & 2 & 0 & 0 & 1\end{array}\right]
  4. Apply \frac{1}{3}R_{1} to make the leading entry of row 1 equal to 1.
    \sim_R \left[\begin{array}{ccc|ccc}1 & \frac{2}{3} & \frac{4}{3} & 0 & \frac{1}{3} & 0 \\ 0 & 1 & -3 & 1 & 0 & 0 \\ 6 & -1 & 2 & 0 & 0 & 1\end{array}\right]
  5. Apply R_{3}-6R_{1} to make the other entries of column 1 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & \frac{2}{3} & \frac{4}{3} & 0 & \frac{1}{3} & 0 \\ 0 & 1 & -3 & 1 & 0 & 0 \\ 0 & -5 & -6 & 0 & -2 & 1\end{array}\right]
  6. Apply R_{1}-\frac{2}{3}R_{2}, R_{3}+5R_{2} to make the other entries of column 2 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & \frac{10}{3} & -\frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 1 & -3 & 1 & 0 & 0 \\ 0 & 0 & -21 & 5 & -2 & 1\end{array}\right]
  7. Apply -\frac{1}{21}R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & \frac{10}{3} & -\frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 1 & -3 & 1 & 0 & 0 \\ 0 & 0 & 1 & -\frac{5}{21} & \frac{2}{21} & -\frac{1}{21}\end{array}\right]
  8. Apply R_{1}-\frac{10}{3}R_{3}, R_{2}+3R_{3} to make the other entries of column 3 zero.
    \sim_R \left[\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{8}{63} & \frac{1}{63} & \frac{10}{63} \\ 0 & 1 & 0 & \frac{2}{7} & \frac{2}{7} & -\frac{1}{7} \\ 0 & 0 & 1 & -\frac{5}{21} & \frac{2}{21} & -\frac{1}{21}\end{array}\right]
  9. The left block is I, so the right block is A^{-1}.
    A^{-1}=\begin{bmatrix}\frac{8}{63} & \frac{1}{63} & \frac{10}{63} \\ \frac{2}{7} & \frac{2}{7} & -\frac{1}{7} \\ -\frac{5}{21} & \frac{2}{21} & -\frac{1}{21}\end{bmatrix}
Verification
  1. **Verify AA^{-1}=I.**
    AA^{-1}=\begin{bmatrix}0 & 1 & -3 \\ 3 & 2 & 4 \\ 6 & -1 & 2\end{bmatrix}\begin{bmatrix}\frac{8}{63} & \frac{1}{63} & \frac{10}{63} \\ \frac{2}{7} & \frac{2}{7} & -\frac{1}{7} \\ -\frac{5}{21} & \frac{2}{21} & -\frac{1}{21}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  2. **Verify A^{-1}A=I.**
    A^{-1}A=\begin{bmatrix}\frac{8}{63} & \frac{1}{63} & \frac{10}{63} \\ \frac{2}{7} & \frac{2}{7} & -\frac{1}{7} \\ -\frac{5}{21} & \frac{2}{21} & -\frac{1}{21}\end{bmatrix}\begin{bmatrix}0 & 1 & -3 \\ 3 & 2 & 4 \\ 6 & -1 & 2\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
A^{-1}=\begin{bmatrix}\frac{8}{63} & \frac{1}{63} & \frac{10}{63} \\ \frac{2}{7} & \frac{2}{7} & -\frac{1}{7} \\ -\frac{5}{21} & \frac{2}{21} & -\frac{1}{21}\end{bmatrix}

Book answer. (i) (1/2)[-12, -5, -3; -4, -1, -1; 2, 1, 1]; (ii) (1/6)[-2, 0, 2; 19, -15, -16; -6, 6, 6]; (iii) (1/12)[0, -6, 6; 3, -9, 6; 2, -4, 6]; (iv) [-8, 5, 2; -18, 18, -9; 15, -6, 3]. All agree, except that in (iv) the key omits the factor \frac1{27} in front of the matrix.

Note. Answer key, part (iv): the inverse is \frac{1}{27}\begin{bmatrix}-8&5&2\\-18&18&-9\\15&-6&3\end{bmatrix}. The key prints the matrix without the factor \frac1{27} (it does print \frac12, \frac16, \frac1{12} in the other parts). Multiplying out AA^{-1} shows the factor is required.

Exercise 2.6p. 70–71

Linear systems and applications. Homogeneous systems need |A|=0 for non-trivial solutions; non-homogeneous systems are solved by Gauss elimination, Cramer's rule or X=A^{-1}B (all require the ranks or |A| to be checked first); matrices also encode messages and move curves.

1
Solve the following system of homogeneous linear equations for non-trivial solution if exists.
Solution

Method (section 2.6.2). Write AX=O. If |A|\neq0 only the trivial solution x_1=x_2=x_3=0 exists. If |A|=0 there are infinitely many solutions: reduce A, let the free variable be t and express the others in terms of t.

(i) \begin{aligned}2x_1-3x_2+4x_3&=0 \\ x_1-2x_2+3x_3&=0 \\ 4x_1+x_2-6x_3&=0\end{aligned}
  1. A homogeneous system can be written as AX=O, where A is the matrix of coefficients. It always has the trivial solution X=O; a non-trivial solution exists only if |A|=0 (section 2.6.2).
    A=\begin{bmatrix}2 & -3 & 4 \\ 1 & -2 & 3 \\ 4 & 1 & -6\end{bmatrix},\quad X=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},\quad O=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=2\begin{vmatrix}-2 & 3 \\ 1 & -6\end{vmatrix}-(-3)\begin{vmatrix}1 & 3 \\ 4 & -6\end{vmatrix}+4\begin{vmatrix}1 & -2 \\ 4 & 1\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-2 & 3 \\ 1 & -6\end{vmatrix}=(-2)(-6)-(3)(1)=9 \\ \begin{vmatrix}1 & 3 \\ 4 & -6\end{vmatrix}=(1)(-6)-(3)(4)=-18 \\ \begin{vmatrix}1 & -2 \\ 4 & 1\end{vmatrix}=(1)(1)-(-2)(4)=9 \end{aligned}
  4. Substitute these values and add.
    |A|=2(9)-(-3)(-18)+4(9)=0
  5. |A|=0, so non-trivial solutions exist. Reduce the coefficient matrix by row operations (the zero right side stays zero).
  6. Interchange R_{1} and R_{2} (R_{12}) to get a leading entry 1 in position (1,1).
    \sim_R \begin{bmatrix}1 & -2 & 3 \\ 2 & -3 & 4 \\ 4 & 1 & -6\end{bmatrix}
  7. Apply R_{2}-2R_{1}, R_{3}-4R_{1} to make the other entries of column 1 zero.
    \sim_R \begin{bmatrix}1 & -2 & 3 \\ 0 & 1 & -2 \\ 0 & 9 & -18\end{bmatrix}
  8. Apply R_{1}+2R_{2}, R_{3}-9R_{2} to make the other entries of column 2 zero.
    \sim_R \begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}
  9. The last row is zero, so only two independent equations remain. Read them from the reduced matrix.
    \begin{aligned} x_1-x_3&=0 \\ x_2-2x_3&=0 \end{aligned}
  10. The variable x_3 is free: let x_3=t where t is any real number, and solve for the other two.
    \begin{aligned} x_1&=1t \\ x_2&=2t \\ x_3&=t \end{aligned}
  11. Check with t=1: substitute (1,2,1) into the original three equations; each gives 0.
(x_1,x_2,x_3)=(t,\ 2t,\ t),\ t\in\mathbb R
(ii) \begin{aligned}2x_1-3x_2+4x_3&=0 \\ x_1+x_2+x_3&=0 \\ x_1-4x_2+3x_3&=0\end{aligned}
  1. A homogeneous system can be written as AX=O, where A is the matrix of coefficients. It always has the trivial solution X=O; a non-trivial solution exists only if |A|=0 (section 2.6.2).
    A=\begin{bmatrix}2 & -3 & 4 \\ 1 & 1 & 1 \\ 1 & -4 & 3\end{bmatrix},\quad X=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},\quad O=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=2\begin{vmatrix}1 & 1 \\ -4 & 3\end{vmatrix}-(-3)\begin{vmatrix}1 & 1 \\ 1 & 3\end{vmatrix}+4\begin{vmatrix}1 & 1 \\ 1 & -4\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}1 & 1 \\ -4 & 3\end{vmatrix}=(1)(3)-(1)(-4)=7 \\ \begin{vmatrix}1 & 1 \\ 1 & 3\end{vmatrix}=(1)(3)-(1)(1)=2 \\ \begin{vmatrix}1 & 1 \\ 1 & -4\end{vmatrix}=(1)(-4)-(1)(1)=-5 \end{aligned}
  4. Substitute these values and add.
    |A|=2(7)-(-3)(2)+4(-5)=0
  5. |A|=0, so non-trivial solutions exist. Reduce the coefficient matrix by row operations (the zero right side stays zero).
  6. Interchange R_{1} and R_{2} (R_{12}) to get a leading entry 1 in position (1,1).
    \sim_R \begin{bmatrix}1 & 1 & 1 \\ 2 & -3 & 4 \\ 1 & -4 & 3\end{bmatrix}
  7. Apply R_{2}-2R_{1}, R_{3}-R_{1} to make the other entries of column 1 zero.
    \sim_R \begin{bmatrix}1 & 1 & 1 \\ 0 & -5 & 2 \\ 0 & -5 & 2\end{bmatrix}
  8. Apply -\frac{1}{5}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & -\frac{2}{5} \\ 0 & -5 & 2\end{bmatrix}
  9. Apply R_{1}-R_{2}, R_{3}+5R_{2} to make the other entries of column 2 zero.
    \sim_R \begin{bmatrix}1 & 0 & \frac{7}{5} \\ 0 & 1 & -\frac{2}{5} \\ 0 & 0 & 0\end{bmatrix}
  10. The last row is zero, so only two independent equations remain. Read them from the reduced matrix.
    \begin{aligned} x_1+\frac{7}{5}x_3&=0 \\ x_2-\frac{2}{5}x_3&=0 \end{aligned}
  11. The variable x_3 is free: let x_3=t where t is any real number, and solve for the other two.
    \begin{aligned} x_1&=-\frac{7}{5}t \\ x_2&=\frac{2}{5}t \\ x_3&=t \end{aligned}
  12. Check with t=1: substitute (-\frac{7}{5},\frac{2}{5},1) into the original three equations; each gives 0.
(x_1,x_2,x_3)=(-\frac{7}{5}t,\ \frac{2}{5}t,\ t),\ t\in\mathbb R
(iii) \begin{aligned}x_1+x_2-3x_3&=0 \\ 3x_1-2x_2+x_3&=0 \\ 4x_1-x_2-2x_3&=0\end{aligned}
  1. A homogeneous system can be written as AX=O, where A is the matrix of coefficients. It always has the trivial solution X=O; a non-trivial solution exists only if |A|=0 (section 2.6.2).
    A=\begin{bmatrix}1 & 1 & -3 \\ 3 & -2 & 1 \\ 4 & -1 & -2\end{bmatrix},\quad X=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},\quad O=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=1\begin{vmatrix}-2 & 1 \\ -1 & -2\end{vmatrix}-1\begin{vmatrix}3 & 1 \\ 4 & -2\end{vmatrix}+(-3)\begin{vmatrix}3 & -2 \\ 4 & -1\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-2 & 1 \\ -1 & -2\end{vmatrix}=(-2)(-2)-(1)(-1)=5 \\ \begin{vmatrix}3 & 1 \\ 4 & -2\end{vmatrix}=(3)(-2)-(1)(4)=-10 \\ \begin{vmatrix}3 & -2 \\ 4 & -1\end{vmatrix}=(3)(-1)-(-2)(4)=5 \end{aligned}
  4. Substitute these values and add.
    |A|=1(5)-1(-10)+(-3)(5)=0
  5. |A|=0, so non-trivial solutions exist. Reduce the coefficient matrix by row operations (the zero right side stays zero).
  6. Apply R_{2}-3R_{1}, R_{3}-4R_{1} to make the other entries of column 1 zero.
    \sim_R \begin{bmatrix}1 & 1 & -3 \\ 0 & -5 & 10 \\ 0 & -5 & 10\end{bmatrix}
  7. Apply -\frac{1}{5}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & 1 & -3 \\ 0 & 1 & -2 \\ 0 & -5 & 10\end{bmatrix}
  8. Apply R_{1}-R_{2}, R_{3}+5R_{2} to make the other entries of column 2 zero.
    \sim_R \begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}
  9. The last row is zero, so only two independent equations remain. Read them from the reduced matrix.
    \begin{aligned} x_1-x_3&=0 \\ x_2-2x_3&=0 \end{aligned}
  10. The variable x_3 is free: let x_3=t where t is any real number, and solve for the other two.
    \begin{aligned} x_1&=1t \\ x_2&=2t \\ x_3&=t \end{aligned}
  11. Check with t=1: substitute (1,2,1) into the original three equations; each gives 0.
(x_1,x_2,x_3)=(t,\ 2t,\ t),\ t\in\mathbb R
(iv) \begin{aligned}5x_1+6x_2-7x_3&=0 \\ 2x_1-x_2+x_3&=0 \\ x_1+2x_2+2x_3&=0\end{aligned}
  1. A homogeneous system can be written as AX=O, where A is the matrix of coefficients. It always has the trivial solution X=O; a non-trivial solution exists only if |A|=0 (section 2.6.2).
    A=\begin{bmatrix}5 & 6 & -7 \\ 2 & -1 & 1 \\ 1 & 2 & 2\end{bmatrix},\quad X=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},\quad O=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=5\begin{vmatrix}-1 & 1 \\ 2 & 2\end{vmatrix}-6\begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix}+(-7)\begin{vmatrix}2 & -1 \\ 1 & 2\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-1 & 1 \\ 2 & 2\end{vmatrix}=(-1)(2)-(1)(2)=-4 \\ \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix}=(2)(2)-(1)(1)=3 \\ \begin{vmatrix}2 & -1 \\ 1 & 2\end{vmatrix}=(2)(2)-(-1)(1)=5 \end{aligned}
  4. Substitute these values and add.
    |A|=5(-4)-6(3)+(-7)(5)=-73
  5. |A|=-73\neq0, so A^{-1} exists and X=A^{-1}O=O. The system has only the trivial solution, so no non-trivial solution exists.
    x_1=0,\ x_2=0,\ x_3=0
x_1=x_2=x_3=0\ \text{only (no non-trivial solution)}

Book answer. (i) (x3, 2x3, x3); (ii) (-7/5 x3, 2/5 x3, x3); (iii) (x3, 2x3, x3); (iv) does not exist. All agree (with x_3=t).

2
Find the value of \lambda for which the following system of homogeneous linear equations may have non-trivial solution. Also solve the system for value of \lambda.
Solution

Method. A homogeneous system has a non-trivial solution only when |A|=0. Find |A| as an expression in \lambda, solve |A|=0, then substitute each value of \lambda and solve AX=O with one free variable t.

(i) \begin{aligned}2x_1-\lambda x_2+x_3&=0 \\ 2x_1+3x_2-x_3&=0 \\ 3x_1-2x_2+4x_3&=0\end{aligned}
Find \lambda
  1. A homogeneous system has a non-trivial solution only if |A|=0, so find |A| in terms of \lambda. The parameter \lambda appears only in a_{12}, so expand along column 2.
    |A|=\begin{vmatrix}2 & -\lambda & 1 \\ 2 & 3 & -1 \\ 3 & -2 & 4\end{vmatrix}
  2. Column 2 is (-\lambda,3,-2). The signs for column 2 are -,+,-.
    |A|=-(-\lambda)\begin{vmatrix}2&-1\\3&4\end{vmatrix}+3\begin{vmatrix}2&1\\3&4\end{vmatrix}-(-2)\begin{vmatrix}2&1\\2&-1\end{vmatrix}
  3. Evaluate the three 2\times2 determinants: (2)(4)-(-1)(3)=11, (2)(4)-(1)(3)=5 and (2)(-1)-(1)(2)=-4.
    |A|=\lambda(11)+3(5)+2(-4)=11\lambda+7
  4. Set |A|=0 for a non-trivial solution.
    11\lambda+7=0\;\Rightarrow\;\lambda=-\frac{7}{11}
Solve the system for this \lambda
  1. Substitute \lambda=-\dfrac{7}{11} (so -\lambda=\dfrac{7}{11}) and solve AX=O.
    A=\begin{bmatrix}2 & \frac{7}{11} & 1 \\ 2 & 3 & -1 \\ 3 & -2 & 4\end{bmatrix}
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=2\begin{vmatrix}3 & -1 \\ -2 & 4\end{vmatrix}-\frac{7}{11}\begin{vmatrix}2 & -1 \\ 3 & 4\end{vmatrix}+1\begin{vmatrix}2 & 3 \\ 3 & -2\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}3 & -1 \\ -2 & 4\end{vmatrix}=(3)(4)-(-1)(-2)=10 \\ \begin{vmatrix}2 & -1 \\ 3 & 4\end{vmatrix}=(2)(4)-(-1)(3)=11 \\ \begin{vmatrix}2 & 3 \\ 3 & -2\end{vmatrix}=(2)(-2)-(3)(3)=-13 \end{aligned}
  4. Substitute these values and add.
    |A|=2(10)-\frac{7}{11}(11)+1(-13)=0
  5. |A|=0, so non-trivial solutions exist. Reduce the coefficient matrix by row operations (the zero right side stays zero).
  6. Apply \frac{1}{2}R_{1} to make the leading entry of row 1 equal to 1.
    \sim_R \begin{bmatrix}1 & \frac{7}{22} & \frac{1}{2} \\ 2 & 3 & -1 \\ 3 & -2 & 4\end{bmatrix}
  7. Apply R_{2}-2R_{1}, R_{3}-3R_{1} to make the other entries of column 1 zero.
    \sim_R \begin{bmatrix}1 & \frac{7}{22} & \frac{1}{2} \\ 0 & \frac{26}{11} & -2 \\ 0 & -\frac{65}{22} & \frac{5}{2}\end{bmatrix}
  8. Apply \frac{11}{26}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & \frac{7}{22} & \frac{1}{2} \\ 0 & 1 & -\frac{11}{13} \\ 0 & -\frac{65}{22} & \frac{5}{2}\end{bmatrix}
  9. Apply R_{1}-\frac{7}{22}R_{2}, R_{3}+\frac{65}{22}R_{2} to make the other entries of column 2 zero.
    \sim_R \begin{bmatrix}1 & 0 & \frac{10}{13} \\ 0 & 1 & -\frac{11}{13} \\ 0 & 0 & 0\end{bmatrix}
  10. The last row is zero, so only two independent equations remain. Read them from the reduced matrix.
    \begin{aligned} x_1+\frac{10}{13}x_3&=0 \\ x_2-\frac{11}{13}x_3&=0 \end{aligned}
  11. The variable x_3 is free: let x_3=t where t is any real number, and solve for the other two.
    \begin{aligned} x_1&=-\frac{10}{13}t \\ x_2&=\frac{11}{13}t \\ x_3&=t \end{aligned}
  12. Check with t=1: substitute (-\frac{10}{13},\frac{11}{13},1) into the original three equations; each gives 0.
\lambda=-\dfrac{7}{11},\quad (x_1,x_2,x_3)=(-\frac{10}{13}t,\ \frac{11}{13}t,\ t),\ t\in\mathbb R
(ii) \begin{aligned}x_1-4x_2+3x_3&=0 \\ 2x_1+\lambda x_2+x_3&=0 \\ x_1-2x_2+\lambda x_3&=0\end{aligned}
Find \lambda
  1. Find |A| in terms of \lambda. Expand along row 1 (it contains no \lambda).
    |A|=\begin{vmatrix}1 & -4 & 3 \\ 2 & \lambda & 1 \\ 1 & -2 & \lambda\end{vmatrix}
  2. Use the signs +,-,+ for row 1.
    |A|=1\begin{vmatrix}\lambda&1\\-2&\lambda\end{vmatrix}-(-4)\begin{vmatrix}2&1\\1&\lambda\end{vmatrix}+3\begin{vmatrix}2&\lambda\\1&-2\end{vmatrix}
  3. Evaluate the three 2\times2 determinants: \lambda\cdot\lambda-(1)(-2)=\lambda^2+2, 2\lambda-1 and (2)(-2)-\lambda(1)=-4-\lambda.
    |A|=(\lambda^2+2)+4(2\lambda-1)+3(-4-\lambda)
  4. Expand and collect.
    |A|=\lambda^2+2+8\lambda-4-12-3\lambda=\lambda^2+5\lambda-14
  5. Set |A|=0 and factorise.
    \lambda^2+5\lambda-14=0\;\Rightarrow\;(\lambda-2)(\lambda+7)=0\;\Rightarrow\;\lambda=2\ \text{or}\ \lambda=-7
Solution for \lambda=2
  1. **Case \lambda=2.** Substitute and solve AX=O.
    A=\begin{bmatrix}1 & -4 & 3 \\ 2 & 2 & 1 \\ 1 & -2 & 2\end{bmatrix}
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=1\begin{vmatrix}2 & 1 \\ -2 & 2\end{vmatrix}-(-4)\begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix}+3\begin{vmatrix}2 & 2 \\ 1 & -2\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}2 & 1 \\ -2 & 2\end{vmatrix}=(2)(2)-(1)(-2)=6 \\ \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix}=(2)(2)-(1)(1)=3 \\ \begin{vmatrix}2 & 2 \\ 1 & -2\end{vmatrix}=(2)(-2)-(2)(1)=-6 \end{aligned}
  4. Substitute these values and add.
    |A|=1(6)-(-4)(3)+3(-6)=0
  5. |A|=0, so non-trivial solutions exist. Reduce the coefficient matrix by row operations (the zero right side stays zero).
  6. Apply R_{2}-2R_{1}, R_{3}-R_{1} to make the other entries of column 1 zero.
    \sim_R \begin{bmatrix}1 & -4 & 3 \\ 0 & 10 & -5 \\ 0 & 2 & -1\end{bmatrix}
  7. Apply \frac{1}{10}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \begin{bmatrix}1 & -4 & 3 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 2 & -1\end{bmatrix}
  8. Apply R_{1}+4R_{2}, R_{3}-2R_{2} to make the other entries of column 2 zero.
    \sim_R \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{bmatrix}
  9. The last row is zero, so only two independent equations remain. Read them from the reduced matrix.
    \begin{aligned} x_1+x_3&=0 \\ x_2-\frac{1}{2}x_3&=0 \end{aligned}
  10. The variable x_3 is free: let x_3=t where t is any real number, and solve for the other two.
    \begin{aligned} x_1&=-1t \\ x_2&=\frac{1}{2}t \\ x_3&=t \end{aligned}
  11. Check with t=1: substitute (-1,\frac{1}{2},1) into the original three equations; each gives 0.
Solution for \lambda=-7
  1. **Case \lambda=-7.** Substitute and solve AX=O.
    A=\begin{bmatrix}1 & -4 & 3 \\ 2 & -7 & 1 \\ 1 & -2 & -7\end{bmatrix}
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=1\begin{vmatrix}-7 & 1 \\ -2 & -7\end{vmatrix}-(-4)\begin{vmatrix}2 & 1 \\ 1 & -7\end{vmatrix}+3\begin{vmatrix}2 & -7 \\ 1 & -2\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-7 & 1 \\ -2 & -7\end{vmatrix}=(-7)(-7)-(1)(-2)=51 \\ \begin{vmatrix}2 & 1 \\ 1 & -7\end{vmatrix}=(2)(-7)-(1)(1)=-15 \\ \begin{vmatrix}2 & -7 \\ 1 & -2\end{vmatrix}=(2)(-2)-(-7)(1)=3 \end{aligned}
  4. Substitute these values and add.
    |A|=1(51)-(-4)(-15)+3(3)=0
  5. |A|=0, so non-trivial solutions exist. Reduce the coefficient matrix by row operations (the zero right side stays zero).
  6. Apply R_{2}-2R_{1}, R_{3}-R_{1} to make the other entries of column 1 zero.
    \sim_R \begin{bmatrix}1 & -4 & 3 \\ 0 & 1 & -5 \\ 0 & 2 & -10\end{bmatrix}
  7. Apply R_{1}+4R_{2}, R_{3}-2R_{2} to make the other entries of column 2 zero.
    \sim_R \begin{bmatrix}1 & 0 & -17 \\ 0 & 1 & -5 \\ 0 & 0 & 0\end{bmatrix}
  8. The last row is zero, so only two independent equations remain. Read them from the reduced matrix.
    \begin{aligned} x_1-17x_3&=0 \\ x_2-5x_3&=0 \end{aligned}
  9. The variable x_3 is free: let x_3=t where t is any real number, and solve for the other two.
    \begin{aligned} x_1&=17t \\ x_2&=5t \\ x_3&=t \end{aligned}
  10. Check with t=1: substitute (17,5,1) into the original three equations; each gives 0.
\begin{aligned}\lambda=2:&\ (x_1,x_2,x_3)=(-t,\ \frac{1}{2}t,\ t)\\ \lambda=-7:&\ (x_1,x_2,x_3)=(17t,\ 5t,\ t),\ t\in\mathbb R\end{aligned}

Book answer. (i) λ = -7/11, (-10/13 x3, 11/13 x3, x3); (ii) λ = 2, (-x3, 1/2 x3, x3) or λ = -7, (17 x3, 5 x3, x3). All agree.

3
Solve the following system of linear equations by Gauss elimination method.
Solution

Method (section 2.6.8, Example 33). Reduce the augmented matrix A_b to echelon form (leading 1s, zeros below them). If \operatorname{Rank}A=\operatorname{Rank}A_b=3 read off z, then y, then x by back-substitution. If \operatorname{Rank}A\neq\operatorname{Rank}A_b the system is inconsistent.

(i) \begin{aligned} 2x+3y+4z&=2 \\ 2x+y+z&=5 \\ 3x-2y+z&=-3 \end{aligned}
  1. Write the system and its augmented matrix A_b=[A\,|\,B].
    \begin{aligned} 2x+3y+4z&=2 \\ 2x+y+z&=5 \\ 3x-2y+z&=-3 \end{aligned}\qquad A_b=\left[\begin{array}{ccc|c}2 & 3 & 4 & 2 \\ 2 & 1 & 1 & 5 \\ 3 & -2 & 1 & -3\end{array}\right]
  2. Apply \frac{1}{2}R_{1} to make the leading entry of row 1 equal to 1.
    \sim_R \left[\begin{array}{ccc|c}1 & \frac{3}{2} & 2 & 1 \\ 2 & 1 & 1 & 5 \\ 3 & -2 & 1 & -3\end{array}\right]
  3. Apply R_{2}-2R_{1}, R_{3}-3R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \left[\begin{array}{ccc|c}1 & \frac{3}{2} & 2 & 1 \\ 0 & -2 & -3 & 3 \\ 0 & -\frac{13}{2} & -5 & -6\end{array}\right]
  4. Apply -\frac{1}{2}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \left[\begin{array}{ccc|c}1 & \frac{3}{2} & 2 & 1 \\ 0 & 1 & \frac{3}{2} & -\frac{3}{2} \\ 0 & -\frac{13}{2} & -5 & -6\end{array}\right]
  5. Apply R_{3}+\frac{13}{2}R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \left[\begin{array}{ccc|c}1 & \frac{3}{2} & 2 & 1 \\ 0 & 1 & \frac{3}{2} & -\frac{3}{2} \\ 0 & 0 & \frac{19}{4} & -\frac{63}{4}\end{array}\right]
  6. Apply \frac{4}{19}R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \left[\begin{array}{ccc|c}1 & \frac{3}{2} & 2 & 1 \\ 0 & 1 & \frac{3}{2} & -\frac{3}{2} \\ 0 & 0 & 1 & -\frac{63}{19}\end{array}\right]
  7. \operatorname{Rank}A=\operatorname{Rank}A_b=3, the number of unknowns, so the system has a unique solution (Key Fact, section 2.6.8). Read the equations from the rows and solve from the bottom upwards (back-substitution).
  8. Row 3 gives z directly.
    0x+0y+z=-\frac{63}{19}\;\Rightarrow\;z=-\frac{63}{19}
  9. Row 2 gives y after substituting z.
    y+\frac{3}{2}z=-\frac{3}{2}\;\Rightarrow\;y=-\frac{3}{2}-(\frac{3}{2})(-\frac{63}{19})=\frac{66}{19}
  10. Row 1 gives x after substituting y and z.
    x+(\frac{3}{2})y+(2)z=1\;\Rightarrow\;x=1-(\frac{3}{2})(\frac{66}{19})-(2)(-\frac{63}{19})=\frac{46}{19}
  11. Check in the original equations (left sides evaluated with these values reproduce the right sides).
    \begin{aligned} (2)(\frac{46}{19}) + (3)(\frac{66}{19}) + (4)(-\frac{63}{19})&=2 \\ (2)(\frac{46}{19}) + (1)(\frac{66}{19}) + (1)(-\frac{63}{19})&=5 \\ (3)(\frac{46}{19}) + (-2)(\frac{66}{19}) + (1)(-\frac{63}{19})&=-3 \end{aligned}
x=\frac{46}{19},\ y=\frac{66}{19},\ z=-\frac{63}{19}
(ii) \begin{aligned} 5x-2y+z&=2 \\ 2x+2y+6z&=1 \\ 3x-4y-5z&=3 \end{aligned}
  1. Write the system and its augmented matrix A_b=[A\,|\,B].
    \begin{aligned} 5x-2y+z&=2 \\ 2x+2y+6z&=1 \\ 3x-4y-5z&=3 \end{aligned}\qquad A_b=\left[\begin{array}{ccc|c}5 & -2 & 1 & 2 \\ 2 & 2 & 6 & 1 \\ 3 & -4 & -5 & 3\end{array}\right]
  2. Apply \frac{1}{5}R_{1} to make the leading entry of row 1 equal to 1.
    \sim_R \left[\begin{array}{ccc|c}1 & -\frac{2}{5} & \frac{1}{5} & \frac{2}{5} \\ 2 & 2 & 6 & 1 \\ 3 & -4 & -5 & 3\end{array}\right]
  3. Apply R_{2}-2R_{1}, R_{3}-3R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \left[\begin{array}{ccc|c}1 & -\frac{2}{5} & \frac{1}{5} & \frac{2}{5} \\ 0 & \frac{14}{5} & \frac{28}{5} & \frac{1}{5} \\ 0 & -\frac{14}{5} & -\frac{28}{5} & \frac{9}{5}\end{array}\right]
  4. Apply \frac{5}{14}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \left[\begin{array}{ccc|c}1 & -\frac{2}{5} & \frac{1}{5} & \frac{2}{5} \\ 0 & 1 & 2 & \frac{1}{14} \\ 0 & -\frac{14}{5} & -\frac{28}{5} & \frac{9}{5}\end{array}\right]
  5. Apply R_{3}+\frac{14}{5}R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \left[\begin{array}{ccc|c}1 & -\frac{2}{5} & \frac{1}{5} & \frac{2}{5} \\ 0 & 1 & 2 & \frac{1}{14} \\ 0 & 0 & 0 & 2\end{array}\right]
  6. The last row reads 0x+0y+0z=2, which is impossible. In the echelon form \operatorname{Rank}A=2 but \operatorname{Rank}A_b=3.
  7. A system with \operatorname{Rank}A\neq\operatorname{Rank}A_b is inconsistent (section 2.6.6), so it has no solution.
\text{No solution (inconsistent system)}
(iii) \begin{aligned} 2x+z&=2 \\ 2y-z&=3 \\ x+3y&=5 \end{aligned}
  1. Write the system and its augmented matrix A_b=[A\,|\,B].
    \begin{aligned} 2x+z&=2 \\ 2y-z&=3 \\ x+3y&=5 \end{aligned}\qquad A_b=\left[\begin{array}{ccc|c}2 & 0 & 1 & 2 \\ 0 & 2 & -1 & 3 \\ 1 & 3 & 0 & 5\end{array}\right]
  2. Interchange R_{1} and R_{3} (R_{13}) to get a leading entry 1 in position (1,1).
    \sim_R \left[\begin{array}{ccc|c}1 & 3 & 0 & 5 \\ 0 & 2 & -1 & 3 \\ 2 & 0 & 1 & 2\end{array}\right]
  3. Apply R_{3}-2R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \left[\begin{array}{ccc|c}1 & 3 & 0 & 5 \\ 0 & 2 & -1 & 3 \\ 0 & -6 & 1 & -8\end{array}\right]
  4. Apply \frac{1}{2}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \left[\begin{array}{ccc|c}1 & 3 & 0 & 5 \\ 0 & 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & -6 & 1 & -8\end{array}\right]
  5. Apply R_{3}+6R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \left[\begin{array}{ccc|c}1 & 3 & 0 & 5 \\ 0 & 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & 0 & -2 & 1\end{array}\right]
  6. Apply -\frac{1}{2}R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \left[\begin{array}{ccc|c}1 & 3 & 0 & 5 \\ 0 & 1 & -\frac{1}{2} & \frac{3}{2} \\ 0 & 0 & 1 & -\frac{1}{2}\end{array}\right]
  7. \operatorname{Rank}A=\operatorname{Rank}A_b=3, the number of unknowns, so the system has a unique solution (Key Fact, section 2.6.8). Read the equations from the rows and solve from the bottom upwards (back-substitution).
  8. Row 3 gives z directly.
    0x+0y+z=-\frac{1}{2}\;\Rightarrow\;z=-\frac{1}{2}
  9. Row 2 gives y after substituting z.
    y-\frac{1}{2}z=\frac{3}{2}\;\Rightarrow\;y=\frac{3}{2}-(-\frac{1}{2})(-\frac{1}{2})=\frac{5}{4}
  10. Row 1 gives x after substituting y and z.
    x+(3)y+(0)z=5\;\Rightarrow\;x=5-(3)(\frac{5}{4})-(0)(-\frac{1}{2})=\frac{5}{4}
  11. Check in the original equations (left sides evaluated with these values reproduce the right sides).
    \begin{aligned} (2)(\frac{5}{4}) + (0)(\frac{5}{4}) + (1)(-\frac{1}{2})&=2 \\ (0)(\frac{5}{4}) + (2)(\frac{5}{4}) + (-1)(-\frac{1}{2})&=3 \\ (1)(\frac{5}{4}) + (3)(\frac{5}{4}) + (0)(-\frac{1}{2})&=5 \end{aligned}
x=\frac{5}{4},\ y=\frac{5}{4},\ z=-\frac{1}{2}
(iv) \begin{aligned} x+2y+5z&=4 \\ 3x-2y+2z&=3 \\ 5x-8y-4z&=1 \end{aligned}
  1. Write the system and its augmented matrix A_b=[A\,|\,B].
    \begin{aligned} x+2y+5z&=4 \\ 3x-2y+2z&=3 \\ 5x-8y-4z&=1 \end{aligned}\qquad A_b=\left[\begin{array}{ccc|c}1 & 2 & 5 & 4 \\ 3 & -2 & 2 & 3 \\ 5 & -8 & -4 & 1\end{array}\right]
  2. Apply R_{2}-3R_{1}, R_{3}-5R_{1} to make the entries below the leading 1 in column 1 zero.
    \sim_R \left[\begin{array}{ccc|c}1 & 2 & 5 & 4 \\ 0 & -8 & -13 & -9 \\ 0 & -18 & -29 & -19\end{array}\right]
  3. Apply -\frac{1}{8}R_{2} to make the leading entry of row 2 equal to 1.
    \sim_R \left[\begin{array}{ccc|c}1 & 2 & 5 & 4 \\ 0 & 1 & \frac{13}{8} & \frac{9}{8} \\ 0 & -18 & -29 & -19\end{array}\right]
  4. Apply R_{3}+18R_{2} to make the entries below the leading 1 in column 2 zero.
    \sim_R \left[\begin{array}{ccc|c}1 & 2 & 5 & 4 \\ 0 & 1 & \frac{13}{8} & \frac{9}{8} \\ 0 & 0 & \frac{1}{4} & \frac{5}{4}\end{array}\right]
  5. Apply 4R_{3} to make the leading entry of row 3 equal to 1.
    \sim_R \left[\begin{array}{ccc|c}1 & 2 & 5 & 4 \\ 0 & 1 & \frac{13}{8} & \frac{9}{8} \\ 0 & 0 & 1 & 5\end{array}\right]
  6. \operatorname{Rank}A=\operatorname{Rank}A_b=3, the number of unknowns, so the system has a unique solution (Key Fact, section 2.6.8). Read the equations from the rows and solve from the bottom upwards (back-substitution).
  7. Row 3 gives z directly.
    0x+0y+z=5\;\Rightarrow\;z=5
  8. Row 2 gives y after substituting z.
    y+\frac{13}{8}z=\frac{9}{8}\;\Rightarrow\;y=\frac{9}{8}-(\frac{13}{8})(5)=-7
  9. Row 1 gives x after substituting y and z.
    x+(2)y+(5)z=4\;\Rightarrow\;x=4-(2)(-7)-(5)(5)=-7
  10. Check in the original equations (left sides evaluated with these values reproduce the right sides).
    \begin{aligned} (1)(-7) + (2)(-7) + (5)(5)&=4 \\ (3)(-7) + (-2)(-7) + (2)(5)&=3 \\ (5)(-7) + (-8)(-7) + (-4)(5)&=1 \end{aligned}
x=-7,\ y=-7,\ z=5

Book answer. (i) the key prints only 66/19 and -63/19 (the value of x is missing); (ii) no solution; (iii) 5/4, 5/4, -1/2; (iv) -7, -7, 5. All printed values agree.

Note. Answer key, part (i): only y=\frac{66}{19} and z=-\frac{63}{19} are printed; the first value is missing. The full solution is x=\frac{46}{19}, y=\frac{66}{19}, z=-\frac{63}{19} (checked by substitution).

4
Solve the following system of linear equations by using Cramer's rule.
Solution

Method (section 2.6.8). For AX=B with |A|\neq0: x_1=\dfrac{|A_1|}{|A|}, x_2=\dfrac{|A_2|}{|A|}, x_3=\dfrac{|A_3|}{|A|}, where A_j is A with its jth column replaced by the column B of constants.

(i) \begin{aligned} x_1+x_2+2x_3&=8 \\ -x_1-2x_2+3x_3&=1 \\ 3x_1-7x_2+4x_3&=10 \end{aligned}
  1. Write the system as AX=B.
    A=\begin{bmatrix}1 & 1 & 2 \\ -1 & -2 & 3 \\ 3 & -7 & 4\end{bmatrix},\quad X=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},\quad B=\begin{bmatrix}8 \\ 1 \\ 10\end{bmatrix}
  2. Cramer's rule needs |A|\neq0. Find |A| first.
  3. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=1\begin{vmatrix}-2 & 3 \\ -7 & 4\end{vmatrix}-1\begin{vmatrix}-1 & 3 \\ 3 & 4\end{vmatrix}+2\begin{vmatrix}-1 & -2 \\ 3 & -7\end{vmatrix}
  4. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-2 & 3 \\ -7 & 4\end{vmatrix}=(-2)(4)-(3)(-7)=13 \\ \begin{vmatrix}-1 & 3 \\ 3 & 4\end{vmatrix}=(-1)(4)-(3)(3)=-13 \\ \begin{vmatrix}-1 & -2 \\ 3 & -7\end{vmatrix}=(-1)(-7)-(-2)(3)=13 \end{aligned}
  5. Substitute these values and add.
    |A|=1(13)-1(-13)+2(13)=52
  6. |A|=52\neq0, so there is a unique solution x_j=\dfrac{|A_j|}{|A|}, where A_j is A with column j replaced by B.
  7. **Column 1 replaced by B.**
    A_{1}=\begin{bmatrix}8 & 1 & 2 \\ 1 & -2 & 3 \\ 10 & -7 & 4\end{bmatrix}
  8. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A_{1}|=8\begin{vmatrix}-2 & 3 \\ -7 & 4\end{vmatrix}-1\begin{vmatrix}1 & 3 \\ 10 & 4\end{vmatrix}+2\begin{vmatrix}1 & -2 \\ 10 & -7\end{vmatrix}
  9. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-2 & 3 \\ -7 & 4\end{vmatrix}=(-2)(4)-(3)(-7)=13 \\ \begin{vmatrix}1 & 3 \\ 10 & 4\end{vmatrix}=(1)(4)-(3)(10)=-26 \\ \begin{vmatrix}1 & -2 \\ 10 & -7\end{vmatrix}=(1)(-7)-(-2)(10)=13 \end{aligned}
  10. Substitute these values and add.
    |A_{1}|=8(13)-1(-26)+2(13)=156
  11. Divide by |A|.
    x_{1}=\frac{|A_{1}|}{|A|}=\frac{156}{52}=3
  12. **Column 2 replaced by B.**
    A_{2}=\begin{bmatrix}1 & 8 & 2 \\ -1 & 1 & 3 \\ 3 & 10 & 4\end{bmatrix}
  13. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A_{2}|=1\begin{vmatrix}1 & 3 \\ 10 & 4\end{vmatrix}-8\begin{vmatrix}-1 & 3 \\ 3 & 4\end{vmatrix}+2\begin{vmatrix}-1 & 1 \\ 3 & 10\end{vmatrix}
  14. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}1 & 3 \\ 10 & 4\end{vmatrix}=(1)(4)-(3)(10)=-26 \\ \begin{vmatrix}-1 & 3 \\ 3 & 4\end{vmatrix}=(-1)(4)-(3)(3)=-13 \\ \begin{vmatrix}-1 & 1 \\ 3 & 10\end{vmatrix}=(-1)(10)-(1)(3)=-13 \end{aligned}
  15. Substitute these values and add.
    |A_{2}|=1(-26)-8(-13)+2(-13)=52
  16. Divide by |A|.
    x_{2}=\frac{|A_{2}|}{|A|}=\frac{52}{52}=1
  17. **Column 3 replaced by B.**
    A_{3}=\begin{bmatrix}1 & 1 & 8 \\ -1 & -2 & 1 \\ 3 & -7 & 10\end{bmatrix}
  18. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A_{3}|=1\begin{vmatrix}-2 & 1 \\ -7 & 10\end{vmatrix}-1\begin{vmatrix}-1 & 1 \\ 3 & 10\end{vmatrix}+8\begin{vmatrix}-1 & -2 \\ 3 & -7\end{vmatrix}
  19. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-2 & 1 \\ -7 & 10\end{vmatrix}=(-2)(10)-(1)(-7)=-13 \\ \begin{vmatrix}-1 & 1 \\ 3 & 10\end{vmatrix}=(-1)(10)-(1)(3)=-13 \\ \begin{vmatrix}-1 & -2 \\ 3 & -7\end{vmatrix}=(-1)(-7)-(-2)(3)=13 \end{aligned}
  20. Substitute these values and add.
    |A_{3}|=1(-13)-1(-13)+8(13)=104
  21. Divide by |A|.
    x_{3}=\frac{|A_{3}|}{|A|}=\frac{104}{52}=2
  22. Check: substituting the three values in the equations reproduces the right sides.
x_1=3,\ x_2=1,\ x_3=2
(ii) \begin{aligned} 2x_1+2x_2+x_3&=0 \\ -2x_1+5x_2+2x_3&=1 \\ 8x_1+x_2+4x_3&=-1 \end{aligned}
  1. Write the system as AX=B.
    A=\begin{bmatrix}2 & 2 & 1 \\ -2 & 5 & 2 \\ 8 & 1 & 4\end{bmatrix},\quad X=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},\quad B=\begin{bmatrix}0 \\ 1 \\ -1\end{bmatrix}
  2. Cramer's rule needs |A|\neq0. Find |A| first.
  3. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=2\begin{vmatrix}5 & 2 \\ 1 & 4\end{vmatrix}-2\begin{vmatrix}-2 & 2 \\ 8 & 4\end{vmatrix}+1\begin{vmatrix}-2 & 5 \\ 8 & 1\end{vmatrix}
  4. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}5 & 2 \\ 1 & 4\end{vmatrix}=(5)(4)-(2)(1)=18 \\ \begin{vmatrix}-2 & 2 \\ 8 & 4\end{vmatrix}=(-2)(4)-(2)(8)=-24 \\ \begin{vmatrix}-2 & 5 \\ 8 & 1\end{vmatrix}=(-2)(1)-(5)(8)=-42 \end{aligned}
  5. Substitute these values and add.
    |A|=2(18)-2(-24)+1(-42)=42
  6. |A|=42\neq0, so there is a unique solution x_j=\dfrac{|A_j|}{|A|}, where A_j is A with column j replaced by B.
  7. **Column 1 replaced by B.**
    A_{1}=\begin{bmatrix}0 & 2 & 1 \\ 1 & 5 & 2 \\ -1 & 1 & 4\end{bmatrix}
  8. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A_{1}|=0\begin{vmatrix}5 & 2 \\ 1 & 4\end{vmatrix}-2\begin{vmatrix}1 & 2 \\ -1 & 4\end{vmatrix}+1\begin{vmatrix}1 & 5 \\ -1 & 1\end{vmatrix}
  9. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}5 & 2 \\ 1 & 4\end{vmatrix}=(5)(4)-(2)(1)=18 \\ \begin{vmatrix}1 & 2 \\ -1 & 4\end{vmatrix}=(1)(4)-(2)(-1)=6 \\ \begin{vmatrix}1 & 5 \\ -1 & 1\end{vmatrix}=(1)(1)-(5)(-1)=6 \end{aligned}
  10. Substitute these values and add.
    |A_{1}|=0(18)-2(6)+1(6)=-6
  11. Divide by |A|.
    x_{1}=\frac{|A_{1}|}{|A|}=\frac{-6}{42}=-\frac{1}{7}
  12. **Column 2 replaced by B.**
    A_{2}=\begin{bmatrix}2 & 0 & 1 \\ -2 & 1 & 2 \\ 8 & -1 & 4\end{bmatrix}
  13. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A_{2}|=2\begin{vmatrix}1 & 2 \\ -1 & 4\end{vmatrix}-0\begin{vmatrix}-2 & 2 \\ 8 & 4\end{vmatrix}+1\begin{vmatrix}-2 & 1 \\ 8 & -1\end{vmatrix}
  14. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}1 & 2 \\ -1 & 4\end{vmatrix}=(1)(4)-(2)(-1)=6 \\ \begin{vmatrix}-2 & 2 \\ 8 & 4\end{vmatrix}=(-2)(4)-(2)(8)=-24 \\ \begin{vmatrix}-2 & 1 \\ 8 & -1\end{vmatrix}=(-2)(-1)-(1)(8)=-6 \end{aligned}
  15. Substitute these values and add.
    |A_{2}|=2(6)-0(-24)+1(-6)=6
  16. Divide by |A|.
    x_{2}=\frac{|A_{2}|}{|A|}=\frac{6}{42}=\frac{1}{7}
  17. **Column 3 replaced by B.**
    A_{3}=\begin{bmatrix}2 & 2 & 0 \\ -2 & 5 & 1 \\ 8 & 1 & -1\end{bmatrix}
  18. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A_{3}|=2\begin{vmatrix}5 & 1 \\ 1 & -1\end{vmatrix}-2\begin{vmatrix}-2 & 1 \\ 8 & -1\end{vmatrix}+0\begin{vmatrix}-2 & 5 \\ 8 & 1\end{vmatrix}
  19. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}5 & 1 \\ 1 & -1\end{vmatrix}=(5)(-1)-(1)(1)=-6 \\ \begin{vmatrix}-2 & 1 \\ 8 & -1\end{vmatrix}=(-2)(-1)-(1)(8)=-6 \\ \begin{vmatrix}-2 & 5 \\ 8 & 1\end{vmatrix}=(-2)(1)-(5)(8)=-42 \end{aligned}
  20. Substitute these values and add.
    |A_{3}|=2(-6)-2(-6)+0(-42)=0
  21. Divide by |A|.
    x_{3}=\frac{|A_{3}|}{|A|}=\frac{0}{42}=0
  22. Check: substituting the three values in the equations reproduces the right sides.
x_1=-\frac{1}{7},\ x_2=\frac{1}{7},\ x_3=0
(iii) \begin{aligned} -2x_2+3x_3&=1 \\ 3x_1+6x_2-3x_3&=-2 \\ 6x_1+6x_2+3x_3&=5 \end{aligned}
  1. Write the system as AX=B.
    A=\begin{bmatrix}0 & -2 & 3 \\ 3 & 6 & -3 \\ 6 & 6 & 3\end{bmatrix},\quad X=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},\quad B=\begin{bmatrix}1 \\ -2 \\ 5\end{bmatrix}
  2. Cramer's rule needs |A|\neq0. Find |A| first.
  3. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=0\begin{vmatrix}6 & -3 \\ 6 & 3\end{vmatrix}-(-2)\begin{vmatrix}3 & -3 \\ 6 & 3\end{vmatrix}+3\begin{vmatrix}3 & 6 \\ 6 & 6\end{vmatrix}
  4. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}6 & -3 \\ 6 & 3\end{vmatrix}=(6)(3)-(-3)(6)=36 \\ \begin{vmatrix}3 & -3 \\ 6 & 3\end{vmatrix}=(3)(3)-(-3)(6)=27 \\ \begin{vmatrix}3 & 6 \\ 6 & 6\end{vmatrix}=(3)(6)-(6)(6)=-18 \end{aligned}
  5. Substitute these values and add.
    |A|=0(36)-(-2)(27)+3(-18)=0
  6. |A|=0, so A is singular and Cramer's rule (like the inverse method) cannot be used.
  7. Remark: reducing the augmented matrix gives \operatorname{Rank}A=2 and \operatorname{Rank}A_b=3. Since the ranks differ, the system is in fact inconsistent and has no solution.
|A|=0:\ \text{Cramer's rule cannot be applied}
(iv) \begin{aligned} 2x_1+x_2+3x_3&=1 \\ x_1-2x_2+x_3&=2 \\ 3x_1-4x_2-x_3&=4 \end{aligned}
  1. Write the system as AX=B.
    A=\begin{bmatrix}2 & 1 & 3 \\ 1 & -2 & 1 \\ 3 & -4 & -1\end{bmatrix},\quad X=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},\quad B=\begin{bmatrix}1 \\ 2 \\ 4\end{bmatrix}
  2. Cramer's rule needs |A|\neq0. Find |A| first.
  3. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=2\begin{vmatrix}-2 & 1 \\ -4 & -1\end{vmatrix}-1\begin{vmatrix}1 & 1 \\ 3 & -1\end{vmatrix}+3\begin{vmatrix}1 & -2 \\ 3 & -4\end{vmatrix}
  4. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-2 & 1 \\ -4 & -1\end{vmatrix}=(-2)(-1)-(1)(-4)=6 \\ \begin{vmatrix}1 & 1 \\ 3 & -1\end{vmatrix}=(1)(-1)-(1)(3)=-4 \\ \begin{vmatrix}1 & -2 \\ 3 & -4\end{vmatrix}=(1)(-4)-(-2)(3)=2 \end{aligned}
  5. Substitute these values and add.
    |A|=2(6)-1(-4)+3(2)=22
  6. |A|=22\neq0, so there is a unique solution x_j=\dfrac{|A_j|}{|A|}, where A_j is A with column j replaced by B.
  7. **Column 1 replaced by B.**
    A_{1}=\begin{bmatrix}1 & 1 & 3 \\ 2 & -2 & 1 \\ 4 & -4 & -1\end{bmatrix}
  8. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A_{1}|=1\begin{vmatrix}-2 & 1 \\ -4 & -1\end{vmatrix}-1\begin{vmatrix}2 & 1 \\ 4 & -1\end{vmatrix}+3\begin{vmatrix}2 & -2 \\ 4 & -4\end{vmatrix}
  9. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-2 & 1 \\ -4 & -1\end{vmatrix}=(-2)(-1)-(1)(-4)=6 \\ \begin{vmatrix}2 & 1 \\ 4 & -1\end{vmatrix}=(2)(-1)-(1)(4)=-6 \\ \begin{vmatrix}2 & -2 \\ 4 & -4\end{vmatrix}=(2)(-4)-(-2)(4)=0 \end{aligned}
  10. Substitute these values and add.
    |A_{1}|=1(6)-1(-6)+3(0)=12
  11. Divide by |A|.
    x_{1}=\frac{|A_{1}|}{|A|}=\frac{12}{22}=\frac{6}{11}
  12. **Column 2 replaced by B.**
    A_{2}=\begin{bmatrix}2 & 1 & 3 \\ 1 & 2 & 1 \\ 3 & 4 & -1\end{bmatrix}
  13. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A_{2}|=2\begin{vmatrix}2 & 1 \\ 4 & -1\end{vmatrix}-1\begin{vmatrix}1 & 1 \\ 3 & -1\end{vmatrix}+3\begin{vmatrix}1 & 2 \\ 3 & 4\end{vmatrix}
  14. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}2 & 1 \\ 4 & -1\end{vmatrix}=(2)(-1)-(1)(4)=-6 \\ \begin{vmatrix}1 & 1 \\ 3 & -1\end{vmatrix}=(1)(-1)-(1)(3)=-4 \\ \begin{vmatrix}1 & 2 \\ 3 & 4\end{vmatrix}=(1)(4)-(2)(3)=-2 \end{aligned}
  15. Substitute these values and add.
    |A_{2}|=2(-6)-1(-4)+3(-2)=-14
  16. Divide by |A|.
    x_{2}=\frac{|A_{2}|}{|A|}=\frac{-14}{22}=-\frac{7}{11}
  17. **Column 3 replaced by B.**
    A_{3}=\begin{bmatrix}2 & 1 & 1 \\ 1 & -2 & 2 \\ 3 & -4 & 4\end{bmatrix}
  18. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A_{3}|=2\begin{vmatrix}-2 & 2 \\ -4 & 4\end{vmatrix}-1\begin{vmatrix}1 & 2 \\ 3 & 4\end{vmatrix}+1\begin{vmatrix}1 & -2 \\ 3 & -4\end{vmatrix}
  19. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-2 & 2 \\ -4 & 4\end{vmatrix}=(-2)(4)-(2)(-4)=0 \\ \begin{vmatrix}1 & 2 \\ 3 & 4\end{vmatrix}=(1)(4)-(2)(3)=-2 \\ \begin{vmatrix}1 & -2 \\ 3 & -4\end{vmatrix}=(1)(-4)-(-2)(3)=2 \end{aligned}
  20. Substitute these values and add.
    |A_{3}|=2(0)-1(-2)+1(2)=4
  21. Divide by |A|.
    x_{3}=\frac{|A_{3}|}{|A|}=\frac{4}{22}=\frac{2}{11}
  22. Check: substituting the three values in the equations reproduces the right sides.
x_1=\frac{6}{11},\ x_2=-\frac{7}{11},\ x_3=\frac{2}{11}

Book answer. (i) 3, 1, 2; (ii) -1/7, 1/7, 0; (iii) not possible as A is singular; (iv) 6/11, -7/11, 2/11. All agree.

5
Solve the following system of linear equations by matrix inversion method.
Solution

Method (section 2.6.8). Write AX=B. If |A|\neq0 then A^{-1} exists and X=A^{-1}B. Find A^{-1} by the adjoint method: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A).

(i) \begin{aligned} 5x+3y+z&=6 \\ 2x+y+3z&=19 \\ x+2y+4z&=25 \end{aligned}
  1. Write the system as AX=B. The solution is X=A^{-1}B provided |A|\neq0 (section 2.6.8).
    A=\begin{bmatrix}5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}6 \\ 19 \\ 25\end{bmatrix}
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=5\begin{vmatrix}1 & 3 \\ 2 & 4\end{vmatrix}-3\begin{vmatrix}2 & 3 \\ 1 & 4\end{vmatrix}+1\begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}1 & 3 \\ 2 & 4\end{vmatrix}=(1)(4)-(3)(2)=-2 \\ \begin{vmatrix}2 & 3 \\ 1 & 4\end{vmatrix}=(2)(4)-(3)(1)=5 \\ \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix}=(2)(2)-(1)(1)=3 \end{aligned}
  4. Substitute these values and add. So |A|=-22.
    |A|=5(-2)-3(5)+1(3)=-22
  5. |A|\neq0, so A is non-singular and A^{-1} exists. Find all cofactors.
    \begin{aligned} A_{11}=(-1)^{2}\begin{vmatrix}1 & 3 \\ 2 & 4\end{vmatrix}=+\left[(1)(4)-(3)(2)\right]=-2 \\ A_{12}=(-1)^{3}\begin{vmatrix}2 & 3 \\ 1 & 4\end{vmatrix}=-\left[(2)(4)-(3)(1)\right]=-5 \\ A_{13}=(-1)^{4}\begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix}=+\left[(2)(2)-(1)(1)\right]=3 \\ A_{21}=(-1)^{3}\begin{vmatrix}3 & 1 \\ 2 & 4\end{vmatrix}=-\left[(3)(4)-(1)(2)\right]=-10 \\ A_{22}=(-1)^{4}\begin{vmatrix}5 & 1 \\ 1 & 4\end{vmatrix}=+\left[(5)(4)-(1)(1)\right]=19 \\ A_{23}=(-1)^{5}\begin{vmatrix}5 & 3 \\ 1 & 2\end{vmatrix}=-\left[(5)(2)-(3)(1)\right]=-7 \\ A_{31}=(-1)^{4}\begin{vmatrix}3 & 1 \\ 1 & 3\end{vmatrix}=+\left[(3)(3)-(1)(1)\right]=8 \\ A_{32}=(-1)^{5}\begin{vmatrix}5 & 1 \\ 2 & 3\end{vmatrix}=-\left[(5)(3)-(1)(2)\right]=-13 \\ A_{33}=(-1)^{6}\begin{vmatrix}5 & 3 \\ 2 & 1\end{vmatrix}=+\left[(5)(1)-(3)(2)\right]=-1 \end{aligned}
  6. The adjoint is the transpose of the cofactor matrix: the cofactor A_{ij} goes to row j, column i.
    \operatorname{adj}(A)=\begin{bmatrix}-2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1\end{bmatrix}
  7. Divide by the determinant: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A).
    A^{-1}=\frac{1}{-22}\begin{bmatrix}-2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1\end{bmatrix}=\begin{bmatrix}\frac{1}{11} & \frac{5}{11} & -\frac{4}{11} \\ \frac{5}{22} & -\frac{19}{22} & \frac{13}{22} \\ -\frac{3}{22} & \frac{7}{22} & \frac{1}{22}\end{bmatrix}
  8. Check: AA^{-1} should be I.
    AA^{-1}=\begin{bmatrix}5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{bmatrix}\begin{bmatrix}\frac{1}{11} & \frac{5}{11} & -\frac{4}{11} \\ \frac{5}{22} & -\frac{19}{22} & \frac{13}{22} \\ -\frac{3}{22} & \frac{7}{22} & \frac{1}{22}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  9. Multiply: X=A^{-1}B.
    X=\begin{bmatrix}\frac{1}{11} & \frac{5}{11} & -\frac{4}{11} \\ \frac{5}{22} & -\frac{19}{22} & \frac{13}{22} \\ -\frac{3}{22} & \frac{7}{22} & \frac{1}{22}\end{bmatrix}\begin{bmatrix}6 \\ 19 \\ 25\end{bmatrix}=\begin{bmatrix}\frac{1}{11} \\ -\frac{3}{11} \\ \frac{70}{11}\end{bmatrix}
  10. State the solution and check it in the original equations.
    x=\frac{1}{11},\ y=-\frac{3}{11},\ z=\frac{70}{11}
x=\frac{1}{11},\ y=-\frac{3}{11},\ z=\frac{70}{11}
(ii) \begin{aligned} x+2y-3z&=5 \\ 2x-3y+2z&=1 \\ -x+2y-5z&=-3 \end{aligned}
  1. Write the system as AX=B. The solution is X=A^{-1}B provided |A|\neq0 (section 2.6.8).
    A=\begin{bmatrix}1 & 2 & -3 \\ 2 & -3 & 2 \\ -1 & 2 & -5\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}5 \\ 1 \\ -3\end{bmatrix}
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=1\begin{vmatrix}-3 & 2 \\ 2 & -5\end{vmatrix}-2\begin{vmatrix}2 & 2 \\ -1 & -5\end{vmatrix}+(-3)\begin{vmatrix}2 & -3 \\ -1 & 2\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-3 & 2 \\ 2 & -5\end{vmatrix}=(-3)(-5)-(2)(2)=11 \\ \begin{vmatrix}2 & 2 \\ -1 & -5\end{vmatrix}=(2)(-5)-(2)(-1)=-8 \\ \begin{vmatrix}2 & -3 \\ -1 & 2\end{vmatrix}=(2)(2)-(-3)(-1)=1 \end{aligned}
  4. Substitute these values and add. So |A|=24.
    |A|=1(11)-2(-8)+(-3)(1)=24
  5. |A|\neq0, so A is non-singular and A^{-1} exists. Find all cofactors.
    \begin{aligned} A_{11}=(-1)^{2}\begin{vmatrix}-3 & 2 \\ 2 & -5\end{vmatrix}=+\left[(-3)(-5)-(2)(2)\right]=11 \\ A_{12}=(-1)^{3}\begin{vmatrix}2 & 2 \\ -1 & -5\end{vmatrix}=-\left[(2)(-5)-(2)(-1)\right]=8 \\ A_{13}=(-1)^{4}\begin{vmatrix}2 & -3 \\ -1 & 2\end{vmatrix}=+\left[(2)(2)-(-3)(-1)\right]=1 \\ A_{21}=(-1)^{3}\begin{vmatrix}2 & -3 \\ 2 & -5\end{vmatrix}=-\left[(2)(-5)-(-3)(2)\right]=4 \\ A_{22}=(-1)^{4}\begin{vmatrix}1 & -3 \\ -1 & -5\end{vmatrix}=+\left[(1)(-5)-(-3)(-1)\right]=-8 \\ A_{23}=(-1)^{5}\begin{vmatrix}1 & 2 \\ -1 & 2\end{vmatrix}=-\left[(1)(2)-(2)(-1)\right]=-4 \\ A_{31}=(-1)^{4}\begin{vmatrix}2 & -3 \\ -3 & 2\end{vmatrix}=+\left[(2)(2)-(-3)(-3)\right]=-5 \\ A_{32}=(-1)^{5}\begin{vmatrix}1 & -3 \\ 2 & 2\end{vmatrix}=-\left[(1)(2)-(-3)(2)\right]=-8 \\ A_{33}=(-1)^{6}\begin{vmatrix}1 & 2 \\ 2 & -3\end{vmatrix}=+\left[(1)(-3)-(2)(2)\right]=-7 \end{aligned}
  6. The adjoint is the transpose of the cofactor matrix: the cofactor A_{ij} goes to row j, column i.
    \operatorname{adj}(A)=\begin{bmatrix}11 & 4 & -5 \\ 8 & -8 & -8 \\ 1 & -4 & -7\end{bmatrix}
  7. Divide by the determinant: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A).
    A^{-1}=\frac{1}{24}\begin{bmatrix}11 & 4 & -5 \\ 8 & -8 & -8 \\ 1 & -4 & -7\end{bmatrix}=\begin{bmatrix}\frac{11}{24} & \frac{1}{6} & -\frac{5}{24} \\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} \\ \frac{1}{24} & -\frac{1}{6} & -\frac{7}{24}\end{bmatrix}
  8. Check: AA^{-1} should be I.
    AA^{-1}=\begin{bmatrix}1 & 2 & -3 \\ 2 & -3 & 2 \\ -1 & 2 & -5\end{bmatrix}\begin{bmatrix}\frac{11}{24} & \frac{1}{6} & -\frac{5}{24} \\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} \\ \frac{1}{24} & -\frac{1}{6} & -\frac{7}{24}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  9. Multiply: X=A^{-1}B.
    X=\begin{bmatrix}\frac{11}{24} & \frac{1}{6} & -\frac{5}{24} \\ \frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} \\ \frac{1}{24} & -\frac{1}{6} & -\frac{7}{24}\end{bmatrix}\begin{bmatrix}5 \\ 1 \\ -3\end{bmatrix}=\begin{bmatrix}\frac{37}{12} \\ \frac{7}{3} \\ \frac{11}{12}\end{bmatrix}
  10. State the solution and check it in the original equations.
    x=\frac{37}{12},\ y=\frac{7}{3},\ z=\frac{11}{12}
x=\frac{37}{12},\ y=\frac{7}{3},\ z=\frac{11}{12}
(iii) \begin{aligned} -x+3y-5z&=0 \\ 2x+4y-6z&=1 \\ x-2y+3z&=3 \end{aligned}
  1. Write the system as AX=B. The solution is X=A^{-1}B provided |A|\neq0 (section 2.6.8).
    A=\begin{bmatrix}-1 & 3 & -5 \\ 2 & 4 & -6 \\ 1 & -2 & 3\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}0 \\ 1 \\ 3\end{bmatrix}
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=(-1)\begin{vmatrix}4 & -6 \\ -2 & 3\end{vmatrix}-3\begin{vmatrix}2 & -6 \\ 1 & 3\end{vmatrix}+(-5)\begin{vmatrix}2 & 4 \\ 1 & -2\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}4 & -6 \\ -2 & 3\end{vmatrix}=(4)(3)-(-6)(-2)=0 \\ \begin{vmatrix}2 & -6 \\ 1 & 3\end{vmatrix}=(2)(3)-(-6)(1)=12 \\ \begin{vmatrix}2 & 4 \\ 1 & -2\end{vmatrix}=(2)(-2)-(4)(1)=-8 \end{aligned}
  4. Substitute these values and add. So |A|=4.
    |A|=(-1)(0)-3(12)+(-5)(-8)=4
  5. |A|\neq0, so A is non-singular and A^{-1} exists. Find all cofactors.
    \begin{aligned} A_{11}=(-1)^{2}\begin{vmatrix}4 & -6 \\ -2 & 3\end{vmatrix}=+\left[(4)(3)-(-6)(-2)\right]=0 \\ A_{12}=(-1)^{3}\begin{vmatrix}2 & -6 \\ 1 & 3\end{vmatrix}=-\left[(2)(3)-(-6)(1)\right]=-12 \\ A_{13}=(-1)^{4}\begin{vmatrix}2 & 4 \\ 1 & -2\end{vmatrix}=+\left[(2)(-2)-(4)(1)\right]=-8 \\ A_{21}=(-1)^{3}\begin{vmatrix}3 & -5 \\ -2 & 3\end{vmatrix}=-\left[(3)(3)-(-5)(-2)\right]=1 \\ A_{22}=(-1)^{4}\begin{vmatrix}-1 & -5 \\ 1 & 3\end{vmatrix}=+\left[(-1)(3)-(-5)(1)\right]=2 \\ A_{23}=(-1)^{5}\begin{vmatrix}-1 & 3 \\ 1 & -2\end{vmatrix}=-\left[(-1)(-2)-(3)(1)\right]=1 \\ A_{31}=(-1)^{4}\begin{vmatrix}3 & -5 \\ 4 & -6\end{vmatrix}=+\left[(3)(-6)-(-5)(4)\right]=2 \\ A_{32}=(-1)^{5}\begin{vmatrix}-1 & -5 \\ 2 & -6\end{vmatrix}=-\left[(-1)(-6)-(-5)(2)\right]=-16 \\ A_{33}=(-1)^{6}\begin{vmatrix}-1 & 3 \\ 2 & 4\end{vmatrix}=+\left[(-1)(4)-(3)(2)\right]=-10 \end{aligned}
  6. The adjoint is the transpose of the cofactor matrix: the cofactor A_{ij} goes to row j, column i.
    \operatorname{adj}(A)=\begin{bmatrix}0 & 1 & 2 \\ -12 & 2 & -16 \\ -8 & 1 & -10\end{bmatrix}
  7. Divide by the determinant: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A).
    A^{-1}=\frac{1}{4}\begin{bmatrix}0 & 1 & 2 \\ -12 & 2 & -16 \\ -8 & 1 & -10\end{bmatrix}=\begin{bmatrix}0 & \frac{1}{4} & \frac{1}{2} \\ -3 & \frac{1}{2} & -4 \\ -2 & \frac{1}{4} & -\frac{5}{2}\end{bmatrix}
  8. Check: AA^{-1} should be I.
    AA^{-1}=\begin{bmatrix}-1 & 3 & -5 \\ 2 & 4 & -6 \\ 1 & -2 & 3\end{bmatrix}\begin{bmatrix}0 & \frac{1}{4} & \frac{1}{2} \\ -3 & \frac{1}{2} & -4 \\ -2 & \frac{1}{4} & -\frac{5}{2}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  9. Multiply: X=A^{-1}B.
    X=\begin{bmatrix}0 & \frac{1}{4} & \frac{1}{2} \\ -3 & \frac{1}{2} & -4 \\ -2 & \frac{1}{4} & -\frac{5}{2}\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 3\end{bmatrix}=\begin{bmatrix}\frac{7}{4} \\ -\frac{23}{2} \\ -\frac{29}{4}\end{bmatrix}
  10. State the solution and check it in the original equations.
    x=\frac{7}{4},\ y=-\frac{23}{2},\ z=-\frac{29}{4}
x=\frac{7}{4},\ y=-\frac{23}{2},\ z=-\frac{29}{4}
(iv) \dfrac2x+\dfrac3y+\dfrac{10}z=4,\quad \dfrac4x-\dfrac6y+\dfrac5z=1,\quad \dfrac6x+\dfrac9y-\dfrac{20}z=2
  1. The unknowns appear as reciprocals, so put u=\dfrac1x, v=\dfrac1y, w=\dfrac1z. The system becomes linear in u,v,w.
    \begin{aligned}2u+3v+10w&=4\\4u-6v+5w&=1\\6u+9v-20w&=2\end{aligned}
  2. Write the system as AX=B. The solution is X=A^{-1}B provided |A|\neq0 (section 2.6.8).
    A=\begin{bmatrix}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{bmatrix},\quad X=\begin{bmatrix}u \\ v \\ w\end{bmatrix},\quad B=\begin{bmatrix}4 \\ 1 \\ 2\end{bmatrix}
  3. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=2\begin{vmatrix}-6 & 5 \\ 9 & -20\end{vmatrix}-3\begin{vmatrix}4 & 5 \\ 6 & -20\end{vmatrix}+10\begin{vmatrix}4 & -6 \\ 6 & 9\end{vmatrix}
  4. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-6 & 5 \\ 9 & -20\end{vmatrix}=(-6)(-20)-(5)(9)=75 \\ \begin{vmatrix}4 & 5 \\ 6 & -20\end{vmatrix}=(4)(-20)-(5)(6)=-110 \\ \begin{vmatrix}4 & -6 \\ 6 & 9\end{vmatrix}=(4)(9)-(-6)(6)=72 \end{aligned}
  5. Substitute these values and add. So |A|=1200.
    |A|=2(75)-3(-110)+10(72)=1200
  6. |A|\neq0, so A is non-singular and A^{-1} exists. Find all cofactors.
    \begin{aligned} A_{11}=(-1)^{2}\begin{vmatrix}-6 & 5 \\ 9 & -20\end{vmatrix}=+\left[(-6)(-20)-(5)(9)\right]=75 \\ A_{12}=(-1)^{3}\begin{vmatrix}4 & 5 \\ 6 & -20\end{vmatrix}=-\left[(4)(-20)-(5)(6)\right]=110 \\ A_{13}=(-1)^{4}\begin{vmatrix}4 & -6 \\ 6 & 9\end{vmatrix}=+\left[(4)(9)-(-6)(6)\right]=72 \\ A_{21}=(-1)^{3}\begin{vmatrix}3 & 10 \\ 9 & -20\end{vmatrix}=-\left[(3)(-20)-(10)(9)\right]=150 \\ A_{22}=(-1)^{4}\begin{vmatrix}2 & 10 \\ 6 & -20\end{vmatrix}=+\left[(2)(-20)-(10)(6)\right]=-100 \\ A_{23}=(-1)^{5}\begin{vmatrix}2 & 3 \\ 6 & 9\end{vmatrix}=-\left[(2)(9)-(3)(6)\right]=0 \\ A_{31}=(-1)^{4}\begin{vmatrix}3 & 10 \\ -6 & 5\end{vmatrix}=+\left[(3)(5)-(10)(-6)\right]=75 \\ A_{32}=(-1)^{5}\begin{vmatrix}2 & 10 \\ 4 & 5\end{vmatrix}=-\left[(2)(5)-(10)(4)\right]=30 \\ A_{33}=(-1)^{6}\begin{vmatrix}2 & 3 \\ 4 & -6\end{vmatrix}=+\left[(2)(-6)-(3)(4)\right]=-24 \end{aligned}
  7. The adjoint is the transpose of the cofactor matrix: the cofactor A_{ij} goes to row j, column i.
    \operatorname{adj}(A)=\begin{bmatrix}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{bmatrix}
  8. Divide by the determinant: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A).
    A^{-1}=\frac{1}{1200}\begin{bmatrix}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{bmatrix}=\begin{bmatrix}\frac{1}{16} & \frac{1}{8} & \frac{1}{16} \\ \frac{11}{120} & -\frac{1}{12} & \frac{1}{40} \\ \frac{3}{50} & 0 & -\frac{1}{50}\end{bmatrix}
  9. Check: AA^{-1} should be I.
    AA^{-1}=\begin{bmatrix}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{bmatrix}\begin{bmatrix}\frac{1}{16} & \frac{1}{8} & \frac{1}{16} \\ \frac{11}{120} & -\frac{1}{12} & \frac{1}{40} \\ \frac{3}{50} & 0 & -\frac{1}{50}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  10. Multiply: X=A^{-1}B.
    X=\begin{bmatrix}\frac{1}{16} & \frac{1}{8} & \frac{1}{16} \\ \frac{11}{120} & -\frac{1}{12} & \frac{1}{40} \\ \frac{3}{50} & 0 & -\frac{1}{50}\end{bmatrix}\begin{bmatrix}4 \\ 1 \\ 2\end{bmatrix}=\begin{bmatrix}\frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5}\end{bmatrix}
  11. Return to the original unknowns: x=\dfrac1u, y=\dfrac1v, z=\dfrac1w.
    x=\frac{1}{\frac{1}{2}}=2,\quad y=\frac{1}{\frac{1}{3}}=3,\quad z=\frac{1}{\frac{1}{5}}=5
x=2,\ y=3,\ z=5

Book answer. (i) 1/11, -3/11, 70/11; (ii) 37/12, 7/3, 11/12; (iii) 7/4, -23/2, -29/4; (iv) 2, 3, 5. All agree.

6
If A=\begin{bmatrix}3&2&1\\4&-1&2\\7&3&-3\end{bmatrix}; find A^{-1} and hence solve the system of equations. 3x+4y+7z=14;\ 2x-y+3z=4;\ x+2y-3z=0.
Solution
  1. Find A^{-1} by the adjoint method.
  2. Expand along row 1. The signs of the cofactors follow the pattern +\ -\ + (or -\ +\ -) and each element multiplies the 2\times2 determinant left when its row and column are deleted.
    |A|=3\begin{vmatrix}-1 & 2 \\ 3 & -3\end{vmatrix}-2\begin{vmatrix}4 & 2 \\ 7 & -3\end{vmatrix}+1\begin{vmatrix}4 & -1 \\ 7 & 3\end{vmatrix}
  3. Evaluate each 2\times2 determinant as (top left)(bottom right) minus (top right)(bottom left).
    \begin{aligned} \begin{vmatrix}-1 & 2 \\ 3 & -3\end{vmatrix}=(-1)(-3)-(2)(3)=-3 \\ \begin{vmatrix}4 & 2 \\ 7 & -3\end{vmatrix}=(4)(-3)-(2)(7)=-26 \\ \begin{vmatrix}4 & -1 \\ 7 & 3\end{vmatrix}=(4)(3)-(-1)(7)=19 \end{aligned}
  4. Substitute these values and add. So |A|=62.
    |A|=3(-3)-2(-26)+1(19)=62
  5. |A|\neq0, so A is non-singular and A^{-1} exists. Find all cofactors.
    \begin{aligned} A_{11}=(-1)^{2}\begin{vmatrix}-1 & 2 \\ 3 & -3\end{vmatrix}=+\left[(-1)(-3)-(2)(3)\right]=-3 \\ A_{12}=(-1)^{3}\begin{vmatrix}4 & 2 \\ 7 & -3\end{vmatrix}=-\left[(4)(-3)-(2)(7)\right]=26 \\ A_{13}=(-1)^{4}\begin{vmatrix}4 & -1 \\ 7 & 3\end{vmatrix}=+\left[(4)(3)-(-1)(7)\right]=19 \\ A_{21}=(-1)^{3}\begin{vmatrix}2 & 1 \\ 3 & -3\end{vmatrix}=-\left[(2)(-3)-(1)(3)\right]=9 \\ A_{22}=(-1)^{4}\begin{vmatrix}3 & 1 \\ 7 & -3\end{vmatrix}=+\left[(3)(-3)-(1)(7)\right]=-16 \\ A_{23}=(-1)^{5}\begin{vmatrix}3 & 2 \\ 7 & 3\end{vmatrix}=-\left[(3)(3)-(2)(7)\right]=5 \\ A_{31}=(-1)^{4}\begin{vmatrix}2 & 1 \\ -1 & 2\end{vmatrix}=+\left[(2)(2)-(1)(-1)\right]=5 \\ A_{32}=(-1)^{5}\begin{vmatrix}3 & 1 \\ 4 & 2\end{vmatrix}=-\left[(3)(2)-(1)(4)\right]=-2 \\ A_{33}=(-1)^{6}\begin{vmatrix}3 & 2 \\ 4 & -1\end{vmatrix}=+\left[(3)(-1)-(2)(4)\right]=-11 \end{aligned}
  6. The adjoint is the transpose of the cofactor matrix: the cofactor A_{ij} goes to row j, column i.
    \operatorname{adj}(A)=\begin{bmatrix}-3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11\end{bmatrix}
  7. Divide by the determinant: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A).
    A^{-1}=\frac{1}{62}\begin{bmatrix}-3 & 9 & 5 \\ 26 & -16 & -2 \\ 19 & 5 & -11\end{bmatrix}=\begin{bmatrix}-\frac{3}{62} & \frac{9}{62} & \frac{5}{62} \\ \frac{13}{31} & -\frac{8}{31} & -\frac{1}{31} \\ \frac{19}{62} & \frac{5}{62} & -\frac{11}{62}\end{bmatrix}
  8. Check: AA^{-1} should be I.
    AA^{-1}=\begin{bmatrix}3 & 2 & 1 \\ 4 & -1 & 2 \\ 7 & 3 & -3\end{bmatrix}\begin{bmatrix}-\frac{3}{62} & \frac{9}{62} & \frac{5}{62} \\ \frac{13}{31} & -\frac{8}{31} & -\frac{1}{31} \\ \frac{19}{62} & \frac{5}{62} & -\frac{11}{62}\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
  9. Now the system. Write the coefficients of 3x+4y+7z=14, 2x-y+3z=4, x+2y-3z=0 as rows. The coefficient matrix is the transpose of the given A, because the rows of the system are the columns of A.
    A^t=\begin{bmatrix}3 & 4 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}14 \\ 4 \\ 0\end{bmatrix}
  10. So the system is A^tX=B and X=(A^t)^{-1}B. The inverse of a transpose is the transpose of the inverse, (A^t)^{-1}=(A^{-1})^t, so we can use the A^{-1} just found.
    (A^t)^{-1}=(A^{-1})^t=\begin{bmatrix}-\frac{3}{62} & \frac{13}{31} & \frac{19}{62} \\ \frac{9}{62} & -\frac{8}{31} & \frac{5}{62} \\ \frac{5}{62} & -\frac{1}{31} & -\frac{11}{62}\end{bmatrix}
  11. Multiply by B.
    X=\begin{bmatrix}-\frac{3}{62} & \frac{13}{31} & \frac{19}{62} \\ \frac{9}{62} & -\frac{8}{31} & \frac{5}{62} \\ \frac{5}{62} & -\frac{1}{31} & -\frac{11}{62}\end{bmatrix}\begin{bmatrix}14 \\ 4 \\ 0\end{bmatrix}=\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}
  12. Check: 3(1)+4(1)+7(1)=14, 2(1)-1+3(1)=4, 1+2(1)-3(1)=0.
A^{-1}=\begin{bmatrix}-\frac{3}{62} & \frac{9}{62} & \frac{5}{62} \\ \frac{13}{31} & -\frac{8}{31} & -\frac{1}{31} \\ \frac{19}{62} & \frac{5}{62} & -\frac{11}{62}\end{bmatrix},\quad x=1,\ y=1,\ z=1

Book answer. A^{-1} = [-3/62, 9/62, 5/62; 13/31, -8/31, -1/31; 19/62, 5/62, -11/62]; x = y = z = 1. Agrees.

Note. The system's coefficient matrix is A^t, not A (the equations list the columns of A as rows). Using X=A^{-1}B directly would give wrong values; the correct route is X=(A^{-1})^tB.

7
Determine the value of \lambda for which the following system has no solution, unique solution or infinitely many solutions. x+2y-3z=4;\ 3x-y+5z=2;\ 4x+y+(\lambda^2-14)z=\lambda+2
Solution
Reduce the augmented matrix
  1. Write the augmented matrix A_b of the system x+2y-3z=4,\ 3x-y+5z=2,\ 4x+y+(\lambda^2-14)z=\lambda+2.
    A_b=\left[\begin{array}{ccc|c}1&2&-3&4\\3&-1&5&2\\4&1&\lambda^2-14&\lambda+2\end{array}\right]
  2. Make zeros below the leading 1 in column 1: R_2-3R_1 and R_3-4R_1. For R_3: 1-8=-7, (\lambda^2-14)+12=\lambda^2-2, (\lambda+2)-16=\lambda-14.
    \sim_R\left[\begin{array}{ccc|c}1&2&-3&4\\0&-7&14&-10\\0&-7&\lambda^2-2&\lambda-14\end{array}\right]
  3. Make a zero in the position (3,2): R_3-R_2.
    \sim_R\left[\begin{array}{ccc|c}1&2&-3&4\\0&-7&14&-10\\0&0&\lambda^2-16&\lambda-4\end{array}\right]
  4. The last row is the equation (\lambda^2-16)z=\lambda-4. Factorise the coefficient: \lambda^2-16=(\lambda-4)(\lambda+4).
    (\lambda-4)(\lambda+4)\,z=\lambda-4
  5. The first two rows always give \operatorname{Rank}A\ge2, so the answer depends only on the last row. Consider the three cases below.
No solution
  1. **Case \lambda=-4.** The last row becomes 0\cdot z=-8, since \lambda-4=-8 and \lambda^2-16=0.
    \left[\begin{array}{ccc|c}1&2&-3&4\\0&-7&14&-10\\0&0&0&-8\end{array}\right]
  2. This is impossible. \operatorname{Rank}A=2 but \operatorname{Rank}A_b=3, so the system is inconsistent: no solution.
Infinitely many solutions
  1. **Case \lambda=4.** The last row becomes 0\cdot z=0, which is true for every z.
    \left[\begin{array}{ccc|c}1&2&-3&4\\0&-7&14&-10\\0&0&0&0\end{array}\right]
  2. \operatorname{Rank}A=\operatorname{Rank}A_b=2, which is less than the number of unknowns 3, so there are infinitely many solutions. Let z=t and use rows 2 and 1.
    -7y+14t=-10\;\Rightarrow\;y=\frac{10}{7}+2t,\qquad x=4-2y+3t=\frac{8}{7}-t
  3. Check with t=0: \left(\frac87,\frac{10}7,0\right) gives \frac87+\frac{20}7=4, \frac{24}7-\frac{10}7=2 and \frac{32}7+\frac{10}7=6=\lambda+2.
Unique solution
  1. **Case \lambda\neq4,\ \lambda\neq-4.** The coefficient (\lambda-4)(\lambda+4) is not zero, so divide the last equation by it.
    z=\frac{\lambda-4}{(\lambda-4)(\lambda+4)}=\frac{1}{\lambda+4}
  2. \operatorname{Rank}A=\operatorname{Rank}A_b=3, the number of unknowns, so the solution is unique. Back-substitute: y=\dfrac{10+14z}{7} and x=4-2y+3z.
    y=\frac{10}{7}+\frac{2}{\lambda+4},\qquad x=\frac{8}{7}-\frac{1}{\lambda+4}
Value of \lambdaRanksResult
\lambda=-4\operatorname{Rank}A=2,\ \operatorname{Rank}A_b=3no solution
\lambda=4\operatorname{Rank}A=\operatorname{Rank}A_b=2<3infinitely many solutions
\lambda\neq\pm4\operatorname{Rank}A=\operatorname{Rank}A_b=3unique solution
\lambda=-4:\text{ no solution};\quad \lambda=4:\text{ infinitely many};\quad \lambda\neq\pm4:\text{ unique}

Book answer. λ = ±4: no solution; λ ≠ ±4: unique solution. The no-solution value agrees only for λ = -4; the key does not give the case of infinitely many solutions.

Note. Answer key: the key lumps \lambda=\pm4 together as 'no solution'. At \lambda=4 the third equation is the sum of the first two (4x+y+2z=6 is (x+2y-3z)+(3x-y+5z)=4+2), so the system is consistent with infinitely many solutions, as shown above. Only \lambda=-4 gives no solution.

8
Show that the system of equations 2x-y+3z=\alpha;\ 3x+y-5z=\beta;\ -5x-5y+21z=\gamma is inconsistent if \gamma\neq2\alpha-3\beta.
Solution
  1. Write the augmented matrix. The third equation is -5x-5y+21z=\gamma.
    A_b=\left[\begin{array}{ccc|c}2&-1&3&\alpha\\3&1&-5&\beta\\-5&-5&21&\gamma\end{array}\right]
  2. Make zeros in column 1 below R_1 without fractions: R_2\to2R_2-3R_1 and R_3\to2R_3+5R_1. For R_2: 6-6=0, 2+3=5, -10-9=-19, 2\beta-3\alpha. For R_3: -10+10=0, -10-5=-15, 42+15=57, 2\gamma+5\alpha.
    \sim_R\left[\begin{array}{ccc|c}2&-1&3&\alpha\\0&5&-19&2\beta-3\alpha\\0&-15&57&2\gamma+5\alpha\end{array}\right]
  3. Apply R_3\to R_3+3R_2. Then -15+15=0, 57-57=0 and the right side is (2\gamma+5\alpha)+3(2\beta-3\alpha)=2\gamma+6\beta-4\alpha.
    \sim_R\left[\begin{array}{ccc|c}2&-1&3&\alpha\\0&5&-19&2\beta-3\alpha\\0&0&0&2(\gamma-2\alpha+3\beta)\end{array}\right]
  4. The last row says 0x+0y+0z=2(\gamma-2\alpha+3\beta). If \gamma\neq2\alpha-3\beta the right side is not zero, which is impossible.
  5. Ranks: the first two rows are independent (for example \begin{vmatrix}2&-1\\3&1\end{vmatrix}=5\neq0), so \operatorname{Rank}A=2. When \gamma\neq2\alpha-3\beta the last row of A_b is non-zero, so \operatorname{Rank}A_b=3.
    \operatorname{Rank}A=2\neq3=\operatorname{Rank}A_b
  6. Since \operatorname{Rank}A\neq\operatorname{Rank}A_b, the system is inconsistent whenever \gamma\neq2\alpha-3\beta.
\gamma\neq2\alpha-3\beta\;\Rightarrow\;\operatorname{Rank}A=2\neq\operatorname{Rank}A_b=3\ \text{(inconsistent)}

Note. The sign of the y term in the third equation is smudged in the book; -5y is the sign that makes the stated condition \gamma\neq2\alpha-3\beta correct (twice the first equation minus three times the second has left side -5x-5y+21z).

9
By making use of matrix of order 2 by 2 and 3 by 3 encode and decode the following words: a. PAKISTAN, b. ISLAMABAD, c. COLLEGE
Solution

Method (section 2.7, the cryptography example). Give each letter its number (A=1 to Z=26, a blank is 0). Cut the message into row matrices of the same length as the order of a chosen non-singular encoding matrix K. Post-multiply each row by K to encode; the receiver post-multiplies by K^{-1} to decode. Any non-singular key may be chosen; we use K_2=\begin{bmatrix}2&1\\1&1\end{bmatrix} (determinant 1, so K_2^{-1} has whole-number entries) and the book's K_3=\begin{bmatrix}1&-1&1\\2&-1&0\\1&0&0\end{bmatrix}.

ABCDEFGHIJKLM
12345678910111213
NOPQRSTUVWXYZ
14151617181920212223242526
a. PAKISTAN
Using a 2\times2 key
  1. Key matrix of order 2\times2 (non-singular, |K|=1).
    K=\begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix},\qquad K^{-1}=\begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}
  2. Replace each letter by its number (A=1,\dots,Z=26, blank =0).
    \text{PAKISTAN}\ \to\ 16, 1, 11, 9, 19, 20, 1, 14
  3. Cut the numbers into pieces of length 2, each a row matrix.
  4. Encode: post-multiply each row by the key matrix K. For example the first row.
    [16\ \ 1]\begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}=[33\ \ 17]
  5. Decode: post-multiply each coded row by K^{-1} (the receiver's key).
    K^{-1}=\begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}
LettersUncoded rowCoded row (row \times K)Decoded row (coded \times K^{-1})
PA[16\ \ 1][33\ \ 17][16\ \ 1]
KI[11\ \ 9][31\ \ 20][11\ \ 9]
ST[19\ \ 20][58\ \ 39][19\ \ 20]
AN[1\ \ 14][16\ \ 15][1\ \ 14]

The decoded numbers give back the letters PAKISTAN, so the receiver reads PAKISTAN.

Using a 3\times3 key
  1. Key matrix of order 3\times3 (non-singular, |K|=1).
    K=\begin{bmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix},\qquad K^{-1}=\begin{bmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{bmatrix}
  2. Replace each letter by its number (A=1,\dots,Z=26, blank =0).
    \text{PAKISTAN}\ \to\ 16, 1, 11, 9, 19, 20, 1, 14
  3. Cut the numbers into pieces of length 3, each a row matrix (the last piece is filled with 1 zero, i.e. blank, as in the book's example).
  4. Encode: post-multiply each row by the key matrix K. For example the first row.
    [16\ \ 1\ \ 11]\begin{bmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix}=[29\ \ -17\ \ 16]
  5. Decode: post-multiply each coded row by K^{-1} (the receiver's key).
    K^{-1}=\begin{bmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{bmatrix}
LettersUncoded rowCoded row (row \times K)Decoded row (coded \times K^{-1})
PAK[16\ \ 1\ \ 11][29\ \ -17\ \ 16][16\ \ 1\ \ 11]
IST[9\ \ 19\ \ 20][67\ \ -28\ \ 9][9\ \ 19\ \ 20]
AN(blank)[1\ \ 14\ \ 0][29\ \ -15\ \ 1][1\ \ 14\ \ 0]

The decoded numbers give back the letters PAKISTAN, so the receiver reads PAKISTAN.

\text{decoded: PAKISTAN}
b. ISLAMABAD
Using a 2\times2 key
  1. Key matrix of order 2\times2 (non-singular, |K|=1).
    K=\begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix},\qquad K^{-1}=\begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}
  2. Replace each letter by its number (A=1,\dots,Z=26, blank =0).
    \text{ISLAMABAD}\ \to\ 9, 19, 12, 1, 13, 1, 2, 1, 4
  3. Cut the numbers into pieces of length 2, each a row matrix (the last piece is filled with 1 zero, i.e. blank, as in the book's example).
  4. Encode: post-multiply each row by the key matrix K. For example the first row.
    [9\ \ 19]\begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}=[37\ \ 28]
  5. Decode: post-multiply each coded row by K^{-1} (the receiver's key).
    K^{-1}=\begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}
LettersUncoded rowCoded row (row \times K)Decoded row (coded \times K^{-1})
IS[9\ \ 19][37\ \ 28][9\ \ 19]
LA[12\ \ 1][25\ \ 13][12\ \ 1]
MA[13\ \ 1][27\ \ 14][13\ \ 1]
BA[2\ \ 1][5\ \ 3][2\ \ 1]
D(blank)[4\ \ 0][8\ \ 4][4\ \ 0]

The decoded numbers give back the letters ISLAMABAD, so the receiver reads ISLAMABAD.

Using a 3\times3 key
  1. Key matrix of order 3\times3 (non-singular, |K|=1).
    K=\begin{bmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix},\qquad K^{-1}=\begin{bmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{bmatrix}
  2. Replace each letter by its number (A=1,\dots,Z=26, blank =0).
    \text{ISLAMABAD}\ \to\ 9, 19, 12, 1, 13, 1, 2, 1, 4
  3. Cut the numbers into pieces of length 3, each a row matrix.
  4. Encode: post-multiply each row by the key matrix K. For example the first row.
    [9\ \ 19\ \ 12]\begin{bmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix}=[59\ \ -28\ \ 9]
  5. Decode: post-multiply each coded row by K^{-1} (the receiver's key).
    K^{-1}=\begin{bmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{bmatrix}
LettersUncoded rowCoded row (row \times K)Decoded row (coded \times K^{-1})
ISL[9\ \ 19\ \ 12][59\ \ -28\ \ 9][9\ \ 19\ \ 12]
AMA[1\ \ 13\ \ 1][28\ \ -14\ \ 1][1\ \ 13\ \ 1]
BAD[2\ \ 1\ \ 4][8\ \ -3\ \ 2][2\ \ 1\ \ 4]

The decoded numbers give back the letters ISLAMABAD, so the receiver reads ISLAMABAD.

\text{decoded: ISLAMABAD}
c. COLLEGE
Using a 2\times2 key
  1. Key matrix of order 2\times2 (non-singular, |K|=1).
    K=\begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix},\qquad K^{-1}=\begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}
  2. Replace each letter by its number (A=1,\dots,Z=26, blank =0).
    \text{COLLEGE}\ \to\ 3, 15, 12, 12, 5, 7, 5
  3. Cut the numbers into pieces of length 2, each a row matrix (the last piece is filled with 1 zero, i.e. blank, as in the book's example).
  4. Encode: post-multiply each row by the key matrix K. For example the first row.
    [3\ \ 15]\begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}=[21\ \ 18]
  5. Decode: post-multiply each coded row by K^{-1} (the receiver's key).
    K^{-1}=\begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}
LettersUncoded rowCoded row (row \times K)Decoded row (coded \times K^{-1})
CO[3\ \ 15][21\ \ 18][3\ \ 15]
LL[12\ \ 12][36\ \ 24][12\ \ 12]
EG[5\ \ 7][17\ \ 12][5\ \ 7]
E(blank)[5\ \ 0][10\ \ 5][5\ \ 0]

The decoded numbers give back the letters COLLEGE, so the receiver reads COLLEGE.

Using a 3\times3 key
  1. Key matrix of order 3\times3 (non-singular, |K|=1).
    K=\begin{bmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix},\qquad K^{-1}=\begin{bmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{bmatrix}
  2. Replace each letter by its number (A=1,\dots,Z=26, blank =0).
    \text{COLLEGE}\ \to\ 3, 15, 12, 12, 5, 7, 5
  3. Cut the numbers into pieces of length 3, each a row matrix (the last piece is filled with 2 zeros, i.e. blanks, as in the book's example).
  4. Encode: post-multiply each row by the key matrix K. For example the first row.
    [3\ \ 15\ \ 12]\begin{bmatrix}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix}=[45\ \ -18\ \ 3]
  5. Decode: post-multiply each coded row by K^{-1} (the receiver's key).
    K^{-1}=\begin{bmatrix}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{bmatrix}
LettersUncoded rowCoded row (row \times K)Decoded row (coded \times K^{-1})
COL[3\ \ 15\ \ 12][45\ \ -18\ \ 3][3\ \ 15\ \ 12]
LEG[12\ \ 5\ \ 7][29\ \ -17\ \ 12][12\ \ 5\ \ 7]
E(blank)(blank)[5\ \ 0\ \ 0][5\ \ -5\ \ 5][5\ \ 0\ \ 0]

The decoded numbers give back the letters COLLEGE, so the receiver reads COLLEGE.

\text{decoded: COLLEGE}

Note. The question leaves the encoding matrices to the student, so there is no printed answer. Your own non-singular keys give different coded rows, but decoding must always return the original numbers.

10
Write the transformation matrix and the transformed equation of the curve x^2+y^2-9=0 when it is moved 5 units below.
Solution
  1. Moving the curve 5 units below lowers every point by 5: the new coordinates are x'=x and y'=y-5. The inverse relations are x=x' and y=y'+5.
    x'=x,\qquad y'=y-5
  2. Transformation matrix. In the book's form (Example 36), write the change as a matrix equation: the point is shifted by the vector (0,-5).
    \begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}0\\-5\end{bmatrix}
  3. The same shift as one product, using a 3\times3 matrix acting on (x,y,1) (the third row keeps the 1).
    \begin{bmatrix}x'\\y'\\1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&-5\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}=\begin{bmatrix}x\\y-5\\1\end{bmatrix}
  4. Transformed equation. Replace x by x' and y by y'+5 in the given equation.
    x'^2+(y'+5)^2-9=0
  5. Expand (y'+5)^2=y'^2+10y'+25 and collect.
    x'^2+y'^2+10y'+25-9=0\;\Rightarrow\;x'^2+y'^2+10y'+16=0
The circle of radius 3 centred at the origin (dashed) is moved 5 units down: its centre goes from (0,0) to (0,-5).
\begin{bmatrix}1&0&0\\0&1&-5\\0&0&1\end{bmatrix};\quad x'^2+y'^2+10y'+16=0

Book answer. Matrix [1, 0, 0; 0, 1, -5; 0, 0, 1]; x'^2 + y'^2 + 10y' + 16 = 0. Agrees.

Note. The book's Example 36 shifts a line up by 1 but prints the shift vector as (0,-1) in the matrix equation, although its working uses y'=y+1. Moving up is a shift of +1 in y, moving down is -1; here the shift for 5 units below is -5, as in the answer key.

11
Write the transformation matrix and the transformed equation of the curve x^2+4x-3y=8 when it is moved to 2 units on the left.
Solution
  1. Moving the curve 2 units to the left decreases every x by 2: x'=x-2 and y'=y. Hence x=x'+2 and y=y'.
    x'=x-2,\qquad y'=y
  2. Transformation matrix.
    \begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}-2\\0\end{bmatrix},\qquad \begin{bmatrix}x'\\y'\\1\end{bmatrix}=\begin{bmatrix}1&0&-2\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}
  3. Transformed equation. Replace x by x'+2 and y by y'.
    (x'+2)^2+4(x'+2)-3y'=8
  4. Expand: (x'+2)^2=x'^2+4x'+4 and 4(x'+2)=4x'+8.
    x'^2+4x'+4+4x'+8-3y'-8=0
  5. Collect like terms.
    x'^2+8x'-3y'+4=0
The parabola x^2+4x-3y=8 (dashed, vertex (-2,-4)) and the same parabola moved 2 units left (vertex (-4,-4)).
\begin{bmatrix}1&0&-2\\0&1&0\\0&0&1\end{bmatrix};\quad x'^2+8x'-3y'+4=0

Book answer. Matrix [1, 0, -2; 0, 1, 0; 0, 0, 1]; x'^2 + 8x' - 3y' + 4 = 0. Agrees.

12
Write the transformation matrix and the transformed equation of the curve 2x^2-5y^2=10 when it is moved 1 unit to the right.
Solution
  1. Moving the curve 1 unit to the right increases every x by 1: x'=x+1 and y'=y. Hence x=x'-1 and y=y'.
    x'=x+1,\qquad y'=y
  2. Transformation matrix.
    \begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}1\\0\end{bmatrix},\qquad \begin{bmatrix}x'\\y'\\1\end{bmatrix}=\begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}
  3. Transformed equation. Replace x by x'-1 and y by y'.
    2(x'-1)^2-5y'^2=10
  4. Expand (x'-1)^2=x'^2-2x'+1.
    2x'^2-4x'+2-5y'^2-10=0
  5. Collect like terms.
    2x'^2-5y'^2-4x'-8=0
Both branches of the hyperbola 2x^2-5y^2=10 (dashed) and the same hyperbola moved 1 unit to the right.
\begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix};\quad 2x'^2-5y'^2-4x'-8=0

Book answer. Matrix [1, 0, 1; 0, 1, 0; 0, 0, 1]; 2x'^2 - 5y'^2 - 4x' - 8 = 0. Agrees.

13
Write the transformation matrix and the transformed equation of the line 2x=7y-3 when it is moved 2 units above and 3 units to the right.
Solution
  1. Moving 2 units above adds 2 to every y, and moving 3 units to the right adds 3 to every x: x'=x+3 and y'=y+2. Hence x=x'-3 and y=y'-2.
    x'=x+3,\qquad y'=y+2
  2. Transformation matrix.
    \begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}3\\2\end{bmatrix},\qquad \begin{bmatrix}x'\\y'\\1\end{bmatrix}=\begin{bmatrix}1&0&3\\0&1&2\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}
  3. Transformed equation. Replace x by x'-3 and y by y'-2 in 2x=7y-3.
    2(x'-3)=7(y'-2)-3
  4. Expand both sides.
    2x'-6=7y'-14-3=7y'-17
  5. Bring every term to the left side.
    2x'-7y'-6+17=0\;\Rightarrow\;2x'-7y'+11=0
The point (2,1) of the line 2x=7y-3 moves to (5,3) on the new line: the whole line is shifted 3 right and 2 up, so the two lines are parallel.
\begin{bmatrix}1&0&3\\0&1&2\\0&0&1\end{bmatrix};\quad 2x'-7y'+11=0

Book answer. Matrix [1, 0, 3; 0, 1, 2; 0, 0, 1]; 2x' - 7y' + 11 = 0. Agrees.

Review Exercise 2p. 72

Mixed review of matrices and determinants: orders of products, cofactors and the adjoint, properties of determinants, and the rank conditions for systems of equations.

1 (i)
If order of A is m\times n and order of B is n\times p then order of AB is:
  1. An\times p
  2. Bm\times pAnswer
  3. Cp\times m
  4. Dn\times n
Why

AB is defined because the number of columns of A (n) equals the number of rows of B (n). The product has the rows of A and the columns of B: order m\times p (section 2.2.4).

Book answer. b. Agrees.

1 (ii)
If A is a row matrix of order 1\times n then order of A^tA is:
  1. A1\times n
  2. Bn\times1
  3. C1\times1
  4. Dn\times nAnswer
Why

A^t has order n\times1 (rows and columns interchanged). So A^tA is (n\times1)(1\times n): the inner orders 1 and 1 match and the product has order n\times n.

Book answer. d. Agrees.

1 (iii)
For an element a_{ij} of a square matrix A:
  1. Aa_{ij}=(-1)^{i+j}A_{ij}
  2. Ba_{ij}=(-1)^{i+j}M_{ij}
  3. C\dfrac{A_{ij}}{M_{ij}}=(-1)^{i+j}Answer
  4. Da_{ij}=M_{ij}
Why

The cofactor of a_{ij} is defined by A_{ij}=(-1)^{i+j}M_{ij} (section 2.3.4), where M_{ij} is the minor. Dividing both sides by M_{ij} gives \dfrac{A_{ij}}{M_{ij}}=(-1)^{i+j}, option C.

Book answer. d.

Note. Answer key: the key gives (d), but a_{ij}=M_{ij} is false in general (for example a_{11}=2 and M_{11} for the matrix \begin{bmatrix}2&1\\3&4\end{bmatrix} is 4). The relation A_{ij}=(-1)^{i+j}M_{ij} gives (c).

1 (iv)
If A is any matrix then A and A^t are always conformable for:
  1. AAddition
  2. BmultiplicationAnswer
  3. Csubtraction
  4. Dall of these
Why

If A is m\times n then A^t is n\times m. Multiplication AA^t needs columns of A (n) equal to rows of A^t (n): always true, giving an m\times m matrix.

Book answer. b. Agrees.

1 (v)
If A is a square matrix of order 3\times3 and |A|=3 then value of |\operatorname{adj}A| is:
  1. A3
  2. B\dfrac13
  3. C9Answer
  4. D6
Why

From A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A we get \operatorname{adj}A=|A|\,A^{-1}. Taking determinants of a 3\times3 matrix, a scalar k in front of the matrix comes out as k^3 (Property 4 applied to each of the three rows), and |A^{-1}|=\dfrac{1}{|A|}.

|\operatorname{adj}A|=|A|^3\cdot\frac{1}{|A|}=|A|^2=3^2=9

Book answer. c. Agrees.

1 (vi)
For the square matrix A of order 3\times3 with |A|=9; A_{21}=2; A_{22}=3; A_{23}=-1; a_{21}=1; a_{23}=2, the value of a_{22} is:
  1. A2
  2. B3Answer
  3. C9
  4. D-1
Why

Expand |A| along row 2: |A|=a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23} (section 2.3.5).

9=(1)(2)+a_{22}(3)+(2)(-1)=3a_{22}\;\Rightarrow\;a_{22}=3

Book answer. b. Agrees.

1 (vii)
System of homogeneous linear equations has non-trivial solution if:
  1. A|A|>0
  2. B|A|<0
  3. C|A|=0Answer
  4. D|A|\neq0
Why

A homogeneous system AX=O always has the trivial solution X=O. If |A|\neq0 then A^{-1} exists and X=A^{-1}O=O is the only solution. A non-trivial solution is possible only when |A|=0 (section 2.6.2).

Book answer. d.

Note. Answer key: the key gives (d) |A|\neq0, but the book itself says (Key Fact, section 2.6.2) that the condition for a non-trivial solution is |A|=0. The correct option is (c).

1 (viii)
For non-homogeneous system of equations; the system is inconsistent if:
  1. A\operatorname{Rank}A=\operatorname{Rank}A_b
  2. B\operatorname{Rank}A\neq\operatorname{Rank}A_bAnswer
  3. C\operatorname{Rank}A< no. of variables
  4. D\operatorname{Rank}A_b> no. of variables
Why

By the consistency criterion (section 2.6.6) a system is consistent if \operatorname{Rank}A=\operatorname{Rank}A_b and inconsistent if \operatorname{Rank}A\neq\operatorname{Rank}A_b.

Book answer. c.

Note. Answer key: the key gives (c), but \operatorname{Rank}A< number of variables says nothing about consistency (a consistent system with infinitely many solutions has exactly this property). The criterion in the book is \operatorname{Rank}A\neq\operatorname{Rank}A_b, option (b).

1 (ix)
For a system of non-homogeneous equations with three variables system will have unique solution if:
  1. A\operatorname{Rank}A<3
  2. B\operatorname{Rank}A_b<3
  3. C\operatorname{Rank}A=\operatorname{Rank}A_b=3Answer
  4. D\operatorname{Rank}A=\operatorname{Rank}A_b<3
Why

Unique solution: \operatorname{Rank}A=\operatorname{Rank}A_b= number of unknowns =3 (section 2.6.6, Key Fact in Example 33).

Book answer. c. Agrees.

1 (x)
A system of non-homogeneous equation having infinite many solutions can be solved by using:
  1. AInversion method
  2. BCramer's rule
  3. CGauss-Jordan methodAnswer
  4. Dall of these
Why

Infinitely many solutions occur only when \operatorname{Rank}A=\operatorname{Rank}A_b< number of unknowns; for a square system this means |A|=0. The inversion method and Cramer's rule need A^{-1} or |A|\neq0 (the book's Remember box in section 2.6.8), so they cannot be used. Row reduction (Gauss elimination or Gauss-Jordan) works for every system, as in Example 34.

Book answer. d.

Note. Answer key: the key gives (d), but matrix inversion and Cramer's rule both require |A|\neq0, which means a unique solution. For a system with infinitely many solutions only the row-reduction method (c) applies.

2
For the matrix A=\begin{bmatrix}1&2&0\\-3&4&9\\2&1&6\end{bmatrix}; find A_{13}, A_{23} and A_{33}; hence find |A|.
Solution
  1. Find the three cofactors A_{13}, A_{23}, A_{33}: delete the row and column of the element and attach the sign (-1)^{i+j}.
    \begin{aligned} A_{13}=(-1)^{4}\begin{vmatrix}-3 & 4 \\ 2 & 1\end{vmatrix}=+\left[(-3)(1)-(4)(2)\right]=-11 \\ A_{23}=(-1)^{5}\begin{vmatrix}1 & 2 \\ 2 & 1\end{vmatrix}=-\left[(1)(1)-(2)(2)\right]=3 \\ A_{33}=(-1)^{6}\begin{vmatrix}1 & 2 \\ -3 & 4\end{vmatrix}=+\left[(1)(4)-(2)(-3)\right]=10 \end{aligned}
  2. Hence expand |A| along column 3, where a_{13}=0, a_{23}=9, a_{33}=6: |A|=a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}.
    |A|=0(-11)+9(3)+6(10)=87
  3. Check by expanding along row 1.
    |A|=1\begin{vmatrix}4&9\\1&6\end{vmatrix}-2\begin{vmatrix}-3&9\\2&6\end{vmatrix}+0=1(24-9)-2(-18-18)=15+72=87
A_{13}=-11,\ A_{23}=3,\ A_{33}=10,\ |A|=87

Book answer. -11, 3, 10; 87. Agrees.

3
Without expanding show that \begin{vmatrix}a+1&l&l\\l&a+1&l\\l&l&a+1\end{vmatrix}=(a+1+2l)(a+1-l)^2.
Solution
  1. Write t=a+1 to shorten the work, and call the determinant \Delta.
    \Delta=\begin{vmatrix}t&l&l\\l&t&l\\l&l&t\end{vmatrix}
  2. Apply R_1\to R_1+R_2+R_3 (Property 7). Every entry of the new first row is t+2l.
    \Delta=\begin{vmatrix}t+2l&t+2l&t+2l\\l&t&l\\l&l&t\end{vmatrix}
  3. Take the common factor t+2l out of R_1 (Property 4).
    \Delta=(t+2l)\begin{vmatrix}1&1&1\\l&t&l\\l&l&t\end{vmatrix}
  4. Apply C_2\to C_2-C_1 and C_3\to C_3-C_1 to make zeros in the first row.
    \Delta=(t+2l)\begin{vmatrix}1&0&0\\l&t-l&0\\l&0&t-l\end{vmatrix}
  5. The determinant is triangular, so it is the product of the diagonal elements 1\cdot(t-l)(t-l) (Property 8).
    \Delta=(t+2l)(t-l)^2
  6. Put back t=a+1.
    \Delta=(a+1+2l)(a+1-l)^2
\Delta=(a+1+2l)(a+1-l)^2

Book answer. No printed answer (proof).

4
Find the value of \lambda so that the following system has infinite many solutions. 2x-3y+z=1;\ x-2y+\lambda z=2;\ 3y+z=-1
Solution
The value of \lambda that makes |A|=0
  1. Write the augmented matrix and A with \lambda in it. A system with infinitely many solutions must have \operatorname{Rank}A=\operatorname{Rank}A_b<3, so first |A| must be 0.
    A=\begin{bmatrix}2&-3&1\\1&-2&\lambda\\0&3&1\end{bmatrix}
  2. Expand |A| along column 1.
    |A|=2\begin{vmatrix}-2&\lambda\\3&1\end{vmatrix}-1\begin{vmatrix}-3&1\\3&1\end{vmatrix}+0=2(-2-3\lambda)-(-3-3)
  3. Simplify.
    |A|=-4-6\lambda+6=2-6\lambda
  4. Set |A|=0.
    2-6\lambda=0\;\Rightarrow\;\lambda=\frac13
Is the system consistent for this \lambda?
  1. Now check whether the system really has solutions for \lambda=\frac13. Reduce the augmented matrix. Swap R_1,R_2 to get a leading 1, then R_2-2R_1.
    \left[\begin{array}{ccc|c}2&-3&1&1\\1&-2&\frac13&2\\0&3&1&-1\end{array}\right]\sim_R\left[\begin{array}{ccc|c}1&-2&\frac13&2\\0&1&\frac13&-3\\0&3&1&-1\end{array}\right]
  2. Apply R_3-3R_2: 3-3=0, 1-1=0 and -1+9=8.
    \sim_R\left[\begin{array}{ccc|c}1&-2&\frac13&2\\0&1&\frac13&-3\\0&0&0&8\end{array}\right]
  3. The last row says 0=8, which is impossible: \operatorname{Rank}A=2 but \operatorname{Rank}A_b=3, so for \lambda=\frac13 the system is inconsistent.
  4. For any \lambda\neq\frac13, |A|\neq0 and the solution is unique. So, with the constants as printed, there is no value of \lambda giving infinitely many solutions.
Likely intended question
  1. What the book most likely intends. If the third equation were 3y+z=-9 (so that the constants are consistent with the first two equations) the last row would be 0=0 and the system would have infinitely many solutions at \lambda=\frac13.
    \operatorname{Rank}A=\operatorname{Rank}A_b=2<3
\lambda=\dfrac13\ (|A|=0);\ \text{with the printed constants the system is inconsistent there}

Book answer. 1/3. Agrees with the value that makes |A|=0.

Note. With the constants as printed the system is inconsistent at \lambda=\frac13 (shown above), so no value of \lambda gives infinitely many solutions. The book's answer \frac13 is the value for which |A|=0, the necessary condition; the question is probably misprinted (for example 3y+z=-9 instead of -1).

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 31 mark
A system of linear equations is said to be inconsistent if it has:
  1. ANo solutionAnswer
  2. BAt least one solution
  3. CInfinitely many solutions
  4. DExactly two solutions
Why

An inconsistent system has no solution at all (\operatorname{Rank}A\neq\operatorname{Rank}A_b). A system with at least one solution (B), including infinitely many (C), is consistent.

Option D cannot occur: a linear system has no solution, one solution or infinitely many, never exactly two.

2025 Annual I, Set B, MCQ 41 mark
If A is a 3\times 3 matrix, then |3A| results
  1. A3|A|
  2. B6|A|
  3. C9|A|
  4. D27|A|Answer
Why

Multiplying a matrix by 3 multiplies every row by 3. Taking the common factor 3 out of each of the three rows, one at a time, gives |3A|=3\cdot3\cdot3\,|A|=3^3|A|=27|A|. In general |kA|=k^n|A| for an n\times n matrix.

Option A multiplies only one row by 3. Option C is 3^2, which is the factor for a 2\times2 matrix, and B has no basis.

2026 Annual I, Set B, MCQ 81 mark
Rank of matrix A=\begin{bmatrix}1&2&-1\\0&1&7\\0&0&0\end{bmatrix} is:
  1. A1
  2. B2Answer
  3. C3
  4. D4
Why

The matrix is already in echelon form: the leading entry of row 1 is 1, the leading entry of row 2 is 1 and lies to the right of it, and the zero row is at the bottom. The rank is the number of non-zero rows, which is 2.

Option C counts all three rows, including the zero row. Option D is impossible, since a 3\times3 matrix cannot have rank above 3. Option A would need two rows to vanish.

2026 Annual I, Set B, MCQ 91 mark
A consistent system MUST have at least:
  1. AAt least one solutionAnswer
  2. BInfinite solutions
  3. CNo solution
  4. DMaximum two solutions
Why

A system of linear equations is consistent if it has at least one solution. When \operatorname{Rank}A=\operatorname{Rank}A_b, the solution is either unique (rank equals the number of unknowns) or there are infinitely many (rank less than the number of unknowns).

Option B is wrong because a consistent system may have exactly one solution. Option C describes an inconsistent system. Option D cannot happen: a linear system never has exactly two solutions.

2026 Annual I, Set B, Q.2 (ii) OR4 marks
Find the rank of matrix: \begin{bmatrix}1&2&3\\2&4&6\\1&1&1\end{bmatrix}.
Solution
  1. Reduce to echelon form with row operations. The first row already has leading entry 1.
    A=\begin{bmatrix}1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1\end{bmatrix}
  2. Make zeros below the leading 1: R_2-2R_1 and R_3-R_1.
    \sim_R \begin{bmatrix}1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -2\end{bmatrix}
  3. Interchange rows 2 and 3 (R_{23}) so that the zero row is at the bottom.
    \sim_R \begin{bmatrix}1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & 0 & 0\end{bmatrix}
  4. Multiply row 2 by -1 to make its leading entry 1.
    \sim_R \begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0\end{bmatrix}
  5. This is in echelon form. Count the non-zero rows: there are 2.
    \operatorname{Rank}(A)=2
\operatorname{Rank}(A)=2

Note. Quick check: row 2 is twice row 1, so the rows are dependent and |A|=0. Row 3 is not a multiple of row 1, so the rank is exactly 2, not 1.

2026 Annual I, Set B, Q.2 (v)4 marks
If A=\begin{bmatrix}1&2&3\\2&4&6\\3&6&9\end{bmatrix} then find A_{13}, A_{23}, A_{31} and A_{33}.
Solution
  1. A_{ij} is the cofactor of a_{ij}: A_{ij}=(-1)^{i+j}M_{ij}, where the minor M_{ij} is the determinant left after deleting row i and column j.
  2. A_{13}: delete row 1 and column 3.
    A_{13}=(-1)^{1+3}\begin{vmatrix}2 & 4 \\ 3 & 6\end{vmatrix}=+(12-12)=0
  3. A_{23}: delete row 2 and column 3.
    A_{23}=(-1)^{2+3}\begin{vmatrix}1 & 2 \\ 3 & 6\end{vmatrix}=-(6-6)=0
  4. A_{31}: delete row 3 and column 1.
    A_{31}=(-1)^{3+1}\begin{vmatrix}2 & 3 \\ 4 & 6\end{vmatrix}=+(12-12)=0
  5. A_{33}: delete row 3 and column 3.
    A_{33}=(-1)^{3+3}\begin{vmatrix}1 & 2 \\ 2 & 4\end{vmatrix}=+(4-4)=0
A_{13}=A_{23}=A_{31}=A_{33}=0

Note. All four are zero because the rows of A are multiples of one another (R_2=2R_1, R_3=3R_1), so every 2\times2 determinant formed from them vanishes. The sign factor does not matter, so the answer is also 0 if the question means minors.

2025 Annual I, Set B, Q.2 (vii)4 marks
Verify that \begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}=(b-a)(c-a)(c-b)
Solution
  1. Make zeros in column 1 below the top entry: R_2-R_1 and R_3-R_1 (this does not change the value of the determinant).
    \begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{vmatrix}=\begin{vmatrix}1 & a & a^2 \\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2\end{vmatrix}
  2. Take the common factor b-a out of row 2 and c-a out of row 3, using b^2-a^2=(b-a)(b+a) and c^2-a^2=(c-a)(c+a).
    =(b-a)(c-a)\begin{vmatrix}1 & a & a^2 \\ 0 & 1 & b+a \\ 0 & 1 & c+a\end{vmatrix}
  3. Expand along column 1: only the top entry 1 survives.
    =(b-a)(c-a)\left[1\cdot\left((c+a)-(b+a)\right)\right]=(b-a)(c-a)(c-b)
  4. This equals the right side, so the result is verified.
\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}=(b-a)(c-a)(c-b)

Note. Alternative: expand along row 1 and simplify; the row-operation method is shorter and gives the factors directly.

2025 Annual I, Set B, Q.2 (ix) OR4 marks
Find rank of the matrix \begin{bmatrix}1&1&0&-2\\2&0&2&2\\4&1&3&1\end{bmatrix}.
Solution
  1. Reduce to echelon form. Row 1 already has leading entry 1.
    A=\begin{bmatrix}1 & 1 & 0 & -2 \\ 2 & 0 & 2 & 2 \\ 4 & 1 & 3 & 1\end{bmatrix}
  2. Make zeros below it: R_2-2R_1 and R_3-4R_1.
    \sim_R \begin{bmatrix}1 & 1 & 0 & -2 \\ 0 & -2 & 2 & 6 \\ 0 & -3 & 3 & 9\end{bmatrix}
  3. Make the leading entry of row 2 equal to 1: -\tfrac12R_2.
    \sim_R \begin{bmatrix}1 & 1 & 0 & -2 \\ 0 & 1 & -1 & -3 \\ 0 & -3 & 3 & 9\end{bmatrix}
  4. Make the entry below it zero: R_3+3R_2.
    \sim_R \begin{bmatrix}1 & 1 & 0 & -2 \\ 0 & 1 & -1 & -3 \\ 0 & 0 & 0 & 0\end{bmatrix}
  5. The matrix is in echelon form with 2 non-zero rows.
    \operatorname{Rank}(A)=2
\operatorname{Rank}(A)=2

Note. The matrix is 3\times4, so its rank cannot exceed 3. Row 3 equals R_1+\tfrac32R_2, which is why it vanishes.

2026 Annual I, Set B, Q.2 (xii)4 marks
Find the inverse of matrix if possible: \begin{bmatrix}0&3&4\\-4&2&6\\2&0&5\end{bmatrix}.
Solution
  1. Adjoint method: A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A) exists only if |A|\neq0. Expand |A| along row 1.
    |A|=0\begin{vmatrix}2 & 6 \\ 0 & 5\end{vmatrix}-3\begin{vmatrix}-4 & 6 \\ 2 & 5\end{vmatrix}+4\begin{vmatrix}-4 & 2 \\ 2 & 0\end{vmatrix}
  2. Evaluate each 2\times2 determinant: (-4)(5)-(6)(2)=-32 and (-4)(0)-(2)(2)=-4.
    |A|=0-3(-32)+4(-4)=96-16=80
  3. |A|=80\neq0, so A is non-singular and the inverse exists. Find the nine cofactors.
    \begin{aligned} A_{11}&=+(10-0)=10\\ A_{12}&=-(-20-12)=32\\ A_{13}&=+(0-4)=-4 \\ A_{21}&=-(15-0)=-15\\ A_{22}&=+(0-8)=-8\\ A_{23}&=-(0-6)=6 \\ A_{31}&=+(18-8)=10\\ A_{32}&=-(0+16)=-16\\ A_{33}&=+(0+12)=12 \end{aligned}
  4. The adjoint is the transpose of the matrix of cofactors.
    \operatorname{adj}(A)=\begin{bmatrix}10 & -15 & 10 \\ 32 & -8 & -16 \\ -4 & 6 & 12\end{bmatrix}
  5. Divide by |A|=80 and reduce each fraction.
    A^{-1}=\dfrac1{80}\begin{bmatrix}10 & -15 & 10 \\ 32 & -8 & -16 \\ -4 & 6 & 12\end{bmatrix}=\begin{bmatrix}\tfrac18 & -\tfrac3{16} & \tfrac18 \\ \tfrac25 & -\tfrac1{10} & -\tfrac15 \\ -\tfrac1{20} & \tfrac3{40} & \tfrac3{20}\end{bmatrix}
  6. Check one entry: row 1 of A times column 1 of \operatorname{adj}(A) must give |A|.
    0(10)+3(32)+4(-4)=96-16=80=|A|\;\checkmark
A^{-1}=\begin{bmatrix}\tfrac18 & -\tfrac3{16} & \tfrac18 \\ \tfrac25 & -\tfrac1{10} & -\tfrac15 \\ -\tfrac1{20} & \tfrac3{40} & \tfrac3{20}\end{bmatrix}
2025 Annual I, Set B, Q.2 (xii)4 marks
Solve x+y+z=3; y+z=2; z=2 using Cramer's rule.
Solution
  1. Write the system as AX=B and find |A|. Cramer's rule needs |A|\neq0. The matrix is upper triangular.
    A=\begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix},\qquad |A|=1\cdot1\cdot1=1\neq0
  2. Replace the column of x by the constants 3,2,2 and find |A_x|.
    |A_x|=\begin{vmatrix}3 & 1 & 1 \\ 2 & 1 & 1 \\ 2 & 0 & 1\end{vmatrix}=3(1-0)-1(2-2)+1(0-2)=3-0-2=1
  3. Replace the column of y by the constants and find |A_y|. Rows 2 and 3 are identical, so the determinant is 0.
    |A_y|=\begin{vmatrix}1 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 2 & 1\end{vmatrix}=0
  4. Replace the column of z by the constants and find |A_z| (triangular).
    |A_z|=\begin{vmatrix}1 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 2\end{vmatrix}=1\cdot1\cdot2=2
  5. Apply Cramer's rule.
    x=\dfrac{|A_x|}{|A|}=1,\qquad y=\dfrac{|A_y|}{|A|}=0,\qquad z=\dfrac{|A_z|}{|A|}=2
  6. Check by back-substitution: z=2, then y=2-z=0, then x=3-y-z=1.
x=1,\quad y=0,\quad z=2
2026 Annual I, Set B, Q.48 marks
Solve: 2x+y-z=1; x+2y+z=4; 3x+y+2z=7.
Solution

No method is named, so use Gauss elimination: reduce the augmented matrix A_b to echelon form, then back-substitute.

  1. Write the augmented matrix.
    A_b=\left[\begin{array}{ccc|c}2&1&-1&1\\1&2&1&4\\3&1&2&7\end{array}\right]
  2. Interchange rows 1 and 2 (R_{12}) to get a leading 1 at the top.
    \sim_R\left[\begin{array}{ccc|c}1&2&1&4\\2&1&-1&1\\3&1&2&7\end{array}\right]
  3. Make zeros below it: R_2-2R_1 and R_3-3R_1.
    \sim_R\left[\begin{array}{ccc|c}1&2&1&4\\0&-3&-3&-7\\0&-5&-1&-5\end{array}\right]
  4. Make the leading entry of row 2 equal to 1: -\tfrac13R_2.
    \sim_R\left[\begin{array}{ccc|c}1&2&1&4\\0&1&1&\tfrac73\\0&-5&-1&-5\end{array}\right]
  5. Make the entry below it zero: R_3+5R_2. Here -5+\tfrac{35}{3}=\tfrac{20}{3}.
    \sim_R\left[\begin{array}{ccc|c}1&2&1&4\\0&1&1&\tfrac73\\0&0&4&\tfrac{20}3\end{array}\right]
  6. Make the last leading entry 1: \tfrac14R_3. Now the matrix is in echelon form.
    \sim_R\left[\begin{array}{ccc|c}1&2&1&4\\0&1&1&\tfrac73\\0&0&1&\tfrac53\end{array}\right]
  7. Here \operatorname{Rank}A=\operatorname{Rank}A_b=3 = the number of unknowns, so the solution is unique. Back-substitute from the last row upwards.
    z=\dfrac53,\qquad y=\dfrac73-z=\dfrac73-\dfrac53=\dfrac23,\qquad x=4-2y-z=4-\dfrac43-\dfrac53=1
  8. Check in all three original equations.
    2(1)+\tfrac23-\tfrac53=1,\qquad 1+\tfrac43+\tfrac53=4,\qquad 3+\tfrac23+\tfrac{10}3=7\;\checkmark
x=1,\quad y=\dfrac23,\quad z=\dfrac53

Note. Cramer's rule gives the same result: |A|=12 and x=\dfrac{12}{12}=1, so either method earns full marks. Show the augmented matrix, each row operation and the back-substitution.

2025 Annual I, Set B, Q.58 marks
Solve the system x+2y-3z=4, 2x-3y+4z=5, 3x+4y-5z=6 using the Gaussian elimination method.
Solution
  1. Write the augmented matrix. Row 1 already has a leading 1.
    A_b=\left[\begin{array}{ccc|c}1&2&-3&4\\2&-3&4&5\\3&4&-5&6\end{array}\right]
  2. Make zeros below the leading 1: R_2-2R_1 and R_3-3R_1.
    \sim_R\left[\begin{array}{ccc|c}1&2&-3&4\\0&-7&10&-3\\0&-2&4&-6\end{array}\right]
  3. Make a leading 1 in row 3 using -\tfrac12R_3, which avoids fractions, and interchange rows 2 and 3 (R_{23}).
    \sim_R\left[\begin{array}{ccc|c}1&2&-3&4\\0&1&-2&3\\0&-7&10&-3\end{array}\right]
  4. Make the entry below the second leading 1 zero: R_3+7R_2.
    \sim_R\left[\begin{array}{ccc|c}1&2&-3&4\\0&1&-2&3\\0&0&-4&18\end{array}\right]
  5. Make the last leading entry 1: -\tfrac14R_3. The matrix is in echelon form.
    \sim_R\left[\begin{array}{ccc|c}1&2&-3&4\\0&1&-2&3\\0&0&1&-\tfrac92\end{array}\right]
  6. \operatorname{Rank}A=\operatorname{Rank}A_b=3 = the number of unknowns, so the solution is unique. Back-substitute from the last row.
    \begin{aligned}z&=-\dfrac92\\ y&=3+2z=3-9=-6\\ x&=4-2y+3z=4+12-\dfrac{27}2=\dfrac52\end{aligned}
  7. Check in the original equations.
    \tfrac52-12+\tfrac{27}2=4,\qquad 5+18-18=5,\qquad \tfrac{15}2-24+\tfrac{45}2=6\;\checkmark
x=\dfrac52,\quad y=-6,\quad z=-\dfrac92
2026 Annual I, Set D, MCQ 31 mark
Rank of identity matrix is:
  1. ALess than its order
  2. BEqual to its orderAnswer
  3. CGreater than its order
  4. DUndefined
Why

The identity matrix I_n is already in reduced echelon form: every row has a leading 1, and there is no zero row. So it has n non-zero rows and \operatorname{Rank}I_n=n, its order.

Rank can never exceed the number of rows, so option C is impossible. Option A would need a zero row, and the rank of any matrix is defined, so D is wrong too.

2025 Annual I, Set D, MCQ 31 mark
A system of linear equations is said to be inconsistent if it has:
  1. ANo solutionAnswer
  2. BAt least one solution
  3. CExactly one solution
  4. DExactly two solution
Why

An inconsistent system has no solution at all (\operatorname{Rank}A\neq\operatorname{Rank}A_b). A system with one solution (C) or at least one solution (B) is consistent.

Option D cannot occur: a linear system has no solution, one solution or infinitely many, never exactly two.

2026 Annual I, Set D, MCQ 41 mark
A system is inconsistent if it has:
  1. ANo solutionAnswer
  2. BUnique solution
  3. CInfinite solutions
  4. DRepeated solutions
Why

By definition a system of linear equations with no solution at all is inconsistent; in terms of rank, \operatorname{Rank}A\neq\operatorname{Rank}A_b.

A unique solution (B) and infinitely many solutions (C) both make the system consistent, since at least one solution exists. "Repeated solutions" (D) is not a category for linear systems.

2025 Annual I, Set D, MCQ 41 mark
If \begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=5, then \begin{vmatrix}2a&2b&2c\\3d&3e&3f\\g&h&i\end{vmatrix}=
  1. A6
  2. B10
  3. C15
  4. D30Answer
Why

Take the common factor 2 out of row 1 and the common factor 3 out of row 2. Each factor multiplies the determinant: the new determinant is 2\cdot3\cdot\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=6\cdot5=30.

Option B is 2\cdot5 (row 1 only) and option C is 3\cdot5 (row 2 only). Option A forgets to multiply by the original value 5.

2026 Annual I, Set D, Q.2 (vii)4 marks
Find the rank of matrix: \begin{bmatrix}3&-4&2\\2&3&5\\1&0&1\end{bmatrix}.
Solution
  1. Reduce to echelon form. Interchange rows 1 and 3 (R_{13}) so that a leading 1 comes first.
    A=\begin{bmatrix}3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}\sim_R \begin{bmatrix}1 & 0 & 1 \\ 2 & 3 & 5 \\ 3 & -4 & 2\end{bmatrix}
  2. Make zeros below the leading 1: R_2-2R_1 and R_3-3R_1.
    \sim_R \begin{bmatrix}1 & 0 & 1 \\ 0 & 3 & 3 \\ 0 & -4 & -1\end{bmatrix}
  3. Make the leading entry of row 2 equal to 1: \tfrac13R_2.
    \sim_R \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & -4 & -1\end{bmatrix}
  4. Make the entry below it zero: R_3+4R_2.
    \sim_R \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 3\end{bmatrix}
  5. Make the last leading entry 1: \tfrac13R_3. The matrix is in echelon form with three non-zero rows.
    \sim_R \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix},\qquad \operatorname{Rank}(A)=3
\operatorname{Rank}(A)=3

Note. Check: |A|=-9\neq0 (this is the matrix of Exercise 2.3, Q.5 (ii)), so A is non-singular and its rank is 3.

2025 Annual I, Set D, Q.2 (vii)4 marks
Verify that \begin{vmatrix}a+k&a&a\\a&a+k&a\\a&a&a+k\end{vmatrix}=k^2(3a+k).
Solution
  1. Add columns 2 and 3 to column 1 (C_1+C_2+C_3). This does not change the value of the determinant. Each entry of the new column 1 is 3a+k.
    \begin{vmatrix}a+k & a & a \\ a & a+k & a \\ a & a & a+k\end{vmatrix}=\begin{vmatrix}3a+k & a & a \\ 3a+k & a+k & a \\ 3a+k & a & a+k\end{vmatrix}
  2. Take the common factor 3a+k out of column 1.
    =(3a+k)\begin{vmatrix}1 & a & a \\ 1 & a+k & a \\ 1 & a & a+k\end{vmatrix}
  3. Make zeros below the top entry: R_2-R_1 and R_3-R_1.
    =(3a+k)\begin{vmatrix}1 & a & a \\ 0 & k & 0 \\ 0 & 0 & k\end{vmatrix}
  4. The determinant is triangular, so it is the product of the diagonal entries 1\cdot k\cdot k.
    =(3a+k)\cdot k^2=k^2(3a+k)
  5. This equals the right side, so the result is verified.
\begin{vmatrix}a+k&a&a\\a&a+k&a\\a&a&a+k\end{vmatrix}=k^2(3a+k)

Note. A special case of Review Exercise 2, Q.3: replace a+1 by a+k and l by a in (a+1+2l)(a+1-l)^2.

2026 Annual I, Set D, Q.2 (viii) OR4 marks
Find the inverse of given matrix if possible: \begin{bmatrix}2&1&0\\0&3&2\\2&1&4\end{bmatrix}.
Solution
  1. Adjoint method. First find |A| by expanding along row 1.
    |A|=2\begin{vmatrix}3 & 2 \\ 1 & 4\end{vmatrix}-1\begin{vmatrix}0 & 2 \\ 2 & 4\end{vmatrix}+0\begin{vmatrix}0 & 3 \\ 2 & 1\end{vmatrix}
  2. Evaluate.
    |A|=2(12-2)-1(0-4)+0=20+4=24\neq0
  3. |A|\neq0, so A^{-1} exists. Find the cofactors.
    \begin{aligned} A_{11}&=+(12-2)=10\\ A_{12}&=-(0-4)=4\\ A_{13}&=+(0-6)=-6 \\ A_{21}&=-(4-0)=-4\\ A_{22}&=+(8-0)=8\\ A_{23}&=-(2-2)=0 \\ A_{31}&=+(2-0)=2\\ A_{32}&=-(4-0)=-4\\ A_{33}&=+(6-0)=6 \end{aligned}
  4. The adjoint is the transpose of the cofactor matrix.
    \operatorname{adj}(A)=\begin{bmatrix}10 & -4 & 2 \\ 4 & 8 & -4 \\ -6 & 0 & 6\end{bmatrix}
  5. Divide by 24 and reduce the fractions.
    A^{-1}=\dfrac1{24}\begin{bmatrix}10 & -4 & 2 \\ 4 & 8 & -4 \\ -6 & 0 & 6\end{bmatrix}=\begin{bmatrix}\tfrac5{12} & -\tfrac16 & \tfrac1{12} \\ \tfrac16 & \tfrac13 & -\tfrac16 \\ -\tfrac14 & 0 & \tfrac14\end{bmatrix}
  6. Check one entry: row 1 of A times column 1 of \operatorname{adj}(A) gives |A|.
    2(10)+1(4)+0(-6)=24=|A|\;\checkmark
A^{-1}=\begin{bmatrix}\tfrac5{12} & -\tfrac16 & \tfrac1{12} \\ \tfrac16 & \tfrac13 & -\tfrac16 \\ -\tfrac14 & 0 & \tfrac14\end{bmatrix}

Note. This is Example 12 of the book.

2025 Annual I, Set D, Q.2 (ix) OR4 marks
Find rank of the matrix \begin{bmatrix}1&1&0&-2\\2&0&2&2\\4&1&3&1\end{bmatrix}.
Solution
  1. Reduce to echelon form. Row 1 already has leading entry 1.
    A=\begin{bmatrix}1 & 1 & 0 & -2 \\ 2 & 0 & 2 & 2 \\ 4 & 1 & 3 & 1\end{bmatrix}
  2. Make zeros below it: R_2-2R_1 and R_3-4R_1.
    \sim_R \begin{bmatrix}1 & 1 & 0 & -2 \\ 0 & -2 & 2 & 6 \\ 0 & -3 & 3 & 9\end{bmatrix}
  3. Make the leading entry of row 2 equal to 1: -\tfrac12R_2.
    \sim_R \begin{bmatrix}1 & 1 & 0 & -2 \\ 0 & 1 & -1 & -3 \\ 0 & -3 & 3 & 9\end{bmatrix}
  4. Make the entry below it zero: R_3+3R_2.
    \sim_R \begin{bmatrix}1 & 1 & 0 & -2 \\ 0 & 1 & -1 & -3 \\ 0 & 0 & 0 & 0\end{bmatrix}
  5. The matrix is in echelon form with 2 non-zero rows.
    \operatorname{Rank}(A)=2
\operatorname{Rank}(A)=2

Note. The same matrix appears in Set B (Q.2 (ix) OR). The matrix is 3\times4, so its rank cannot exceed 3; row 3 equals R_1+\tfrac32R_2, which is why it vanishes.

2026 Annual I, Set D, Q.2 (xii)4 marks
For what value of \lambda the given matrix is singular? \begin{bmatrix}1&-1&2\\2&1&\lambda\\3&1&2\end{bmatrix}.
Solution
  1. A matrix is singular when its determinant is zero. Expand |A| along row 1.
    |A|=1\begin{vmatrix}1 & \lambda \\ 1 & 2\end{vmatrix}-(-1)\begin{vmatrix}2 & \lambda \\ 3 & 2\end{vmatrix}+2\begin{vmatrix}2 & 1 \\ 3 & 1\end{vmatrix}
  2. Evaluate the 2\times2 determinants.
    |A|=(2-\lambda)+(4-3\lambda)+2(2-3)=2-\lambda+4-3\lambda-2
  3. Collect terms.
    |A|=4-4\lambda
  4. Singular means |A|=0.
    4-4\lambda=0\;\Rightarrow\;\lambda=1
  5. Check: for \lambda=1 the third row is a combination of the others, R_1=-4R_2+3R_3 gives (-8+9,\,-4+3,\,-4+6)=(1,-1,2), so the rows are dependent.
\lambda=1
2025 Annual I, Set D, Q.2 (xii)4 marks
Solve the system x-y+z=1; x-z=2; z=3 using Cramer's rule.
Solution
  1. Write the system as AX=B and find |A| by expanding along row 3, which has only one non-zero entry.
    A=\begin{bmatrix}1 & -1 & 1 \\ 1 & 0 & -1 \\ 0 & 0 & 1\end{bmatrix},\qquad |A|=1\cdot\begin{vmatrix}1 & -1 \\ 1 & 0\end{vmatrix}=0+1=1\neq0
  2. Replace the column of x by the constants 1,2,3.
    |A_x|=\begin{vmatrix}1 & -1 & 1 \\ 2 & 0 & -1 \\ 3 & 0 & 1\end{vmatrix}=-(-1)\begin{vmatrix}2 & -1 \\ 3 & 1\end{vmatrix}=2+3=5
  3. Replace the column of y by the constants.
    |A_y|=\begin{vmatrix}1 & 1 & 1 \\ 1 & 2 & -1 \\ 0 & 3 & 1\end{vmatrix}=1(2+3)-1(1-0)+1(3-0)=5-1+3=7
  4. Replace the column of z by the constants (expand along row 3).
    |A_z|=\begin{vmatrix}1 & -1 & 1 \\ 1 & 0 & 2 \\ 0 & 0 & 3\end{vmatrix}=3\begin{vmatrix}1 & -1 \\ 1 & 0\end{vmatrix}=3(0+1)=3
  5. Apply Cramer's rule.
    x=\dfrac{5}{1}=5,\qquad y=\dfrac71=7,\qquad z=\dfrac31=3
  6. Check in the original equations.
    5-7+3=1,\qquad 5-3=2,\qquad z=3\;\checkmark
x=5,\quad y=7,\quad z=3
2026 Annual I, Set D, Q.58 marks
Solve the system: x+y+z=6; 2x+3y+z=10; x+2y+3z=11.
Solution

No method is named, so use Gauss elimination on the augmented matrix.

  1. Write the augmented matrix. The first row already has a leading 1.
    A_b=\left[\begin{array}{ccc|c}1&1&1&6\\2&3&1&10\\1&2&3&11\end{array}\right]
  2. Make zeros below the leading 1: R_2-2R_1 and R_3-R_1.
    \sim_R\left[\begin{array}{ccc|c}1&1&1&6\\0&1&-1&-2\\0&1&2&5\end{array}\right]
  3. Row 2 already has leading entry 1. Make the entry below it zero: R_3-R_2.
    \sim_R\left[\begin{array}{ccc|c}1&1&1&6\\0&1&-1&-2\\0&0&3&7\end{array}\right]
  4. Make the last leading entry 1: \tfrac13R_3.
    \sim_R\left[\begin{array}{ccc|c}1&1&1&6\\0&1&-1&-2\\0&0&1&\tfrac73\end{array}\right]
  5. \operatorname{Rank}A=\operatorname{Rank}A_b=3 = the number of unknowns, so the solution is unique. Back-substitute.
    z=\dfrac73,\qquad y=-2+z=\dfrac13,\qquad x=6-y-z=6-\dfrac13-\dfrac73=\dfrac{10}3
  6. Check in the original equations.
    \tfrac{10}3+\tfrac13+\tfrac73=6,\quad \tfrac{20}3+1+\tfrac73=10,\quad \tfrac{10}3+\tfrac23+7=11\;\checkmark
x=\dfrac{10}{3},\quad y=\dfrac13,\quad z=\dfrac73
2025 Annual I, Set D, Q.58 marks
Solve the system x+y+z=6, 2x+3y+z=11, 3x+y+2z=8 using the Gaussian elimination method.
Solution
  1. Write the augmented matrix. Row 1 already has a leading 1.
    A_b=\left[\begin{array}{ccc|c}1&1&1&6\\2&3&1&11\\3&1&2&8\end{array}\right]
  2. Make zeros below the leading 1: R_2-2R_1 and R_3-3R_1.
    \sim_R\left[\begin{array}{ccc|c}1&1&1&6\\0&1&-1&-1\\0&-2&-1&-10\end{array}\right]
  3. Row 2 already has leading entry 1. Make the entry below it zero: R_3+2R_2.
    \sim_R\left[\begin{array}{ccc|c}1&1&1&6\\0&1&-1&-1\\0&0&-3&-12\end{array}\right]
  4. Make the last leading entry 1: -\tfrac13R_3. The matrix is in echelon form.
    \sim_R\left[\begin{array}{ccc|c}1&1&1&6\\0&1&-1&-1\\0&0&1&4\end{array}\right]
  5. \operatorname{Rank}A=\operatorname{Rank}A_b=3 = the number of unknowns, so the solution is unique. Back-substitute from the last row.
    z=4,\qquad y=-1+z=3,\qquad x=6-y-z=6-3-4=-1
  6. Check in the original equations.
    -1+3+4=6,\qquad -2+9+4=11,\qquad -3+3+8=8\;\checkmark
x=-1,\quad y=3,\quad z=4

Exam focus

  1. Solve a 3\times3 system by Gaussian elimination: reduce the augmented matrix to echelon form and back-substitute, then check. (problem)
    The 8-mark Section C question in all four papers: 2026 Set B Q.4, 2026 Set D Q.5, 2025 Set B Q.5 and 2025 Set D Q.5. Exercise 2.6, Q.3 is the practice set.
    Go to answer
  2. Find the rank of a 3\times3 or 3\times4 matrix by reducing it to echelon form. (problem)
    2026 Set B Q.2 (ii) OR, 2026 Set D Q.2 (vii), and 2025 Sets B and D Q.2 (ix) OR (the same 3\times4 matrix), each for 4 marks. Exercise 2.5, Q.2.
    Go to answer
  3. Find the inverse of a 3\times3 matrix by the adjoint method (or say it does not exist). (problem)
    2026 Set B Q.2 (xii) and 2026 Set D Q.2 (viii) OR (Example 12 of the book), 4 marks each. Find |A|, the nine cofactors, the transpose, then divide by |A|. Exercise 2.3, Q.5.
    Go to answer
  4. Solve a system by Cramer's rule, with all four determinants shown. (problem)
    2025 Set B Q.2 (xii) and 2025 Set D Q.2 (xii), 4 marks each. Also a good check on any Section C system. Exercise 2.6, Q.4.
    Go to answer
  5. Verify a determinant identity using row or column operations and common factors, e.g. the Vandermonde determinant, or k^2(3a+k). (problem)
    2025 Set B Q.2 (vii) and 2025 Set D Q.2 (vii). Exercise 2.4 and Review Exercise 2, Q.3 are the practice.
    Go to answer
  6. State the rank and consistency facts: rank of I_n, inconsistent means no solution, a consistent system has at least one solution. (mcq)
    Four MCQs in the 2026 papers (Set B MCQ 8 and 9, Set D MCQ 3 and 4) and two in 2025 (Set B MCQ 3, Set D MCQ 3).
    Go to answer
  7. Use properties of determinants in MCQs: |kA|=k^n|A| and a factor taken out of a row. (mcq)
    2025 Set B MCQ 4 (|3A|=27|A|) and 2025 Set D MCQ 4 (rows multiplied by 2 and 3 give 6\cdot5=30).
    Go to answer
  8. Find cofactors A_{ij} of a matrix, and the value of \lambda that makes a matrix singular (|A|=0). (problem)
    2026 Set B Q.2 (v) (cofactors A_{13},A_{23},A_{31},A_{33}) and 2026 Set D Q.2 (xii) (singular matrix). Exercise 2.3, Q.4 and Review Exercise 2, Q.2.
    Go to answer
  9. Applications of matrices (encryption and decoding, geometric transformations). (problem)
    Not in the 2025 or 2026 new-syllabus papers. The book's Exercise 2.6, Q.9 to Q.13 cover them, so keep them for last.
    Go to answer

Mark this unit as done to fill the Class 11 Mathematics ring.

Next, unit 3
Vectors in space

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