Chapter 5 of 15

Updated 6 August 2026

NCERT Solutions

Matrices — NCERT Solutions

Jammu & Kashmir Board · Class 12 · Mathematics

NCERT Solutions for Matrices, Jammu & Kashmir Board Class 12 Mathematics: 56 textbook questions solved step by step.

119 questions68 flashcards12 formulas & key relations5 concepts

Interactive on Super Tutor

Studying Matrices? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for NCERT solutions and more.

Try Super Tutor — Free

Free trial, no card needed.

An example illustrating two matrices that are equal, emphasizing that they must have the same order and corresponding elements must be identical. — One of 5 illustrations for Matrices in Super Tutor — alongside flashcards, concept maps and practice questions.

56 Questions Solved · 5 Sections

The first 28 solutions are open to read. The other 28 are free with a Super Tutor account.

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.

Free with a Super Tutor account

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

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

Free with a Super Tutor account

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 $p = n$ .

Free with a Super Tutor account

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

Free with a Super Tutor account

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

Free with a Super Tutor account

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 ✓

Free with a Super Tutor account

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 ✓

Free with a Super Tutor account

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

Free with a Super Tutor account

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 ✓

Free with a Super Tutor account

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

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

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

Free with a Super Tutor account

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. ✓

Free with a Super Tutor account

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. ✓

Free with a Super Tutor account

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 $\frac{1}{2}(A-A') = A$ .)

Free with a Super Tutor account

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

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

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

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

Free with a Super Tutor account

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

Free with a Super Tutor account

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

Free with a Super Tutor account

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

Free with a Super Tutor account

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.

Free with a Super Tutor account

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.

Free with a Super Tutor account

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.

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

Free with a Super Tutor account

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:
$4 + 4x = 0 \Rightarrow x = -1$

$\boxed{x = -1}$

Free with a Super Tutor account

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.

Free with a Super Tutor account

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 $\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix}$ :

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

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

Free with a Super Tutor account

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

Free with a Super Tutor account

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): $c = 2$

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

Free with a Super Tutor account

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

Free with a Super Tutor account

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

Free with a Super Tutor account

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

Free with a Super Tutor account

28 more solved questions in Matrices

They are free with a Super Tutor account, along with practice quizzes and flashcards for this chapter. Free to start, no card needed.

Frequently Asked Questions

What are the important topics in Matrices for Jammu & Kashmir Board Class 12 Mathematics?

Key topics in Matrices include Basic Idea, Order, and Elements of a Matrix, Types of Matrices, Equality, Addition, Scalar Multiplication, and Difference, Multiplication of Matrices. Study these first, then practise questions on each for the Jammu & Kashmir Board Class 12 board exam.

Are these NCERT Solutions for Matrices free?

The first 28 of the 56 solutions on this page are open to read. The other 28 are free with a Super Tutor account — signing up is free and needs no card.

How should I revise Matrices for the Jammu & Kashmir Board Class 12 board exam?

Learn the core ideas first, then work through the 119 practice questions on Matrices. Revise definitions regularly and use flashcards for quick recall before the exam.

Sources & Official References

National Education Policy 2020 — education.gov.in

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

More resources for Matrices

All Jammu & Kashmir Board Class 12 Mathematics NCERT Solutions

For serious students

Get the full Matrices chapter — start free.

Quizzes, flashcards, an AI doubt solver and a study plan for Jammu & Kashmir Board Class 12 Mathematics. Free to start, no card needed.

Try Super Tutor Free

Try this chapter free

Matrices — NCERT Solutions

Studying Matrices? Get the full interactive chapter.

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Miscellaneous Exercise on Chapter 3

Frequently Asked Questions

Sources & Official References

More resources for Matrices

Practice Quiz

Important Questions

Revision Notes

Formula Sheet

Chapter Summary

Concept Maps

Study Plan

Flashcards

Syllabus

Get the full Matrices chapter — start free.