Chapter 5 of 15

Updated 6 August 2026

NCERT Solutions

Matrices — NCERT Solutions

Madhya Pradesh Board · Class 12 · Mathematics

NCERT Solutions for Matrices, Madhya Pradesh Board Class 12 Mathematics: 56 textbook questions solved step by step.

119 questions68 flashcards12 formulas & key relations5 concepts

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56 Questions Solved · 5 Sections

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Exercise 3.1

1In the matrix

A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}

, write: (i) The order of the matrix, (ii) The number of elements, (iii) Write the elements

a_{13}, a_{21}, a_{33}, a_{24}, a_{23}

.Show solution

Given: Matrix $A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}$

(i) Order of the matrix:
The matrix has 3 rows and 4 columns.
$\therefore \text{Order} = 3 \times 4$

(ii) Number of elements:
Number of elements $= m \times n = 3 \times 4 = \mathbf{12}$

(iii) Elements:

$a_{13}$ = element in row 1, column 3 $= 19$
$a_{21}$ = element in row 2, column 1 $= 35$
$a_{33}$ = element in row 3, column 3 $= -5$
$a_{24}$ = element in row 2, column 4 $= 12$
$a_{23}$ = element in row 2, column 3 $= \dfrac{5}{2}$

2If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?Show solution

Concept: For a matrix of order $m \times n$ , the number of elements $= m \times n$ .

For 24 elements: We need all pairs $(m, n)$ such that $m \times n = 24$ .

The possible orders are:
$1\times24,\; 24\times1,\; 2\times12,\; 12\times2,\; 3\times8,\; 8\times3,\; 4\times6,\; 6\times4$

For 13 elements: We need $m \times n = 13$ . Since 13 is prime, the only factor pairs are $1 \times 13$ and $13 \times 1$ .

The possible orders are:
$1\times13,\; 13\times1$

3If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?Show solution

Concept: For a matrix of order $m \times n$ , the number of elements $= m \times n$ .

For 18 elements: We need all pairs $(m, n)$ such that $m \times n = 18$ .

The possible orders are:
$1\times18,\; 18\times1,\; 2\times9,\; 9\times2,\; 3\times6,\; 6\times3$

For 5 elements: Since 5 is prime, $m \times n = 5$ gives only:
$1\times5,\; 5\times1$

4Construct a

2 \times 2

matrix,

A = [a_{ij}]

, whose elements are given by: (i)

a_{ij} = \frac{(i+j)^2}{2}

, (ii)

a_{ij} = \frac{i}{j}

, (iii)

a_{ij} = \frac{(i+2j)^2}{2}

Show solution

A $2\times2$ matrix has elements $a_{11}, a_{12}, a_{21}, a_{22}$ where $i,j \in \{1,2\}$ .

(i) $a_{ij} = \dfrac{(i+j)^2}{2}$

$a_{11}=\frac{(1+1)^2}{2}=\frac{4}{2}=2,\quad a_{12}=\frac{(1+2)^2}{2}=\frac{9}{2}$
$a_{21}=\frac{(2+1)^2}{2}=\frac{9}{2},\quad a_{22}=\frac{(2+2)^2}{2}=\frac{16}{2}=8$
$\therefore A = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}$

(ii) $a_{ij} = \dfrac{i}{j}$

$a_{11}=\frac{1}{1}=1,\quad a_{12}=\frac{1}{2},\quad a_{21}=\frac{2}{1}=2,\quad a_{22}=\frac{2}{2}=1$
$\therefore A = \begin{bmatrix} 1 & \frac{1}{2} \\ 2 & 1 \end{bmatrix}$

(iii) $a_{ij} = \dfrac{(i+2j)^2}{2}$

$a_{11}=\frac{(1+2)^2}{2}=\frac{9}{2},\quad a_{12}=\frac{(1+4)^2}{2}=\frac{25}{2}$
$a_{21}=\frac{(2+2)^2}{2}=\frac{16}{2}=8,\quad a_{22}=\frac{(2+4)^2}{2}=\frac{36}{2}=18$
$\therefore A = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}$

5Construct a

3 \times 4

matrix, whose elements are given by: (i)

a_{ij} = \frac{1}{2}|-3i+j|

, (ii)

a_{ij} = 2i - j

Show solution

A $3\times4$ matrix has $i \in \{1,2,3\}$ and $j \in \{1,2,3,4\}$ .

(i) $a_{ij} = \dfrac{1}{2}|-3i+j|$

Computing each element:

$a_{11}=\frac{1}{2}|-3+1|=1,\; a_{12}=\frac{1}{2}|-3+2|=\frac{1}{2},\; a_{13}=\frac{1}{2}|-3+3|=0,\; a_{14}=\frac{1}{2}|-3+4|=\frac{1}{2}$

$a_{21}=\frac{1}{2}|-6+1|=\frac{5}{2},\; a_{22}=\frac{1}{2}|-6+2|=2,\; a_{23}=\frac{1}{2}|-6+3|=\frac{3}{2},\; a_{24}=\frac{1}{2}|-6+4|=1$

$a_{31}=\frac{1}{2}|-9+1|=4,\; a_{32}=\frac{1}{2}|-9+2|=\frac{7}{2},\; a_{33}=\frac{1}{2}|-9+3|=3,\; a_{34}=\frac{1}{2}|-9+4|=\frac{5}{2}$

$\therefore A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix}$

(ii) $a_{ij} = 2i - j$

$a_{11}=2-1=1,\; a_{12}=2-2=0,\; a_{13}=2-3=-1,\; a_{14}=2-4=-2$

$a_{21}=4-1=3,\; a_{22}=4-2=2,\; a_{23}=4-3=1,\; a_{24}=4-4=0$

$a_{31}=6-1=5,\; a_{32}=6-2=4,\; a_{33}=6-3=3,\; a_{34}=6-4=2$

$\therefore A = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}$

6Find the values of

x, y

and

z

from the following equations: (i)

\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}

, (ii)

\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}

, (iii)

\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}

Show solution

Concept: Two matrices are equal if and only if their corresponding elements are equal.

(i) Equating corresponding elements:
$y = 4,\quad z = 3,\quad x = 1$
$\boxed{x=1,\; y=4,\; z=3}$

(ii) Equating corresponding elements:
$x + y = 6 \quad \cdots(1)$
$5 + z = 5 \Rightarrow z = 0 \quad \cdots(2)$
$xy = 8 \quad \cdots(3)$

From (1): $y = 6 - x$ . Substituting in (3):
$x(6-x) = 8 \Rightarrow 6x - x^2 = 8 \Rightarrow x^2 - 6x + 8 = 0$
$(x-4)(x-2) = 0 \Rightarrow x = 4 \text{ or } x = 2$

If $x = 4$ , then $y = 2$ . If $x = 2$ , then $y = 4$ .
$\boxed{x=4,\; y=2,\; z=0 \quad \text{or} \quad x=2,\; y=4,\; z=0}$

(iii) Equating corresponding elements:
$x + y + z = 9 \quad \cdots(1)$
$x + z = 5 \quad \cdots(2)$
$y + z = 7 \quad \cdots(3)$

From (1) and (2): $y = 9 - 5 = 4$

From (3): $z = 7 - y = 7 - 4 = 3$

From (2): $x = 5 - z = 5 - 3 = 2$
$\boxed{x=2,\; y=4,\; z=3}$

7Find the value of

a, b, c

and

d

from the equation:

\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}

Show solution

Given: $\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$

Equating corresponding elements:
$a - b = -1 \quad \cdots(1)$
$2a + c = 5 \quad \cdots(2)$
$2a - b = 0 \quad \cdots(3)$
$3c + d = 13 \quad \cdots(4)$

Subtracting (1) from (3):
$(2a - b) - (a - b) = 0 - (-1)$
$a = 1$

From (1): $b = a + 1 = 1 + 1 = 2$

From (2): $c = 5 - 2a = 5 - 2 = 3$

From (4): $d = 13 - 3c = 13 - 9 = 4$

$\boxed{a = 1,\; b = 2,\; c = 3,\; d = 4}$

A = [a_{ij}]_{m \times n}

is a square matrix, if (A)

m < n

(B)

m > n

(C)

m = n

(D) None of theseShow solution

Correct Answer: (C) $m = n$

A matrix is called a square matrix when the number of rows equals the number of columns, i.e., $m = n$ .

9Which of the given values of

x

and

y

make the following pair of matrices equal:

\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix}

\begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}

? (A)

x = \frac{-1}{3}, y = 7

(B) Not possible to find (C)

y = 7, x = \frac{-2}{3}

(D)

x = \frac{-1}{3}, y = \frac{-2}{3}

Show solution

Correct Answer: (B) Not possible to find

Equating corresponding elements:
$3x + 7 = 0 \Rightarrow x = \frac{-7}{3} \quad \cdots(1)$
$2 - 3x = 4 \Rightarrow 3x = -2 \Rightarrow x = \frac{-2}{3} \quad \cdots(2)$

From (1) and (2), we get two different values of $x$ , which is a contradiction. Hence it is not possible to find values of $x$ and $y$ that satisfy all conditions simultaneously.

10The number of all possible matrices of order

3 \times 3

with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512Show solution

Correct Answer: (D) 512

A $3 \times 3$ matrix has $3 \times 3 = 9$ entries. Each entry can be filled in 2 ways (either 0 or 1).

$\text{Total number of matrices} = 2^9 = 512$

Exercise 3.2

1Let

A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}

B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}

C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}

. Find each of the following: (i) A+B, (ii) A-B, (iii) 3A-C, (iv) AB, (v) BAShow solution

Given: $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$ , $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$

(i) $A + B$ :
$A+B = \begin{bmatrix} 2+1 & 4+3 \\ 3+(-2) & 2+5 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}$

(ii) $A - B$ :
$A-B = \begin{bmatrix} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}$

(iii) $3A - C$ :
$3A = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}$
$3A - C = \begin{bmatrix} 6-(-2) & 12-5 \\ 9-3 & 6-4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}$

(iv) $AB$ :
$AB = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$
$= \begin{bmatrix} 2(1)+4(-2) & 2(3)+4(5) \\ 3(1)+2(-2) & 3(3)+2(5) \end{bmatrix} = \begin{bmatrix} 2-8 & 6+20 \\ 3-4 & 9+10 \end{bmatrix} = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}$

(v) $BA$ :
$BA = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$
$= \begin{bmatrix} 1(2)+3(3) & 1(4)+3(2) \\ -2(2)+5(3) & -2(4)+5(2) \end{bmatrix} = \begin{bmatrix} 2+9 & 4+6 \\ -4+15 & -8+10 \end{bmatrix} = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}$

2Compute the following: (i)

\begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix}

, (ii)

\begin{bmatrix} a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2 \end{bmatrix} + \begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix}

, (iii)

\begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix}

, (iv)

\begin{bmatrix} \cos^2 x & \sin^2 x \\ \sin^2 x & \cos^2 x \end{bmatrix} + \begin{bmatrix} \sin^2 x & \cos^2 x \\ \cos^2 x & \sin^2 x \end{bmatrix}

Show solution

(i)
$\begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}$

(ii)
$\begin{bmatrix} a^2+b^2+2ab & b^2+c^2+2bc \\ a^2+c^2-2ac & a^2+b^2-2ab \end{bmatrix} = \begin{bmatrix} (a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2 \end{bmatrix}$

(iii)
$\begin{bmatrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{bmatrix} = \begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}$

(iv)
$\begin{bmatrix} \cos^2x+\sin^2x & \sin^2x+\cos^2x \\ \sin^2x+\cos^2x & \cos^2x+\sin^2x \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$
(Using $\sin^2x + \cos^2x = 1$ )

3Compute the indicated products: (i)

\begin{bmatrix} a & b \\ -b & a \end{bmatrix}\begin{bmatrix} a & -b \\ b & a \end{bmatrix}

, (ii)

\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \end{bmatrix}

, (iii)

\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}

, (iv)

\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}

, (v)

\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix}

, (vi)

\begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}

Show solution

(i)
$\begin{bmatrix} a & b \\ -b & a \end{bmatrix}\begin{bmatrix} a & -b \\ b & a \end{bmatrix} = \begin{bmatrix} a^2+b^2 & -ab+ab \\ -ab+ab & b^2+a^2 \end{bmatrix} = \begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix}$

(ii)
$\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}_{3\times1}\begin{bmatrix} 2 & 3 & 4 \end{bmatrix}_{1\times3} = \begin{bmatrix} 1\cdot2 & 1\cdot3 & 1\cdot4 \\ 2\cdot2 & 2\cdot3 & 2\cdot4 \\ 3\cdot2 & 3\cdot3 & 3\cdot4 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}$

(iii)
$\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}$
$= \begin{bmatrix} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}$

(iv)
$\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}$

Row 1: $(2+0+12,\; -6+6+0,\; 10+12+20) = (14, 0, 42)$

Row 2: $(3+0+15,\; -9+8+0,\; 15+16+25) = (18, -1, 56)$

Row 3: $(4+0+18,\; -12+10+0,\; 20+20+30) = (22, -2, 70)$

$= \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}$

(v)
$\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix}$

Row 1: $(2-1,\; 0+2,\; 2+1) = (1, 2, 3)$

Row 2: $(3-2,\; 0+4,\; 3+2) = (1, 4, 5)$

Row 3: $(-1-1,\; 0+2,\; -1+1) = (-2, 2, 0)$

$= \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix}$

(vi)
$\begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}$

Row 1: $(6-1+9,\; -9+0+3) = (14, -6)$

Row 2: $(-2+0+6,\; 3+0+2) = (4, 5)$

$= \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$

4If

A = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix}

B = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix}

and

C = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix}

, then compute

(A+B)

and

(B-C)

. Also, verify that

A+(B-C) = (A+B)-C

.Show solution

Computing $A+B$ :
$A+B = \begin{bmatrix} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{bmatrix} = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix}$

Computing $B-C$ :
$B-C = \begin{bmatrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3 \end{bmatrix} = \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix}$

LHS: $A+(B-C)$ :
$= \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$

RHS: $(A+B)-C$ :
$= \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}$

Since LHS = RHS, $A+(B-C) = (A+B)-C$ is verified.

5If

A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}

and

B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}

, then compute

3A - 5B

.Show solution

Computing $3A$ :
$3A = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$

Computing $5B$ :
$5B = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$

Therefore:
$3A - 5B = \begin{bmatrix} 2-2 & 3-3 & 5-5 \\ 1-1 & 2-2 & 4-4 \\ 7-7 & 6-6 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

6Simplify

\cos\theta\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix}

Show solution

$= \begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix}$

$= \begin{bmatrix} \cos^2\theta+\sin^2\theta & \cos\theta\sin\theta-\sin\theta\cos\theta \\ -\sin\theta\cos\theta+\sin\theta\cos\theta & \cos^2\theta+\sin^2\theta \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$

7Find X and Y, if (i)

X+Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}

and

X-Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}

, (ii)

2X+3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}

and

3X+2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}

Show solution

(i) Adding the two equations:
$2X = \begin{bmatrix} 7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$
$X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$

Subtracting:
$2Y = \begin{bmatrix} 7-3 & 0 \\ 2-0 & 5-3 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix}$
$Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$

(ii) Let $2X+3Y = P = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}$ and $3X+2Y = Q = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$ .

Multiply first equation by 2 and second by 3:
$4X + 6Y = \begin{bmatrix} 4 & 6 \\ 8 & 0 \end{bmatrix}$
$9X + 6Y = \begin{bmatrix} 6 & -6 \\ -3 & 15 \end{bmatrix}$

Subtracting:
$5X = \begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix} \Rightarrow X = \frac{1}{5}\begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix}$

Multiply first equation by 3 and second by 2:
$6X + 9Y = \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix}$
$6X + 4Y = \begin{bmatrix} 4 & -4 \\ -2 & 10 \end{bmatrix}$

Subtracting:
$5Y = \begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix} \Rightarrow Y = \frac{1}{5}\begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix}$

8Find X, if

Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}

and

2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}

Show solution

Given: $2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$

$2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}$

$X = \frac{1}{2}\begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}$

9Find

x

and

y

, if

2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}

Show solution

$\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$

$\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$

Equating corresponding elements:
$2 + y = 5 \Rightarrow y = 3$
$2x + 2 = 8 \Rightarrow 2x = 6 \Rightarrow x = 3$

$\boxed{x = 3,\; y = 3}$

10Solve the equation for

x, y, z

and

t

, if

2\begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}

Show solution

$\begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$

$\begin{bmatrix} 2x+3 & 2z-3 \\ 2y & 2t+6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$

Equating corresponding elements:
$2x+3=9 \Rightarrow x=3$
$2z-3=15 \Rightarrow z=9$
$2y=12 \Rightarrow y=6$
$2t+6=18 \Rightarrow t=6$

$\boxed{x=3,\; y=6,\; z=9,\; t=6}$

11If

x\begin{bmatrix} 2 \\ 3 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}

, find the values of

x

and

y

.Show solution

$\begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$

$\begin{bmatrix} 2x-y \\ 3x+y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$

Equating corresponding elements:
$2x - y = 10 \quad \cdots(1)$
$3x + y = 5 \quad \cdots(2)$

Adding (1) and (2):
$5x = 15 \Rightarrow x = 3$

From (1): $y = 2(3) - 10 = -4$

$\boxed{x = 3,\; y = -4}$

12Given

3\begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix}

, find the values of

x, y, z

and

w

.Show solution

$\begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix}$

Equating corresponding elements:
$3x = x+4 \Rightarrow 2x = 4 \Rightarrow x = 2$
$3y = 6+x+y \Rightarrow 2y = 6+2 = 8 \Rightarrow y = 4$
$3z = -1+z+w \Rightarrow 2z - w = -1 \quad \cdots(1)$
$3w = 2w+3 \Rightarrow w = 3$

From (1): $2z - 3 = -1 \Rightarrow z = 1$

$\boxed{x=2,\; y=4,\; z=1,\; w=3}$

13If

F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}

, show that

F(x)F(y) = F(x+y)

.Show solution

Given: $F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Computing $F(x)F(y)$ :
$F(x)F(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Element $(1,1)$ : $\cos x\cos y - \sin x\sin y = \cos(x+y)$

Element $(1,2)$ : $-\cos x\sin y - \sin x\cos y = -\sin(x+y)$

Element $(1,3)$ : $0$

Element $(2,1)$ : $\sin x\cos y + \cos x\sin y = \sin(x+y)$

Element $(2,2)$ : $-\sin x\sin y + \cos x\cos y = \cos(x+y)$

Element $(2,3)$ : $0$

Element $(3,1)$ : $0$ , Element $(3,2)$ : $0$ , Element $(3,3)$ : $1$

$\therefore F(x)F(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(x+y)$

Hence proved.

14Show that (i)

\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \neq \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}

, (ii)

\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}

Show solution

(i) Computing LHS:
$\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 10-3 & 5-4 \\ 12+21 & 6+28 \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$

Computing RHS:
$\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} = \begin{bmatrix} 10+6 & -2+7 \\ 15+24 & -3+28 \end{bmatrix} = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$

Since $\begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix} \neq \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$ , the result is proved.

(ii) Let $P = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$ and $Q = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}$

Computing PQ:
Row 1: $(-1+0+6,\; 1-2+9,\; 0+2+12) = (5, 8, 14)$
Row 2: $(0+0+0,\; 0-1+0,\; 0+1+0) = (0, -1, 1)$
Row 3: $(-1+0+0,\; 1-1+0,\; 0+1+0) = (-1, 0, 1)$
$PQ = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$

Computing QP:
Row 1: $(-1+0+0,\; -2+1+0,\; -3+0+0) = (-1, -1, -3)$
Row 2: $(0+1+1,\; 0-1+1,\; 0+0+0) = (2, 0, 0)$ — let me recalculate:
Row 1: $(-1)(1)+(1)(0)+(0)(1),\; (-1)(2)+(1)(1)+(0)(1),\; (-1)(3)+(1)(0)+(0)(0) = (-1, -1, -3)$
Row 2: $(0)(1)+(-1)(0)+(1)(1),\; (0)(2)+(-1)(1)+(1)(1),\; (0)(3)+(-1)(0)+(1)(0) = (1, 0, 0)$
Row 3: $(2)(1)+(3)(0)+(4)(1),\; (2)(2)+(3)(1)+(4)(1),\; (2)(3)+(3)(0)+(4)(0) = (6, 11, 6)$
$QP = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}$

Since $PQ \neq QP$ , the result is proved.

15Find

A^2 - 5A + 6I

, if

A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}

Show solution

Step 1: Compute $A^2 = A \cdot A$
$A^2 = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}\begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$

Row 1: $(4+0+1,\; 0+0-1,\; 2+0+0) = (5, -1, 2)$
Row 2: $(4+2+3,\; 0+1-3,\; 2+3+0) = (9, -2, 5)$
Row 3: $(2-2+0,\; 0-1+0,\; 1-3+0) = (0, -1, -2)$

$A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$

Step 2: Compute $5A$
$5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}$

Step 3: Compute $6I$
$6I = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

Step 4: $A^2 - 5A + 6I$
$= \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+6 \end{bmatrix} = \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$

16If

A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}

, prove that

A^3 - 6A^2 + 7A + 2I = 0

.Show solution

Step 1: Compute $A^2$
$A^2 = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$

Row 1: $(1+0+4,\; 0+0+0,\; 2+0+6) = (5, 0, 8)$
Row 2: $(0+0+2,\; 0+4+0,\; 0+2+3) = (2, 4, 5)$
Row 3: $(2+0+6,\; 0+0+0,\; 4+0+9) = (8, 0, 13)$

$A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}$

Step 2: Compute $A^3 = A^2 \cdot A$
$A^3 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$

Row 1: $(5+0+16,\; 0+0+0,\; 10+0+24) = (21, 0, 34)$
Row 2: $(2+0+10,\; 0+8+0,\; 4+4+15) = (12, 8, 23)$
Row 3: $(8+0+26,\; 0+0+0,\; 16+0+39) = (34, 0, 55)$

$A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}$

Step 3: Compute $A^3 - 6A^2 + 7A + 2I$
$6A^2 = \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix},\quad 7A = \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix},\quad 2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$

$A^3 - 6A^2 + 7A + 2I = \begin{bmatrix} 21-30+7+2 & 0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0 & 55-78+21+2 \end{bmatrix}$

$= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$

Hence proved.

17If

A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}

and

I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

, find

k

so that

A^2 = kA - 2I

.Show solution

Step 1: Compute $A^2$
$A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$
$= \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$

Step 2: Compute $kA - 2I$
$kA - 2I = k\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}$

Step 3: Equate $A^2 = kA - 2I$
$\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}$

From element $(1,2)$ : $-2k = -2 \Rightarrow k = 1$

Verification: $3k-2 = 1$ ✓, $4k = 4$ ✓, $-2k-2 = -4$ ✓

$\boxed{k = 1}$

18If

A = \begin{bmatrix} 0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix}

and

I

is the identity matrix of order 2, show that

I + A = (I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}

.Show solution

Let $t = \tan\dfrac{\alpha}{2}$ . Then:
$I + A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}, \quad I - A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$

Using the identities:
$\cos\alpha = \frac{1-t^2}{1+t^2}, \quad \sin\alpha = \frac{2t}{1+t^2}$

Computing RHS = $(I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$ :

$= \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}\begin{bmatrix} \frac{1-t^2}{1+t^2} & \frac{-2t}{1+t^2} \\ \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2} \end{bmatrix}$

Element $(1,1)$ : $\dfrac{1-t^2}{1+t^2} + \dfrac{2t^2}{1+t^2} = \dfrac{1+t^2}{1+t^2} = 1$

Element $(1,2)$ : $\dfrac{-2t}{1+t^2} + \dfrac{t(1-t^2)}{1+t^2} = \dfrac{-2t+t-t^3}{1+t^2} = \dfrac{-t(1+t^2)}{1+t^2} = -t$

Element $(2,1)$ : $\dfrac{-t(1-t^2)}{1+t^2} + \dfrac{2t}{1+t^2} = \dfrac{-t+t^3+2t}{1+t^2} = \dfrac{t(1+t^2)}{1+t^2} = t$

Element $(2,2)$ : $\dfrac{2t^2}{1+t^2} + \dfrac{1-t^2}{1+t^2} = \dfrac{1+t^2}{1+t^2} = 1$

$\text{RHS} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = I + A = \text{LHS}$

Hence proved.

19A trust fund has ₹30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) ₹1800, (b) ₹2000

Let the amount invested in the first bond be ₹ $x$ and in the second bond be ₹ $(30000 - x)$ .

Using matrix multiplication, the interest earned is:
$[x \quad (30000-x)]\begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix} = \text{Total Interest}$

(a) Total interest = ₹1800:
$0.05x + 0.07(30000-x) = 1800$
$0.05x + 2100 - 0.07x = 1800$
$-0.02x = -300$
$x = 15000$

∴ ₹15,000 in the first bond and ₹15,000 in the second bond.

(b) Total interest = ₹2000:
$0.05x + 0.07(30000-x) = 2000$
$0.05x + 2100 - 0.07x = 2000$
$-0.02x = -100$
$x = 5000$

∴ ₹5,000 in the first bond and ₹25,000 in the second bond.

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20The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹80, ₹60 and ₹40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Number of books (in dozens converted to units):

Chemistry: $10 \times 12 = 120$
Physics: $8 \times 12 = 96$
Economics: $10 \times 12 = 120$

Represent as row matrix: $A = [120 \quad 96 \quad 120]$

Selling prices as column matrix: $B = \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}$

Total amount = $AB$ :
$AB = [120 \times 80 + 96 \times 60 + 120 \times 40]$
$= [9600 + 5760 + 4800]$
$= [20160]$

$\therefore \text{Total amount} = \mathbf{₹20,160}$

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21The restriction on

n, k

and

p

so that

PY + WY

will be defined are: (A)

k=3, p=n

(B)

k

is arbitrary,

p=2

(C)

p

is arbitrary,

k=3

(D)

k=2, p=3

Correct Answer: (A) $k = 3, p = n$

Given orders: $X: 2\times n$ , $Y: 3\times k$ , $Z: 2\times p$ , $W: n\times 3$ , $P: p\times k$ .

For $PY$ : $P$ is $p\times k$ and $Y$ is $3\times k$ . For multiplication $PY$ to be defined, number of columns of $P$ must equal number of rows of $Y$ : $k = 3$ . Result is $p\times k$ .

For $WY$ : $W$ is $n\times 3$ and $Y$ is $3\times k$ . This is defined (columns of $W$ = rows of $Y$ = 3). Result is $n\times k$ .

For $PY + WY$ : Both must have the same order, so $p = n$ .

Therefore $k = 3$ and $p = n$ .

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22If

n = p

, then the order of the matrix

7X - 5Z

is: (A)

p\times 2

(B)

2\times n

(C)

n\times 3

(D)

p\times n

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Exercise 3.3

1Find the transpose of each of the following matrices: (i)

\begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}

, (ii)

\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}

, (iii)

\begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}

Concept: Transpose of a matrix is obtained by interchanging its rows and columns.

(i) $A = \begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}_{3\times1}$
$A' = \begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}_{1\times3}$

(ii) $A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$
$A' = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$

(iii) $A = \begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}$
$A' = \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}$

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2If

A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}

and

B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}

, then verify that (i)

(A+B)' = A'+B'

, (ii)

(A-B)' = A'-B'

Computing $A'$ and $B'$ :
$A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}, \quad B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}$

(i) Verify $(A+B)' = A'+B'$ :
$A+B = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}$
$\text{LHS: }(A+B)' = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$
$\text{RHS: }A'+B' = \begin{bmatrix} -1-4 & 5+1 & -2+1 \\ 2+1 & 7+2 & 1+3 \\ 3-5 & 9+0 & 1+1 \end{bmatrix} = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$
LHS = RHS ✓

(ii) Verify $(A-B)' = A'-B'$ :
$A-B = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}$
$\text{LHS: }(A-B)' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$
$\text{RHS: }A'-B' = \begin{bmatrix} -1+4 & 5-1 & -2-1 \\ 2-1 & 7-2 & 1-3 \\ 3+5 & 9-0 & 1-1 \end{bmatrix} = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$
LHS = RHS ✓

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3If

A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}

and

B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}

, then verify that (i)

(A+B)' = A'+B'

, (ii)

(A-B)' = A'-B'

Finding $A$ from $A'$ :
$A = (A')' = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}$

Finding $B'$ :
$B' = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}$

(i) Verify $(A+B)' = A'+B'$ :
$A+B = \begin{bmatrix} 3-1 & -1+2 & 0+1 \\ 4+1 & 2+2 & 1+3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix}$
$\text{LHS: }(A+B)' = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$
$\text{RHS: }A'+B' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$
LHS = RHS ✓

(ii) Verify $(A-B)' = A'-B'$ :
$A-B = \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix}$
$\text{LHS: }(A-B)' = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$
$\text{RHS: }A'-B' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$
LHS = RHS ✓

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4If

A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}

and

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

, then find

(A+2B)'

Finding $A$ :
$A = (A')' = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}$

Computing $A + 2B$ :
$2B = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$
$A + 2B = \begin{bmatrix} -2-2 & 1+0 \\ 3+2 & 2+4 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}$

Computing $(A+2B)'$ :
$(A+2B)' = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$

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5For the matrices

A

and

B

, verify that

(AB)' = B'A'

, where (i)

A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}

B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}

, (ii)

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

B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}

(i)
$AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}\begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}$

$\text{LHS: }(AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

$B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix},\quad A' = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$

$\text{RHS: }B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

LHS = RHS ✓

(ii)
$AB = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}$

$\text{LHS: }(AB)' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$

$B' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix},\quad A' = \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}$

$\text{RHS: }B'A' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}\begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$

LHS = RHS ✓

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6If (i)

A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}

, then verify that

A'A = I

. (ii) If

A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}

, then verify that

A'A = I

(i)
$A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$

$= \begin{bmatrix} \cos^2\alpha+\sin^2\alpha & \cos\alpha\sin\alpha-\sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha-\cos\alpha\sin\alpha & \sin^2\alpha+\cos^2\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \checkmark$

(ii)
$A' = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}$

$A'A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}\begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}$

$= \begin{bmatrix} \sin^2\alpha+\cos^2\alpha & \sin\alpha\cos\alpha-\cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha-\sin\alpha\cos\alpha & \cos^2\alpha+\sin^2\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \checkmark$

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7(i) Show that the matrix

A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}

is a symmetric matrix. (ii) Show that the matrix

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

is a skew symmetric matrix.

(i) A matrix is symmetric if $A' = A$ .

$A' = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} = A$

Since $A' = A$ , the matrix is symmetric. ✓

(ii) A matrix is skew symmetric if $A' = -A$ .

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

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

Since $A' = -A$ , the matrix is skew symmetric. ✓

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8For the matrix

A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}

, verify that (i)

(A+A')

is a symmetric matrix, (ii)

(A-A')

is a skew symmetric matrix.

$A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$

(i) $A + A'$ :
$A + A' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$

Let $P = A + A'$ . Then:
$P' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} = P$

Since $P' = P$ , $(A+A')$ is a symmetric matrix. ✓

(ii) $A - A'$ :
$A - A' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

Let $Q = A - A'$ . Then:
$Q' = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = -Q$

Since $Q' = -Q$ , $(A-A')$ is a skew symmetric matrix. ✓

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9Find

\frac{1}{2}(A+A')

and

\frac{1}{2}(A-A')

, when

A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}

$A' = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}$

Computing $\frac{1}{2}(A+A')$ :
$A + A' = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
$\frac{1}{2}(A+A') = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Computing $\frac{1}{2}(A-A')$ :
$A - A' = \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix}$
$\frac{1}{2}(A-A') = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} = A$

(Note: Since $A$ is already skew symmetric, $\frac{1}{2}(A+A') = O$ and $\frac{1}{2}(A-A') = A$ .)

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10Express the following matrices as the sum of a symmetric and a skew symmetric matrix: (i)

\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}

, (ii)

\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}

, (iii)

\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}

, (iv)

\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}

Concept: Any square matrix $A$ can be written as $A = P + Q$ where $P = \frac{1}{2}(A+A')$ is symmetric and $Q = \frac{1}{2}(A-A')$ is skew symmetric.

(i) $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ , $A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$

$P = \frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$

$Q = \frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

$A = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

(ii) $A = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$

Since $A' = A$ , the matrix is already symmetric.
$P = A = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}, \quad Q = O$

(iii) $A = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}$ , $A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}$

$P = \frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix}$

$Q = \frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix}$

$A = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix}$

(iv) $A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}$ , $A' = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}$

$P = \frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}$

$Q = \frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$

$A = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$

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11If A, B are symmetric matrices of same order, then AB – BA is a (A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix

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12If

A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}

, and

A + A' = I

, then the value of

\alpha

is (A)

\frac{\pi}{6}

(B)

\frac{\pi}{3}

(C)

\pi

(D)

\frac{3\pi}{2}

Correct Answer: (B) $\dfrac{\pi}{3}$

$A' = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$

$A + A' = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$

Equating: $2\cos\alpha = 1 \Rightarrow \cos\alpha = \dfrac{1}{2} \Rightarrow \alpha = \dfrac{\pi}{3}$

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Exercise 3.4

1Matrices A and B will be inverse of each other only if (A)

AB = BA

(B)

AB = BA = 0

(C)

AB = 0

BA = I

(D)

AB = BA = I

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Miscellaneous Exercise on Chapter 3

1If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.

Given: $A$ and $B$ are symmetric matrices, so $A' = A$ and $B' = B$ .

To prove: $AB - BA$ is skew symmetric, i.e., $(AB-BA)' = -(AB-BA)$ .

Proof:
$(AB - BA)' = (AB)' - (BA)'$
$= B'A' - A'B' \quad [\text{using } (PQ)' = Q'P']$
$= BA - AB \quad [\text{since } A'=A, B'=B]$
$= -(AB - BA)$

Since $(AB-BA)' = -(AB-BA)$ , the matrix $AB - BA$ is skew symmetric. Hence proved.

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2Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Case 1: A is symmetric, i.e., $A' = A$ .

Let $P = B'AB$ . Then:
$P' = (B'AB)' = B'A'(B')' = B'A'B = B'AB = P$
(since $A' = A$ and $(B')' = B$ )

Since $P' = P$ , $B'AB$ is symmetric.

Case 2: A is skew symmetric, i.e., $A' = -A$ .

$P' = (B'AB)' = B'A'(B')' = B'A'B = B'(-A)B = -B'AB = -P$

Since $P' = -P$ , $B'AB$ is skew symmetric.

Hence proved.

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3Find the values of

x, y, z

if the matrix

A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}

satisfy the equation

A'A = I

Computing $A'$ :
$A' = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}$

Computing $A'A = I$ :
$A'A = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}\begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Element $(1,1)$ : $0 + x^2 + x^2 = 1 \Rightarrow 2x^2 = 1 \Rightarrow x^2 = \dfrac{1}{2} \Rightarrow x = \pm\dfrac{1}{\sqrt{2}}$

Element $(2,2)$ : $4y^2 + y^2 + y^2 = 1 \Rightarrow 6y^2 = 1 \Rightarrow y^2 = \dfrac{1}{6} \Rightarrow y = \pm\dfrac{1}{\sqrt{6}}$

Element $(3,3)$ : $z^2 + z^2 + z^2 = 1 \Rightarrow 3z^2 = 1 \Rightarrow z^2 = \dfrac{1}{3} \Rightarrow z = \pm\dfrac{1}{\sqrt{3}}$

Verification of off-diagonal elements confirms these values are consistent.

$\boxed{x = \pm\frac{1}{\sqrt{2}},\; y = \pm\frac{1}{\sqrt{6}},\; z = \pm\frac{1}{\sqrt{3}}}$

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4For what values of

x

[1\quad 2\quad 1]\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = 0

Step 1: Compute $[1\quad 2\quad 1]\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}$

$= [1+4+1,\; 2+0+0,\; 0+2+2] = [6,\; 2,\; 4]$

Step 2: Multiply by $\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix}$ :

$[6\quad 2\quad 4]\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = 0 + 4 + 4x = 4 + 4x$

Step 3: Set equal to 0:
$4 + 4x = 0 \Rightarrow x = -1$

$\boxed{x = -1}$

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5If

A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

, show that

A^2 - 5A + 7I = 0

Step 1: Compute $A^2$
$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

Step 2: Compute $5A$
$5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$

Step 3: Compute $7I$
$7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

Step 4:
$A^2 - 5A + 7I = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

Hence proved.

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6Find

x

, if

\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = 0

Step 1: Compute $\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$

$= [x+0-2,\; 0-10+0,\; 2x-5-3] = [x-2,\; -10,\; 2x-8]$

Step 2: Multiply by $\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix}$ :

$[x-2,\; -10,\; 2x-8]\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = x(x-2) - 40 + (2x-8) = 0$

$x^2 - 2x - 40 + 2x - 8 = 0$
$x^2 - 48 = 0$
$x^2 = 48$
$x = \pm 4\sqrt{3}$

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7A manufacturer produces three products

x, y, z

which he sells in two markets. Annual sales are indicated below: Market I: 10,000; 2,000; 18,000 and Market II: 6,000; 20,000; 8,000. (a) If unit sale prices of

x, y

and

z

are ₹2.50, ₹1.50 and ₹1.00 respectively, find the total revenue in each market with the help of matrix algebra. (b) If the unit costs of the above three commodities are ₹2.00, ₹1.00 and 50 paise respectively. Find the gross profit.

Sales matrix:
$S = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}$

(a) Revenue:

Price matrix: $P = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$

$\text{Revenue} = SP = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}\begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$

Market I: $10000(2.50) + 2000(1.50) + 18000(1.00) = 25000 + 3000 + 18000 = 46000$

Market II: $6000(2.50) + 20000(1.50) + 8000(1.00) = 15000 + 30000 + 8000 = 53000$

$\text{Revenue} = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix}$

Total revenue: Market I = ₹46,000, Market II = ₹53,000.

(b) Gross Profit:

Cost matrix: $C = \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix}$

$\text{Cost} = SC = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}\begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix}$

Market I: $10000(2) + 2000(1) + 18000(0.5) = 20000 + 2000 + 9000 = 31000$

Market II: $6000(2) + 20000(1) + 8000(0.5) = 12000 + 20000 + 4000 = 36000$

Gross Profit:

Market I: $46000 - 31000 =$ ₹15,000
Market II: $53000 - 36000 =$ ₹17,000

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8Find the matrix X so that

X\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}

Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (since $X$ must be $2\times2$ for the product to be $2\times3$ ).

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$

From row 1:
$a + 4b = -7 \quad \cdots(1)$
$2a + 5b = -8 \quad \cdots(2)$

From (2) $- 2\times$ (1): $-3b = 6 \Rightarrow b = -2$

From (1): $a = -7 + 8 = 1$

From row 2:
$c + 4d = 2 \quad \cdots(3)$
$2c + 5d = 4 \quad \cdots(4)$

From (4) $- 2\times$ (3): $-3d = 0 \Rightarrow d = 0$

From (3): $c = 2$

$\boxed{X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}}$

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9If

A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}

is such that

A^2 = I

, then (A)

1+\alpha^2+\beta\gamma=0

(B)

1-\alpha^2+\beta\gamma=0

(C)

1-\alpha^2-\beta\gamma=0

(D)

1+\alpha^2-\beta\gamma=0

Correct Answer: (C) $1 - \alpha^2 - \beta\gamma = 0$

$A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} = \begin{bmatrix} \alpha^2+\beta\gamma & \alpha\beta-\alpha\beta \\ \alpha\gamma-\alpha\gamma & \beta\gamma+\alpha^2 \end{bmatrix} = \begin{bmatrix} \alpha^2+\beta\gamma & 0 \\ 0 & \alpha^2+\beta\gamma \end{bmatrix}$

For $A^2 = I$ :
$\alpha^2 + \beta\gamma = 1 \Rightarrow 1 - \alpha^2 - \beta\gamma = 0$

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10If the matrix A is both symmetric and skew symmetric, then (A) A is a diagonal matrix (B) A is a zero matrix (C) A is a square matrix (D) None of these

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11If A is square matrix such that

A^2 = A

, then

(I+A)^3 - 7A

is equal to (A) A (B) I-A (C) I (D) 3A

Correct Answer: (C) I

Given: $A^2 = A$ (A is idempotent)

Note: $A^2 = A \Rightarrow A^3 = A^2 \cdot A = A \cdot A = A^2 = A$

Expanding $(I+A)^3$ :
$(I+A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$
$= I + 3A + 3A^2 + A^3$
$= I + 3A + 3A + A \quad [\text{since } A^2 = A^3 = A]$
$= I + 7A$

Therefore:
$(I+A)^3 - 7A = I + 7A - 7A = I$

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Exercise 3.1

Exercise 3.2

Exercise 3.3

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Miscellaneous Exercise on Chapter 3

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