Chapter 3 Matrices

NCERT Solutions Class 12 Maths Chapter 3 Matrices

Class 12 Maths Chapter 3: Matrices introduces students to a new way of representing and solving mathematical problems using rectangular arrays of numbers. This chapter covers types of matrices, matrix operations, transpose of a matrix, and special matrices. Matrices are widely used in solving systems of linear equations and form the base for advanced topics like determinants and linear algebra.

ALLEN provides NCERT Solutions for class 12 Maths chapter 3, which are prepared strictly according to the latest textbook issued by NCERT and fully aligned with the current syllabus prescribed by CBSE. Class 12 Maths Ch 3 NCERT Solutions are prepared by subject experts in a step by step manner so that Class 12 students can understand the concepts related to matrices without confusion. Regular practice of these questions can help improve a student’s accuracy in answer writing, strengthens fundamentals, and also prepares students well for board exams and competitive exams. The explanations are kept simple, clear, and suitable for easy revision.

1.0Key Concepts of Class 12 Maths Chapter 3 Matrices

Class 12 Maths Chapter 3, Matrices, introduces matrices and their basic properties and operations. The main concepts covered in NCERT Solutions for Class 12 Maths Chapter 3 include:

Introduction to Matrices: Understand the meaning, order, elements, and representation of a matrix.
Types of Matrices: Learn about row, column, square, zero, diagonal, scalar, and identity matrices.
Equality of Matrices: Understand the conditions required for two matrices to be equal.
Operations on Matrices: Learn how to perform addition, subtraction, and multiplication of a matrix by a scalar.
Multiplication of Matrices: Understand the rules, conditions, and properties of matrix multiplication.

2.0NCERT Class 12 Maths Chapter 3 Matrices : Detailed Solutions

EXERCISE - 3.1

In the matrix $2353 5 - 2 1 19 \frac{5}{2} - 5 - 7 1217$ , write (i) the order of the matrix (ii) the number of elements (iii) write the elements $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$

Sol.

(i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4. (ii) Since, the order of the matrix is 3 × 4, so there are $3 \times 4 = 12$ elements in it. (iii) Let $2353 5 - 2 1 19 \frac{5}{2} - 5 - 7 1217 = a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33} a_{14} a_{24} a_{34}$

On comparing the corresponding elements, we get $a_{13} = 19; a_{21} = 35; a_{33} = - 5, a_{24} = 12; a_{23} = \frac{5}{2}$

If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements? Sol. We know that if a matrix is of the order $m \times n$ , it has mn elements. Hence, a matrix containing 24 elements can have any one of the following orders: 1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2 or 24 × 1. Similarly, a matrix containing 13 elements can have order 1 × 13 or 13 × 1.
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements? Sol. A matrix containing 18 elements can have any one of the following orders: 1 × 18, 18 × 1.2 × 9, 9 × 2, 3 × 6, 6 × 3 Similarly, a matrix containing 5 elements can have order 1 × 5 or 5 × 1.
Construct $2 \times 2$ matrix, $A = [a_{ij}]$ , whose elements are given by: (i) $a_{ij} = \frac{( i + j ) ^{2}}{2}$ (ii) $a_{ij} = \frac{i}{j}$ (iii) $a_{ij} = \frac{( i + 2 j ) ^{2}}{2}$

Sol.

(i) The order of the given matrix is $2 \times 2$ , so $A = [a_{11} a_{21} a_{12} a_{22}]_{2 \times 2}$ where $a_{ij} = \frac{( i + j ) ^{2}}{2}$ . $∴ a_{11} = \frac{( 1 + 1 ) ^{2}}{2} = 2, a_{12} = \frac{( 1 + 2 ) ^{2}}{2} = \frac{9}{2}$ , $a_{21} = \frac{( 2 + 1 ) ^{2}}{2} = \frac{9}{2}, a_{22} = \frac{( 2 + 2 ) ^{2}}{2} = 8$ Hence, the required matrix is $A = [2 9/2 9/2 8]_{2 \times 2}$ (ii) Here, $A = [a_{11} a_{21} a_{12} a_{22}]_{2 \times 2}$ , where $a_{ij} = \frac{i}{j}$ $∴ a_{11} = \frac{1}{1} = 1, a_{12} = \frac{1}{2}, a_{21} = \frac{2}{1} = 2, a_{22} = \frac{2}{2} = 1$ Hence, the required matrix is $A = [12 1/2 1]_{2 \times 2}$ (iii) Here, $A = [a_{11} a_{21} a_{12} a_{22}]_{2 \times 2}$ , where $a_{ij} = \frac{( i + 2 j ) ^{2}}{2}$ $∴ a_{11} = \frac{( 1 + 2 ) ^{2}}{2} = \frac{9}{2}, a_{12} = \frac{( 1 + 4 ) ^{2}}{2} = \frac{25}{2}$ , $a_{21} = \frac{( 2 + 2 ) ^{2}}{2} = 8, a_{22} = \frac{( 2 + 4 ) ^{2}}{2} = 18$ Hence, the required matrix is $A = [9/2 8 25/2 18]_{2 \times 2}$

Construct a $3 \times 4$ matrix whose elements are given by: (i) $a_{ij} = \frac{1}{2} ∣ - 3 i + j ∣$ (ii) $a_{ij} = 2 i - j$

Sol.

(i) The order of given matrix is $3 \times 4$ , so the required matrix is

$A = a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33} a_{14} a_{24} a_{34}_{3 \times 4},$

where $a_{ij} = \frac{1}{2} ∣ - 3 i + j ∣$

$∴ a_{11} = \frac{1}{2} ∣ - 3 + 1∣ = 1, a_{13} = \frac{1}{2} ∣ - 3 + 3∣ = 0, a_{21} = \frac{1}{2} ∣ - 6 + 1∣ = \frac{5}{2}, a_{23} = \frac{1}{2} ∣ - 6 + 3∣ = \frac{3}{2}, a_{31} = \frac{1}{2} ∣ - 9 + 1∣ = 4, a_{33} = \frac{1}{2} ∣ - 9 + 3∣ = 3, a_{12} = \frac{1}{2} ∣ - 3 + 2∣ = \frac{1}{2}, a_{14} = \frac{1}{2} ∣ - 3 + 4∣ = \frac{1}{2} a_{22} = \frac{1}{2} ∣ - 6 + 2∣ = 2, a_{24} = \frac{1}{2} ∣ - 6 + 4∣ = 1 a_{32} = \frac{1}{2} ∣ - 9 + 2∣ = \frac{7}{2}, a_{34} = \frac{1}{2} ∣ - 9 + 4∣ = \frac{5}{2}$

Hence, the required matrix is

$A = 1 5/2 4 1/2 2 7/2 0 3/2 3 1/2 1 5/2_{3 \times 4}$

Hence, the required matrix is

$A = 135024 - 1 13 - 2 02_{3 \times 4}$

Find the values of x, y and z from the following equations : (i) $[4 x 35] = [y 1 z 5]$ (ii) $[x + y 5 + z 2 xy] = [6528]$ (iii) $x + y + z x + z y + z = 957$

Sol.

(i) Given, $[4 x 35] = [y 1 z 5]$ . By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get $4 = y, 3 = z$ and $x = 1 \Rightarrow x = 1$ , $y = 4$ and $z = 3$ (ii) Here, $A = a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33} a_{14} a_{24} a_{34}_{3 \times 4}$ ,

where $a_{ij} = 2 i - j$

$∴ a_{11} = 2 - 1 = 1, a_{13} = 2 - 3 = - 1, a_{21} = 4 - 1 = 3, a_{23} = 4 - 3 = 1, a_{31} = 6 - 1 = 5, a_{33} = 6 - 3 = 3, a_{12} = 2 - 2 = 0, a_{14} = 2 - 4 = - 2, a_{22} = 4 - 2 = 2, a_{24} = 4 - 4 = 0, a_{32} = 6 - 2 = 4, a_{34} = 6 - 4 = 2$

(ii) Given, $[x + y 5 + z 2 xy] = [6528]$ . By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get

$x + y = 6 5 + z = 5$

and $xy = 8$ From Eq. (2), we get $z = 0$ From Eq. (1), $y = 6 - x$ Substituting this value of y in Eq. (3), we obtain

$\Rightarrow \Rightarrow x (6 - x) = 8 (x - 2) (x - 4) = 0 \Rightarrow \Rightarrow x = 2 or x = 4$

When $x = 2$ , then from Eq. (4), $y = 6 - 2 = 4$ and when $x = 4$ , then from Eq. (4), $y = 6 - 4 = 2$ . So, either $x = 2, y = 4$ and $z = 0$ or $x = 4$ , $y = 2$ and $z = 0$

(iii) Given, $x + y + z x + z y + z = 957$ . By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get

$x + y + z = 9 x + z = 5 y + z = 7$

Subtracting Eq. (2) from Eq. (1), we get $y = 4$ Subtracting Eq. (3) from Eq.(1), we get $x = 2$ Substituting $y = 4$ in Eq. (3), we get $4 + z = 7$ $\Rightarrow z = 7 - 4 = 3$ Hence; $x = 2, y = 4, z = 3$

Find the values of a, b, c and d from the equation $[a - b 2 a - b 2 a + c 3 c + d] = [- 1 0 513]$ . Sol. Given, $[a - b 2 a - b 2 a + c 3 c + d] = [- 1 0 513]$ By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:

$a - b = - 1 2 a - b = 0 2 a + c = 5 and 3 c + d = 13$

Subtracting Eq. (1) from Eq.(2), we get $a = 1$ Putting $a = 1$ in Eq. (1) and Eq.(3), we get $1 - b = - 1$ and $2 + c = 5 \Rightarrow b = 2$ and $c = 3$ Substituting $c = 3$ in Eq. (4), We obtain $3 \times 3 + d = 13$ $\Rightarrow d = 13 - 9 = 4$ Hence; $a = 1, b = 2, c = 3$ and $d = 4$ .

Choose the correct answer in the following questions from 8 to 10.

$A = [a_{ij}]_{m \times n}$ is a square matrix, if ? (A) m < n (B) $m > n$ (C) $m = n$ (D) None of these Sol. (C) Since, in a square matrix, the number of rows is equal to the number of columns; therefore, we must have $m = n$ .
Which of the given values of x and y make the following pairs of matrices equal?

$[3 x + 7 y + 1 5 2 - 3 x], [08 y - 2 4]$

(A) $x = \frac{- 1}{3}, y = 7$ (B) Not possible to find (C) $y = 7, x = \frac{- 2}{3}$ (D) $x = \frac{- 1}{3}, y = \frac{- 2}{3}$ Sol. (B) According to the question,

$[3 x + 7 y + 1 5 2 - 3 x] = [08 y - 2 4]$

By definition of equality of matrices, we have

$3 x + 7 = 0 5 = y - 2 y + 1 = 8 2 - 3 x = 4$

From Eq. (2), $y = 7$ From Eq. (1),

$3 x + 7 = 0 \Rightarrow x = \frac{- 7}{3}$

From Eq. (4),

$2 - 3 x = 4 \Rightarrow x = \frac{- 2}{3}$

'x' can have only one value at a time. Hence, it is not possible to find the values of x and y for which the given matrices are equal.

The number of all possible matrices of order $3 \times 3$ with each entry or 0 or 1 is (A) 27 (B) 18 (C) 81 (D) 512

Sol. (D) We know that a matrix having order of $3 \times 3$ contains 9 elements. Each element can be selected in 2 ways (it can be either 0 or 1). Hence, All the nine entries can be chosen by $2^{9}$ ways $= 512$ ways $∴$ Required number of matrices is 512. So, the correct option is (D). Aliter : Required number of possible matrices $= ($ Number of entries $)^{Number of elements}$ $= (2)^{3 \times 3}$ $= (2)^{9}$ $= 512$

EXERCISE - 3.2

Let $A = [2342], B = [1 - 2 35], C = [- 2 3 54]$ Find each of the following: (i) $A + B$ (ii) A - B (iii) 3A - C (iv) AB (v) BA

Sol.

(i) $A + B = [2342] + [1 - 2 35]$

$= [2 + 1 3 + (- 2) 4 + 3 2 + 5] = [3177]$

(ii)

$A - B = [2342] - [1 - 2 35] = [2 - 1 3 - (- 2) 4 - 3 2 - 5] = [15 1 - 3]$

(iii)

$3 A - C = 3 [2342] - [- 2 3 54] = [69126] - [- 2 3 54] = [8672]$

(iv) $AB = [2342] [1 - 2 35]$

$= [2 \times 1 + 4 \times (- 2) 3 \times 1 + 2 \times (- 2) 2 \times 3 + 4 \times 5 3 \times 3 + 2 \times 5] = [- 6 - 1 2619]$

(v)

$BA = [1 - 2 35] [2342] = [1 \times 2 + 3 \times 3 (- 2) \times 2 + 5 \times 3 1 \times 4 + 3 \times 2 (- 2) \times 4 + 5 \times 2] = [1111102]$

Compute the following:

(i) $[a - b b a] + [a b b a]$ (ii) $[a^{2} + b^{2} a^{2} + c^{2} b^{2} + c^{2} a^{2} + b^{2}] + [2 ab - 2 ac 2 bc - 2 ab]$ (iii) $- 1 82 458 - 6 165 + 1283702654$ (iv) $[cos^{2} x sin^{2} x sin^{2} x cos^{2} x] + [sin^{2} x cos^{2} x cos^{2} x sin^{2} x]$

Sol.

(i) $[a - b b a] + [a b b a] = [a + a - b + b b + b a + a] = [2 a 0 2 b 2 a]$

(ii)

$[a^{2} + b^{2} a^{2} + c^{2} b^{2} + c^{2} a^{2} + b^{2}] + [2 ab - 2 a c 2 b c - 2 ab] = [a^{2} + b^{2} + 2 ab a^{2} + c^{2} - 2 a c b^{2} + c^{2} + 2 b c a^{2} + b^{2} - 2 ab] = [(a + b)^{2} (a - c)^{2} (b + c)^{2} (a - b)^{2}] {∵ and (a + b)^{2} = a^{2} + 2 ab + b^{2} (a - b)^{2} = a^{2} - 2 ab + b^{2}}$

(iii)

$- 1 82 458 - 6 165 + 1283702654 = - 1 + 12 8 + 8 2 + 3 4 + 7 5 + 0 8 + 2 - 6 + 6 16 + 5 5 + 4 = 11165115100219$

(iv)

$[cos^{2} x sin^{2} x sin^{2} x cos^{2} x] + [sin^{2} x cos^{2} x cos^{2} x sin^{2} x] = [cos^{2} x + sin^{2} x sin^{2} x + cos^{2} x sin^{2} x + cos^{2} x cos^{2} x + sin^{2} x] = [1111]$

Compute the indicated products. (i) $[a - b b a] [a b - b a]$ (ii) $123 [234]$ (iii) $[12 - 2 3] [122331]$ (iv) $234345456 103 - 3 20 545$ (v) $23 - 1 121 [1 - 1 0211]$ (vi) $[3 - 1 - 1 0 32] 213 - 3 01$ Sol.

(i)

$[a - b b a] [a b - b a] = [a \times a + b \times b (- b) \times a + a \times b a \times (- b) + b \times a (- b) \times (- b) + a \times a] = [a^{2} + b^{2} 0 0 b^{2} + a^{2}]$

(ii)

$123 [234] = 1 \times 2 2 \times 2 3 \times 2 1 \times 3 2 \times 3 3 \times 3 1 \times 4 2 \times 4 3 \times 4 = 2463694812$

(iii)

$= = [12 - 2 3] [122331] [1 \times 1 + (- 2) \times 2 2 \times 1 + 3 \times 2 1 \times 2 + (- 2) \times 3 2 \times 2 + 3 \times 3 1 \times 3 + (- 2) \times 1 2 \times 3 + 3 \times 1] [- 3 8 - 4 13 19]$

(iv)

$234345456 103 - 3 20 545 = 2 + 0 + 12 3 + 0 + 15 4 + 0 + 18 - 6 + 6 + 0 - 9 + 8 + 0 - 12 + 10 + 0 10 + 12 + 20 15 + 16 + 25 20 + 20 + 30 = 141822 0 - 1 - 2 425670$

(v)

$23 - 1 121 [1 - 1 0211] = 2 - 1 3 - 2 - 1 - 1 0 + 2 0 + 4 0 + 2 2 + 1 3 + 2 - 1 + 1 = 11 - 2 242350$

(vi)

$[3 - 1 - 1 0 32] 213 - 3 01 = [6 - 1 + 9 - 2 + 0 + 6 - 9 + 0 + 3 3 + 0 + 2] = [144 - 6 5]$

$If A = 151 20 - 1 - 3 21, B = 342 - 1 20 253,$

Also, verify that $A + (B - C) = (A + B) - C$ .

Sol.

$A + B = 151 20 - 1 - 3 21 + 342 - 1 20 253 = 493 12 - 1 - 1 74$

and

$B - C = 342 - 1 20 253 - 401 13 - 2 223 = - 1 41 - 2 - 1 2 030$

and

$A + (B - C) = 151 20 - 1 - 3 21 + - 1 41 - 2 - 1 2 030 = 092 0 - 1 1 - 3 51$

∴

$(A + B) - C = 493 12 - 1 - 1 74 - 401 13 - 2 223 = 092 0 - 1 1 - 3 51$

Hence, $A + (B - C) = (A + B) - C$ is verified.

If $A = 2/3 1/3 7/3 1 2/3 2 5/3 4/3 2/3$ and $B = 2/5 1/5 7/5 3/5 2/5 6/5 1 4/5 2/5$ , then compute 3A - 5B.

Sol. 3A - 5B

$= 3 2/3 1/3 7/3 1 2/3 2 5/3 4/3 2/3 - 5 2/5 1/5 7/5 3/5 2/5 6/5 1 4/5 2/5 = 217326542 - 217326542 = 000000000 = O (Zero matrix)$

Simplify

$cos θ [cos θ - sin θ sin θ cos θ] + sin θ [sin θ cos θ - cos θ sin θ]$

Sol. $cos θ [cos θ - sin θ sin θ cos θ] + sin θ [sin θ cos θ - cos θ sin θ]$

$= [cos^{2} θ - cos θ sin θ cos θ sin θ cos^{2} θ] + [sin^{2} θ sin θ cos θ - sin θ cos θ sin^{2} θ] = [cos^{2} θ + sin^{2} θ 0 0 cos^{2} θ + sin^{2} θ] = [1001] (∵ cos^{2} θ + sin^{2} θ = 1)$

Find X and Y if: (i) $X + Y = [7205]$ and $X - Y = [3003]$ (ii) $2 X + 3 Y = [2430]$ and $3 X + 2 Y = [2 - 1 - 2 5]$

Sol. (i) Given, $X + Y = [7205]$ and $X - Y = [3003]$ Adding equation (1) and (2), we get

$2 X = [7205] + [3003] = [10208]$

$\Rightarrow X = \frac{1}{2} [10208] = [5104]$

Subtracting equation (2) from equation (1), we get

$2 Y = [7205] - [3003] = [4202]$

⇒

$Y = \frac{1}{2} [4202] = [2101]$

(ii) Given, $2 X + 3 Y = [2430]$ and $3 X + 2 Y = [2 - 1 - 2 5]$ Multiplying equation (1) by 2, Eq. (2) by 3 and then subtracting, we get

$2 (2 X + 3 Y) - 3 (3 X + 2 Y) = 2 [2430] - 3 [2 - 1 - 2 5]$

$\Rightarrow 4 X + 6 Y - 9 X - 6 Y = [4860] - [6 - 3 - 6 15]$

$\Rightarrow - 5 X = [4 - 6 8 + 3 6 + 6 0 - 15] = [- 2 11 12 - 15] \Rightarrow X = - \frac{1}{5} [- 2 11 12 - 15] = [2/5 - 11/5 - 12/5 3]$

Then, from Eq. (1),

$3 Y = [2430] - 2 X = [2430] - 2 [2/5 - 11/5 - 12/5 3] = [2 - \frac{4}{5} 4 + \frac{22}{5} 3 + \frac{24}{5} 0 - 6] = [6/5 42/5 39/5 - 6]$

$\Rightarrow Y = \frac{1}{3} [6/5 42/5 39/5 - 6] = [2/5 14/5 13/5 - 2]$

Find X , if $Y = [3124]$ and $2 X + Y = [1 - 3 02]$ .

Sol. Given, $2 X + Y = [1 - 3 02]$ and $Y = [3124]$

$\Rightarrow 2 X = [1 - 3 02] - Y = [1 - 3 02] - [3124] = [1 - 3 - 3 - 1 0 - 2 2 - 4] = [- 2 - 4 - 2 - 2]$

$\Rightarrow X = \frac{1}{2} [- 2 - 4 - 2 - 2] = [- 1 - 2 - 1 - 1]$

Find x and y, if $2 [10 3 x] + [y 1 02] = [5168]$ .

Sol. Given, $2 [10 3 x] + [y 1 02] = [5168]$

$\Rightarrow [2 + y 0 + 1 6 + 0 2 x + 2] = [5168]$

By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get

$2 + y = 5$

and $2 x + 2 = 8$ $\Rightarrow y = 5 - 2 = 3$ and $2 x = 8 - 2$ $\Rightarrow y = 3$ and $x = \frac{6}{2} = 3$

Solve the equation for x, y, z and t, if

$2 [x y z t] + 3 [10 - 1 2] = 3 [3456] .$

Sol. Given, $2 [x y z t] + 3 [10 - 1 2] = 3 [3456]$

$\Rightarrow [2 x 2 y 2 z 2 t] + [30 - 3 6] = [9121518] \Rightarrow [2 x + 3 2 y + 0 2 z - 3 2 t + 6] = [9121518]$

By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get

$2 x + 3 = 9, 2 y + 0 = 12, 2 z - 3 = 15 and 2 t + 6 = 18$

$\Rightarrow x = \frac{9 - 3}{2}, y = \frac{12}{2}, z = \frac{15 + 3}{2}$ and $t = \frac{18 - 6}{2}$ $\Rightarrow x = 3, y = 6, z = 9$ and $t = 6$ .

If $x [23] + y [- 1 1] = [105]$ , then find the values of x and y. Sol. Given that $x [23] + y [- 1 1] = [105]$

$\Rightarrow [2 x - y 3 x + y] = [105]$

By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get

$2 x - y = 10$

and $3 x + y = 5$ Adding Eq. (1) and (2), we get :

$5 x = 15$

$\Rightarrow x = 3$ Substituting $x = 3$ in Eq. (1), we get :

$2 \times 3 - y = 10$

$\Rightarrow y = 6 - 10 = - 4$

If $3 [x z y w] = [x - 1 6 2 w] + [4 z + w x + y 3]$ , find the values of x, y, z and w. Sol. Given, $3 [x z y w] = [x - 1 6 2 w] + [4 z + w x + y 3]$

$\Rightarrow [3 x 3 z 3 y 3 w] = [x + 4 - 1 + z + w 6 + x + y 2 w + 3]$

By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get

$\Rightarrow \Rightarrow \Rightarrow 3 x = x + 4 2 x = 4 x = 2 and 3 y = 6 + x + y 2 y = 6 + x$

$\Rightarrow y = \frac{6 + x}{2} = \frac{6 + 2}{2} = \frac{8}{2} = 4$ Now, $3 z = - 1 + z + w, 2 z = - 1 + w$

$\Rightarrow z = \frac{- 1 + w}{2}$

Now, $3 w = 2 w + 3 \Rightarrow w = 3$ Putting the value of w in Eq. (1), we get :

$z = \frac{- 1 + 3}{2} = 1$

Hence, the values of x, y, z and w are 2, 4, 1 and 3, respectively.

If $F (x) = cos x sin x 0 - sin x cos x 0 001$ , then show that

$F (x) . F (y) = F (x + y) .$

Sol. $F (x) F (y) = cos x sin x 0 - sin x cos x 0 001 cos y sin y 0 - sin y cos y 0 001$

$= cos x cos y - sin x sin y sin x cos y + cos x sin y 0 - sin y cos x - sin x cos y - sin x sin y + cos x cos y 0 001$

$= cos (x + y) sin (x + y) 0 - sin (x + y) cos (x + y) 0 001 {∵ cos (A + B) = cos A cos B - sin A sin B sin (A + B) = sin A cos B + sin B cos A}$

$\Rightarrow F (x) . F (y) = F (x + y)$

Show that

(i) $[56 - 1 7] [2314] \neq = [2314] [56 - 1 7]$ (ii) $101211300 - 1 02 1 - 1 3 014 \neq = - 1 02 1 - 1 3 014 101211300$

Sol.

(i) $[56 - 1 7] [2314] = [10 - 3 12 + 21 5 - 4 6 + 28]$

$= [733134] = [2314] [56 - 1 7] = [10 + 6 15 + 24 - 2 + 7 - 3 + 28] = [1639525]$

Hence, $[56 - 1 7] [2314] \neq = [2314] [56 - 1 7]$ (ii) Here, $101211300 - 1 02 1 - 1 3 014$

$= - 1 + 0 + 6 0 + 0 + 0 - 1 + 0 + 0 1 - 2 + 9 0 + (- 1) + 0 1 - 1 + 0 0 + 2 + 12 0 + 1 + 0 0 + 1 + 0 = 50 - 1 8 - 1 0 1411$

and $- 1 02 1 - 1 3 014 101211300$

$= - 1 + 0 + 0 0 + 0 + 1 2 + 0 + 4 - 2 + 1 + 0 0 - 1 + 1 4 + 3 + 4 - 3 + 0 + 0 0 + 0 + 0 6 + 0 + 0 = - 1 16 - 1 011 - 3 06$

$101211300 - 1 02 1 - 1 3 014 \neq = - 1 02 1 - 1 3 014 101211300$

Hence, the required result is verified.

Find $A^{2} - 5 A + 6 I$ if $A = 221 01 - 1 130$ Sol. Here, $A^{2} = A . A = 221 01 - 1 130 221 01 - 1 130$

$= 4 + 0 + 1 4 + 2 + 3 2 - 2 + 0 0 + 0 - 1 0 + 1 - 3 0 - 1 + 0 2 + 0 + 0 2 + 3 + 0 1 - 3 + 0 = 590 - 1 - 2 - 1 25 - 2$

$∴ A^{2} - 5 A + 6 I = 590 - 1 - 2 - 1 25 - 2 - 5 221 01 - 1 130 + 6100 = 590 - 1 - 2 - 1 25 - 2 - 10105 05 - 5 5150 + 600 = 5 - 10 + 6 9 - 10 + 0 0 - 5 + 0 - 1 - 0 + 0 - 2 - 5 + 6 - 1 + 5 + 0 2 - 5 + 0 5 - 15 + 0 - 2 - 0 + 6 = 1 - 1 - 5 - 1 - 1 4 - 3 - 10 4$

If $A = 102020213$ , prove that $A^{3} - 6 A^{2} + 7 A + 2 I = O$ Sol. $A^{2} = A \times A = 102020213102020213$

$= 1 + 0 + 4 0 + 0 + 2 2 + 0 + 6 0 + 0 + 0 0 + 4 + 0 0 + 0 + 0 2 + 0 + 6 0 + 2 + 3 4 + 0 + 9 = 5280408513 A^{3} = A^{2} \cdot A = 5280408513102020213$

$= 5 + 0 + 16 2 + 0 + 10 8 + 0 + 26 0 + 0 + 0 0 + 8 + 0 0 + 0 + 0 10 + 0 + 24 4 + 4 + 15 16 + 0 + 39$

$= 211234080342355$

$∴ A^{3} - 6 A^{2} + 7 A + 2 I = 21123408034255 - 65280408513 + 7102020213 + 2100010001 = 211234080342355 - 3012480240483078 + 7014014014721 + 200020002 = 21 - 30 + 7 + 2 12 - 12 + 0 + 0 34 - 48 + 14 + 0 0 - 0 + 0 + 0 8 - 24 + 14 + 2 0 - 0 + 0 + 0 34 - 48 + 14 + 0 23 - 30 + 7 + 0 55 - 78 + 21 + 2 = 000000000 = O$

If $A = [34 - 2 - 2]$ and $I = [1001]$ , then find k so that $A^{2} = kA - 2 I$ . Sol. Given; $A^{2} = kA - 2 I \Rightarrow A \cdot A = kA - 2 I$

$\Rightarrow [34 - 2 - 2] [34 - 2 - 2] = k [34 - 2 - 2] - 2 [1001] \Rightarrow [9 - 8 12 - 8 - 6 + 4 - 8 + 4] = [3 k 4 k - 2 k - 2 k] - [2002] \Rightarrow [14 - 2 - 4] = [3 k - 2 4 k - 2 k - 2 k - 2]$

By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get.

$3 k - 2 = 1 \Rightarrow k = 1 - 2 k = - 2 \Rightarrow k = 1 4 k = 4 \Rightarrow k = 1 - 4 = - 2 k - 2 \Rightarrow k = 1$

Hence, $k = 1$

If $A = [0 tan \frac{α}{2} - tan \frac{α}{2} 0]$ and I is the identity matrix of order 2, show that $I + A = (I - A) [cos α sin α - sin α cos α]$ Sol. RHS $= (I - A) [cos α sin α - sin α cos α]$

$= ([1001] - [0 tan \frac{α}{2} - tan \frac{α}{2} 0]) [cos α sin α - sin α cos α] = [0 - tan \frac{α}{2} tan \frac{α}{2} 0] [cos α sin α - sin α cos α]$

$= [cos α + sin α tan \frac{α}{2} - cos α tan \frac{α}{2} + sin α - sin α + cos α tan \frac{α}{2} sin α tan \frac{α}{2} + cos α]$

$= [1 - 2 sin^{2} \frac{α}{2} + 2 sin \frac{α}{2} cos \frac{α}{2} tan \frac{α}{2} - (2 cos^{2} \frac{α}{2} - 1) tan \frac{α}{2} + 2 sin \frac{α}{2} cos \frac{α}{2} - 2 sin \frac{α}{2} cos \frac{α}{2} + (2 cos^{2} \frac{α}{2} - 1) tan \frac{α}{2} 2 sin \frac{α}{2} cos \frac{α}{2} tan \frac{α}{2} + 1 - 2 sin^{2} \frac{α}{2}] = [1 - 2 sin^{2} \frac{α}{2} + 2 sin^{2} \frac{α}{2} - 2 sin \frac{α}{2} cos \frac{α}{2} + tan \frac{α}{2} + 2 sin \frac{α}{2} cos \frac{α}{2} - 2 sin \frac{α}{2} cos \frac{α}{2} + 2 sin \frac{α}{2} cos \frac{α}{2} - tan \frac{α}{2} 2 sin \frac{α}{2} + 1 - 2 sin^{2} \frac{α}{2}]$

$= [1 tan \frac{α}{2} - tan \frac{α}{2} 1]$

$I + A = [1001] + [0 tan \frac{α}{2} - tan \frac{α}{2} 0] = [1 tan \frac{α}{2} - tan \frac{α}{2} 1] = [1 tan \frac{α}{2} - tan \frac{α}{2} 1]$

= LHS Hence Proved.

A 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. Sol. Let the amount invested in first type of bonds is ₹x, then that invested in second type of bonds will be ₹(30000-x). (a) According to given condition,

$[x 30000 - x] [\frac{5}{100} \frac{7}{100}] = [1800]$

$\Rightarrow [\frac{5 x}{100} + \frac{( 30000 - x ) 7}{100}] = [1800]$ $\Rightarrow [\frac{5 x + 210000 - 7 x}{100}] = [1800]$ $\Rightarrow 210000 - 2 x = 180000$ $\Rightarrow 30000 = 2 x \Rightarrow x = 15000$ Hence, the amounts invested in the two types of bonds are respectively ₹15000 and $₹ (30000 - 15000) = ₹15000$ (b) According to the given condition,

$[x 30000 - x] [\frac{5}{100} \frac{7}{100}] = [2000]$

$\Rightarrow [\frac{5 x}{100} + \frac{( 30000 - x ) 7}{100}] = [2000]$ $\Rightarrow [\frac{5 x + 210000 - 7 x}{100}] = [2000]$ $\Rightarrow - 2 x + 210000 = 200000$ $\Rightarrow - 2 x = - 10000$ $\Rightarrow x = 5000$ Hence, the amounts invested in two types of bonds are respectively ₹5000 and $₹ (30000 - 5000) = ₹25000$ .

The 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 for one book of each subject respectively. Find the total amount that the bookshop will receive from selling the books, using matrix algebra. Sol. Let $A = [10 \times 12 8 \times 12 10 \times 12]$ and $B = 806040$ Amount received by the bookseller on selling these types of books can be computed by evaluating the product AB. Now, $AB = [12096120] 806040$

$= [120 \times 80 + 96 \times 60 + 120 \times 40]_{1 \times 1} = [9600 + 5760 + 4800]_{1 \times 1} = [20160]_{1 \times 1}$

∴ Amount received by the bookseller = ₹20160

Assume $X, Y, Z, W$ and $P$ are matrices of orders $2 \times n, 3 \times k, 2 \times p, n \times 3$ and $p \times k$ respectively. Choose the correct answer in Q. 21 and Q. 22.

The restrictions 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) $k = 2, p = 3$ Sol. (A) Matrices P and Y are of the orders $p \times k$ and $3 \times k$ respectively. Therefore, matrix PY will be defined if $k = 3$ . Consequently, PY will be of the order $p \times k$ . Matrices W and Y are of the orders n × 3 and $3 \times k$ respectively. Since, the number of columns in W is equal to the number of rows in Y, matrix WY is well-defined and is of the order $n \times k$ . Matrices PY and WY can be added only when their orders are the same. However, PY is of the order $p \times k$ , therefore we must have $p = n$ . Thus, $k = 3$ and $p = n$ are the restrictions on n, k and p so that $PY + WY$ will be defined.
If $n = p$ , then the order of the matrix 7X - 5Z is (A) $p \times 2$ (B) $2 \times n$ (C) $n \times 3$ (D) $p \times n$ Sol. (B) Matrix X is of the order 2 × n. Therefore, matrix 7X is also of the same order. Matrix Z is of the order 2 × p i.e. 2 × n [Since, $n = p$ ] Therefore, matrix 5Z is also of the same order. Now, both the matrices 7X and 5Z are of the order 2 × n. Thus, matrix 7X - 5Z is well-defined and is of the order 2 × n.

EXERCISE - 3.3

Find the transpose of each of the following matrices? (i) $5 1/2 - 1$ (ii) $[12 - 1 3]$ (iii) $- 1 32 553 66 - 1$

Sol.

(i) Let $A = 5 1/2 - 1_{3 \times 1}$ ,then $A^{'} = [5 1/2 - 1]_{1 \times 3}$ (ii) Let $A = [12 - 1 3]_{2 \times 2}$ , then $A^{'} = [1 - 1 23]_{2 \times 2}$ (iii) Let $A = - 1 32 553 66 - 1_{3 \times 3}$ , then

$A^{'} = - 1 56 356 23 - 1_{3 \times 3}$

If $A = - 1 5 - 2 271391$ and $B = - 4 11 123 - 5 01$ , then verify that

$(i) (A + B)^{'} = A^{'} + B^{'} (ii) (A - B)^{'} = A^{'} - B^{'}$

Sol.

(i) Here, $A + B = - 1 5 - 2 271391 + - 4 11 123 - 5 01$

$= - 5 6 - 1 394 - 2 92$

Also $A^{'} + B^{'} = - 1 5 - 2 271391^{'} + - 4 11 123 - 5 01^{'}$

$= - 1 23 579 - 2 11 + - 4 1 - 5 120131 = - 5 3 - 2 699 - 1 42$

From Eq.(1) and (2), it is verified that $(A + B)^{'} = A^{'} + B^{'}$ . (ii) Here, $A - B = - 1 5 - 2 271391 - - 4 11 123 - 5 01$

$= - 1 + 4 5 - 1 - 2 - 1 2 - 1 7 - 2 1 - 3 3 + 5 9 - 0 1 - 1$

$\Rightarrow (A - B)^{'} Also, A^{'} - B^{'} = 318459 - 3 - 2 0 = - 1 23 579 - 2 11 - - 4 1 - 5 120131 = - 1 + 4 2 - 1 3 - (- 5) 5 - 1 7 - 2 9 - 0 - 2 - 1 1 - 3 1 - 1 = 318459 - 3 - 2 0$

From Eqs. (1) and (2), it is verified that $(A - B)^{'} = A^{'} - B^{'}$ . 3. If $A^{'} = 3 - 1 0 421$ and $B = [- 1 1 2213]$ , then verify that

$(i) (A + B)^{'} = A^{'} + B^{'}$

Sol. $A = (A^{'})^{'} = 3 - 1 0 421^{'} = [34 - 1 2 01]$

(i) Here, $A + B = [34 - 1 2 01] + [- 1 1 2213]$

$= [251414]$

$∴ (A + B)^{'} = [251414]^{'} = 211544$

$B^{'} = [- 1 1 2213]^{'} = - 1 21 123, A^{'} = 3 - 1 0 421$

From equation (1) and (2), it is verfied that $(A + B)^{'} = A^{'} + B^{'}$ .

(ii)

$RHS LHS = A^{'} - B^{'} = 3 - 1 0 421 - - 1 21 123 = 3 + 1 - 1 - 2 0 - 1 4 - 1 2 - 2 1 - 3 = 4 - 3 - 1 30 - 2 \dots\dots (1) = (A - B)^{'} = ([34 - 1 2 01] - [- 1 1 2213])^{'} = [43 - 3 0 - 1 - 2]^{'} = 4 - 3 - 1 30 - 2 \dots .. (2)$

From equation (1) and (2), it is verified that $(A - B)^{'} = A^{'} - B^{'}$

If $A^{'} = [- 2 1 32]$ and $B = [- 1 1 02]$ , then find $(A + 2 B)^{'}$ . Sol. Given, $A^{'} = [- 2 1 32]$

$\Rightarrow ∴ \Rightarrow A = (A^{'})^{'} = [- 2 1 32]^{'} = [- 2 3 12] and 2 B = 2 [- 1 1 02] = [- 2 2 04] A + 2 B = [- 2 3 12] + [- 2 2 04] = [- 4 5 16] (A + 2 B)^{'} = [- 4 5 16]^{'} = [- 4 1 56]$

For the matrices A and B, verify that $(AB)^{'} = B^{'} A^{'}$ , where (i) $A = 1 - 4 3, B = [- 1 21]$ (ii) $A = 012, B = [157]$

Sol.

(i) Here, $AB = 1 - 4 3 [- 1 21] = - 1 4 - 3 2 - 8 6 1 - 4 3$

$\Rightarrow (AB)^{'} = - 1 4 - 3 2 - 8 6 1 - 4 3^{'} - 1 21 4 - 8 - 4 - 3 63$

Also, $B^{'} A^{'} = [- 1 21]^{'} 1 - 4 3^{'}$

$= = - 1 21 [1 - 4 3] - 1 21 4 - 8 - 4 - 3 63$

From Eqs. (1) and (2), we find that $(AB)^{'} = B^{'} A^{'}$ (ii) Here, $(AB) = 012 [157] = 01205100714$

$\Rightarrow (AB)^{'} = 01205100714^{'} = 00015721014$

Also, $B^{'} A^{'} = [157]^{'} 012^{'}$

$= 157 [012] = 00015721014$

From Eq. (1) and (2), it is verified that $(AB)^{'} = B^{'} A^{'}$ .

(i) If $A = [cos α - sin α sin α cos α]$ , then verify that $A^{'} A = I$ . (ii) If $A = [sin α - cos α cos α sin α]$ , then verify that $A^{'} A = I$ .

Sol.

(i) Here, $A = [cos α - sin α sin α cos α]$

$A^{'} = [cos α - sin α sin α cos α]^{'} = [cos α sin α - sin α cos α] ∴ A^{'} A = [cos α sin α - sin α cos α] [cos α - sin α sin α cos α] = [(cos α) (cos α) + (sin α) (sin α) (sin α) (cos α) + (cos α) (- sin α) (cos α) (sin α) + (- sin α) (cos α) (sin α) (sin α) + (cos α) (cos α)] = [cos^{2} α + sin^{2} α 0 0 sin^{2} α + cos^{2} α], = [1001] = I [∵ sin^{2} α + cos^{2} α = 1]$

(ii) Here, $A = [sin α - cos α cos α sin α]$

$\Rightarrow A^{'} = [sin α - cos α cos α sin α]^{'} = [sin α cos α - cos α sin α] ∴ A^{'} A = [sin α cos α - cos α sin α] [sin α - cos α cos α sin α] = [(sin α) (sin α) + (- cos α) (- cos α) (sin α) (cos α) + (cos α) (- sin α) (sin α) (cos α) + (- cos α) (sin α) (cos α) (cos α) + (sin α) (sin α)] = [sin^{2} α + cos^{2} α sin α cos α - sin α cos α sin α cos α - cos α sin α cos^{2} α + sin^{2} α] = [1001] = I [∵ sin^{2} α + cos^{2} α = 1]$

Hence, we verified that $A^{'} A = I$

(i) Show that the matrix, $A = 1 - 1 5 - 1 21 513$ is a Symmetric matrix. (ii) Show that the matrix, $A = 0 - 1 1 10 - 1 - 1 10$

is a Skew-symmetric matrix. Sol.

(i) Here, $A = 1 - 1 5 - 1 21 513$

$\Rightarrow A^{'} = 1 - 1 5 - 1 21 513^{'} = 1 - 1 5 - 1 21 513 = A$

$∵ A^{'} = A$ . Hence, $A$ is a symmetric matrix. (ii) Here, $A = 0 - 1 1 10 - 1 - 1 10$

$\Rightarrow A^{'} = 0 - 1 1 10 - 1 - 1 10^{'} = 01 - 1 - 1 01 1 - 1 0 = - 0 - 1 1 10 - 1 - 1 10 = - A ∵ A^{'} = - A .$

Hence, A is skew-symmetric matrix.

For the matrix $A = [1657]$ , verify that: (i) $(A + A^{'})$ is a symmetric matrix. (ii) $(A - A^{'})$ is a skew-symmetric matrix.

Sol.

(i) Given, $A = [1657]$

$A + A^{'} = [1657] + [1657]^{'} = [1657] + [1567] = [2111114]$

$\Rightarrow A + A^{'} = [2111114]$ and $(A + A^{'})^{'} = [2111114]^{'} = [2111114] = A + A^{'}$ $∵ (A + A^{'})^{'} = A + A^{'}$ . So, $(A + A^{'})$ is a symmetric matrix. (ii) $A - A^{'} = [1657] - [1657]^{'}$

$= [1657] - [1567] = [01 - 1 0]$

$\Rightarrow A - A^{'} = [01 - 1 0]$ and $(A - A^{'})^{'}$

$= [01 - 1 0]^{'} = [0 - 1 10] = - [01 - 1 0] = - (A - A^{'})$

$∵ (A - A)^{'} = - (A - A)$

Hence, ( $A - A^{'}$ ) is a skew-symmetric matrix.

Find $\frac{1}{2} (A + A^{'})$ and $\frac{1}{2} (A - A^{'})$ , when

$A = 0 - a - b a 0 - c b c 0 .$

Sol. Given $A = 0 - a - b a 0 - c b c 0$ Now,

$\frac{1}{2} (A + A^{'}) = \frac{1}{2} 0 - a - b a 0 - c b c 0 + 0 - a - b a 0 - c b c 0 = \frac{1}{2} 0 - a - b a 0 - c b c 0 + 0 a b - a 0 c - b - c 0 = 000000000$

and $\frac{1}{2} (A - A^{'})$

$= \frac{1}{2} 0 - a - b a 0 - c b c 0 - 0 - a - b a 0 - c b c 0^{'} = \frac{1}{2} 0 - a - b a 0 - c b c 0 - 0 a b - a 0 c - b - c 0 = \frac{1}{2} 0 - 2 a - 2 b 2 a 0 - 2 c 2 b 2 c 0 = 0 - a - b a 0 - c b c 0$

Express the following matrices as the sum of a symmetric and a skew-symmetric matrix. (i) $[31 5 - 1]$ (ii) $6 - 2 2 - 2 3 - 1 2 - 1 3$ (iii) $3 - 2 - 4 3 - 2 - 5 - 1 12$ (iv) $[1 - 1 52]$

Sol.

(i) A square matrix A can be expressed as sum of a symmetric and skew symmetric matrix.

$A = \frac{1}{2} [A + A^{'}] + \frac{1}{2} [A - A^{'}]$

(P=symmetric matrix)

$(Q = skew symmetric$

Let $A = [31 5 - 1], A^{'} = [35 1 - 1]$ Now, $\frac{1}{2} (A + A^{'}) = \frac{1}{2} ([31 5 - 1] + [35 1 - 1])$

$= \frac{1}{2} [66 6 - 2] = [33 3 - 1] = P (let)$

∵

$P^{'} = [33 3 - 1]^{'} = [33 3 - 1] = P$

Thus, $P = \frac{1}{2} (A + A^{'})$ is a symmetric matrix. Again, $\frac{1}{2} (A - A^{'}) = \frac{1}{2} ([31 5 - 1] - [35 1 - 1])$

$= \frac{1}{2} [0 - 4 40] = [0 - 2 20] = Q (let)$

∵

$Q^{'} = [0 - 2 20]^{'} = [02 - 2 0] = - [0 - 2 20] = - Q$

Thus, $Q = \frac{1}{2} (A - A^{'})$ is a skew-symmetric matrix.

Now,

$P + Q = [33 3 - 1] + [0 - 2 20] = [31 5 - 1] = A$

Hence, matrix A is sum of a symmetric matrix and a Skew-Symmetric matrix.

(ii) A square matrix A can be expressed as sum of a symmetric and skew symmetric matrices.

$A = (P = symmetric matrix) \frac{1}{2} [A + A^{'}] + \frac{1}{( Q = skew symmetric [ A - A ^{'} ]}$

Let $A = 6 - 2 2 - 2 3 - 1 2 - 1 3$ , then

$A^{'} = 6 - 2 2 - 2 3 - 1 2 - 1 3^{'} = 6 - 2 2 - 2 3 - 1 2 - 1 3$

Now,

$A + A^{'} = 6 - 2 2 - 2 3 - 1 2 - 1 3 + 6 - 2 2 - 2 3 - 1 2 - 1 3 = 12 - 4 4 - 4 6 - 2 4 - 2 6$

Let $P = \frac{1}{2} (A + A^{'}) = \frac{1}{2} 12 - 4 4 - 4 6 - 2 4 - 2 6$

$= 6 - 2 2 - 2 3 - 1 2 - 1 3$

$∵ P^{'} = 6 - 2 2 - 2 3 - 1 2 - 1 3^{'} = 6 - 2 2 - 2 3 - 1 2 - 1 3 = P$ Thus, $P = \frac{1}{2} (A + A^{'})$ is a symmetric matrix. Now,

$A - A^{'} = 6 - 2 2 - 2 3 - 1 2 - 1 3 - 6 - 2 2 - 2 3 - 1 2 - 1 3 = 000000000$

Let

$Q ∵ Q^{'} = \frac{1}{2} (A - A^{'}) = \frac{1}{2} 000000000 = 000000000 = 000000000 = - Q$

Thus, $Q = \frac{1}{2} (A - A^{'})$ is a skew-symmetric matrix.

$Now, P + Q = 6 - 2 2 - 2 3 - 1 2 - 1 3 + 000000000 = 6 - 2 2 - 2 3 - 1 2 - 1 3 = A$

Hence, matrix A is sum of a symmetric matrix and a Skew-Symmetric matrix.

(iii) A square matrix A can be expressed as sum of a symmetric and skew symmetric matrices.

$A = \frac{1}{2} [A + A^{'}] + \frac{1}{2} [A - A^{'}]$

(P=symmetric matrix)

$(Q = skew symmetric matrix)$

Let $A = 3 - 2 - 4 3 - 2 - 5 - 1 12$ , then $A^{'} = 33 - 1 - 2 - 2 1 - 4 - 5 2$ Now,

$A + A^{'} P = \frac{1}{2} (A + A^{'}) = 3 - 2 - 4 3 - 2 - 5 - 1 12 + 33 - 1 - 2 - 2 1 - 4 - 5 2 = 61 - 5 1 - 4 - 4 - 5 - 4 4 = \frac{1}{2} 61 - 5 1 - 4 - 4 - 5 - 4 4 = 3 1/2 - 5/2 1/2 - 2 - 2 - 5/2 - 2 2$

$∵ P^{'} = 3 1/2 - 5/2 1/2 - 2 - 2 - 5/2 - 2 2^{'} = 3 1/2 - 5/2 1/2 - 2 - 2 - 5/2 - 2 2 = P$

Thus, $P = \frac{1}{2} (A + A^{'})$ is a symmetric matrix. Now,

$A - A^{'} = 3 - 2 - 4 3 - 2 - 5 - 1 12 - 33 - 1 - 2 - 2 1 - 4 - 5 2 = 0 - 5 - 3 50 - 6 360$

Let, $Q = \frac{1}{2} (A - A^{'}) = \frac{1}{2} 0 - 5 - 3 50 - 6 360^{'}$

$= 0 - 5/2 - 3/2 5/2 0 - 3 3/2 30$

$∵ Q^{'} = 0 - 5/2 - 3/2 5/2 0 - 3 3/2 30^{'}$

$= 0 5/2 3/2 - 5/2 03 - 3/2 - 3 0 = - Q$

Thus, $Q = \frac{1}{2} (A - A^{'})$ is a skew-symmetric matrix.

Now,

$P + Q = 3 1/2 - 5/2 1/2 - 2 - 2 - 5/2 - 2 2 + 0 - 5/2 - 3/2 5/2 0 - 3 3/2 30 = 3 - 2 - 4 3 - 2 - 5 - 1 12 = A$

Hence, matrix A is sum of a symmetric matrix and a Skew-Symmetric matrix.

(iv) A square matrix A can be expressed as sum of a symmetric and skew symmetric matrices.

$A = \frac{1}{2} [A + A^{'}] + \frac{1}{2} [A - A^{'}]$

Let $A = [1 - 1 52]$ Then, $A^{'} = [1 - 1 52]^{'} = [15 - 1 2]$ Now, $A + A^{'} = [1 - 1 52] + [15 - 1 2] = [2444]$ Let $P = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [2444] = [1222]$ $∵ P^{'} = [1222]^{'} = [1222] = P$

Thus, $P = \frac{1}{2} (A + A^{'})$ is a symmetric matrix. Now, $A - A^{'} = [1 - 1 52] - [15 - 1 2] = [0 - 6 60]$ Let $^{'} Q = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [0 - 6 60] = [0 - 3 30]$ $∵ Q^{'} = [0 - 3 30]^{'} = [03 - 3 0] = - Q$ Thus, $Q = \frac{1}{2} (A - A^{'})$ is a skew-symmetric matrix.

$P + Q = [1222] + [0 - 3 30] = [1 - 1 52] = A$

Hence, matrix A is sum of a symmetric matrix and a Skew-symmetric matrix.

Choose the correct answer in the following questions from 11 & 12.

If 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

Sol.

(A) Given, A and B are symmetric matrices. $\Rightarrow A^{'} = A$ and $B^{'} = B$

$∴ (AB - BA)^{'} = (AB)^{'} - (BA)^{'} [∵ (A - B)^{'} = A^{'} - B^{'}] = B^{'} A^{'} - A^{'} B^{'} [∵ (AB)^{'} = B^{'} A^{'}] = BA - AB [∵ A^{'} = A and B^{'} = B] = - (AB - BA)$

$\Rightarrow (AB - BA)^{'} = - (AB - BA)$ Thus, $(AB - BA)$ is a skew-symmetric matrix.

If $A = [cos α sin α - sin α cos α]$ , then $A + A^{'} = I$ , if the value of $α$ is ? (A) $π /6$ (B) $π /3$ (C) $π$ (D) $3 π /2$ Sol. (B) Here, $A = [cos α sin α - sin α cos α]$ and

$A^{'} = [cos α sin α - sin α cos α]^{'} = [cos α - sin α sin α cos α]$

Given, $A + A^{'} = I$

$∴ [cos α sin α - sin α cos α] + [cos α - sin α sin α cos α] = [1001] \Rightarrow [2 cos α 0 0 2 cos α] = [1001]$

Comparing the corresponding elements of the above matrices, we have

$2 cos α = 1$

$\Rightarrow cos α = \frac{1}{2} = cos \frac{π}{3}$

$\Rightarrow α = \frac{π}{3}$

EXERCISE - 3.4

Matrices A and B will be inverse of each other only if (A) $AB = BA$ (B) $AB = 0, BA = I$ (C) $AB = BA = 0$ (D) $AB = BA = I$ Sol. (D) We known that if A is a square matrix of order 'n' and if there exists another square matrix B of the same order 'n', such that $AB = BA = I$ , then B is said to be the inverse of A. In this case, it is clear that A is the inverse of B. Thus, matrices A and B will be inverse of each other only if $AB = BA = I$

MISCELLANEOUS EXERCISE

If A and B are symmetric matrices, prove that $AB - BA$ is a skew-symmetric matrix. Sol. Here, A and B are symmetric matrices, then $A^{'} = A$ and $B^{'} = B$

$Now, (AB - BA)^{'} = (AB)^{'} - (BA)^{'} (∴ (A - B)^{'} = A^{'} - B^{'}) [(AB)^{'} = B^{'} A^{'}] = B^{'} A^{'} - A^{'} B^{'} = BA - AB (∵ B^{'} = B and A^{'} = A) = - (AB - BA)$

$∴ (AB - BA)^{'} = - (AB - BA)$

Thus, $(AB - BA)$ is a skew-symmetric matrix.

Show that the matrix $B^{'} AB$ is symmetric or skew-symmetric according as A is symmetric or skew-symmetric. Sol. We suppose that A is a symmetric matrix, then $A^{'} = A$ . Conside

$(B^{'} AB)^{'} = {B^{'} (AB)}^{'} = (AB)^{'} (B^{'})^{'} [∵ (AB)^{'} = B^{'} A^{'}] = (B^{'} A^{'}) B [∵ (B^{'})^{'} = B, A^{'} = A] = B^{'} (A^{'} B) = B^{'} (AB) ∴ (B^{'} AB)^{'} = B^{'} AB$

Which shows that $B^{'} AB$ is a symmetric matrix. Now, we suppose that A is a skew- symmetric matrix. Then, $A^{'} = - A$ Consider

$(B^{'} AB)^{'} = [B^{'} (AB)]^{'} = (AB)^{'} (B^{'})^{'} [∵ (AB^{'}) = B^{'} A^{'}] = (B^{'} A^{'}) B = B^{'} (- A) B = - B^{'} AB [∵ (A^{'})^{'} = A] [∵ A^{'} = - A] ∴ (B^{'} AB)^{'} = - B^{'} AB,$

which shows that $B^{'} AB$ is a skew-symmetric matrix.

Find the values of x,y and z if the matrix $A = 0 x x 2 y y - y z - z z$ satisfies the equation $A^{'} A = I$ . Sol. Given, $A^{'} A = I$

$\Rightarrow \Rightarrow \Rightarrow 0 2 y z x y - z x - y z 0 x x 2 y y - y z - z z = 100010001 0 + x^{2} + x^{2} 0 + y x - y x 0 - z x + z x 0 + x y - x y 4 y^{2} + y^{2} + y^{2} 2 yz - yz - yz 0 - x z + x z 2 yz - yz - yz z^{2} + z^{2} + z^{2} = 100010001 2 x^{2} 00 0 6 y^{2} 0 00 3 z^{2} = 100010001$

On comparing the corresponding elements, we have

$\Rightarrow \Rightarrow 2 x^{2} = 1, 6 y^{2} = 1, 3 z^{2} = 1 x^{2} = \frac{1}{2}, y^{2} = \frac{1}{6}, z^{2} = \frac{1}{3} x = \pm \frac{1}{2}, y = \pm \frac{1}{6}, z = \pm \frac{1}{3}$

For what value of

$x; [121] 121200012 02 x = O$

Sol. Given $[121] 121200012 02 x = O$

$\Rightarrow [121] 0 + 4 + 0 0 + 0 + x 0 + 0 + 2 x = O$

$\Rightarrow [121] 4 x 2 x = O$

$\Rightarrow \Rightarrow [4 + 2 x + 2 x] = [0] 4 + 4 x = 0 \Rightarrow x = - 1$

If $A = [3 - 1 12]$ , show that $A^{2} - 5 A + 7 I = O$ Sol. Given, $A = [3 - 1 12]$

$Now, A^{2} = A.A = [3 - 1 12] [3 - 1 12] = [9 - 1 - 3 - 2 3 + 2 - 1 + 4] = [8 - 5 53] ∴ A^{2} - 5 A + 7 I = [8 - 5 53] - 5 [3 - 1 12] + 7 [1001] = [8 - 5 53] - [15 - 5 510] + [7007] = [8 - 15 + 7 - 5 + 5 + 0 5 - 5 + 0 3 - 10 + 7] = [0000] = O$

Find x , if $[x - 5 - 1] 102020213 x 41 = O$ Sol. Here, $[x - 5 - 1] 102020213 x 41 = O$

$\Rightarrow [x - 5 - 1] 102020213 x 41 = 0 \Rightarrow [x - 5 -1] x + 0 + 2 0 + 8 + 1 2 x + 0 + 3 = 0 \Rightarrow [x (x + 2) + (- 5) (9) + (- 1) (2 x + 3)] = [0] \Rightarrow [x^{2} - 48] = [0] \Rightarrow x^{2} - 48 = 0 \Rightarrow x = \pm 48 = \pm 43$

A manufacturer produces three products x, y, z which he sells in two markets and the annual sales are indicated below.

Market Products

I 10, 000 2,000 18, 000

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.

Sol. Matrix representing the sales is

$A = [100006000200020000180008000]$

(a) Matrix representing the sale price per unit is B

$= 2.50 1.50 1.00$

⇒ Total revenue in each market is given by the product;

$AB = [100006000200020000180008000] 5/2 3/2 1 = [(10000 \times \frac{5}{2}) + (2000 \times \frac{3}{2}) + (18000 \times 1) (6000 \times \frac{5}{2}) + (20000 \times \frac{3}{2}) + (8000 \times 1)] = [4600053000]$

Hence, total revenue in market I is ₹46000 and that in market II is ₹53000. (b) The matrix representing the cost price per unit is $C = 2.00 1.00 0.50$

∴ Total cost in the two markets is given by the product

$A C = [100006000200020000180008000] 21 1/2 = [(10000 \times 2) + (2000 \times 1) + (18000 \times \frac{1}{2}) (6000 \times 2) + (20000 \times 1) + (8000 \times \frac{1}{2})] = [3100036000]$

Profit in market $I = ₹ (46000 - 31000)$

$= ₹ 15000$

and

$profit in market II = ₹ (53000 - 36000) = ₹17000$

$∴$ The gross profit $= ₹ (15000 + 17000)$ = ₹32000.

Find the matrix $X$ so that

$X [142536] = [- 7 2 - 8 4 - 9 6] .$

Sol. Here, $X [142536] = [- 7 2 - 8 4 - 9 6]$ The matrix given on the RHS of the equation is a $2 \times 3$ matrix and the one given on the LHS of the equation is also a $2 \times 3$ matrix. Therefore, X has to be a $2 \times 2$ matrix. Now, Let $X = [a b c d]$ Therefore; we have

$[a b c d] [142536] = [- 7 2 - 8 4 - 9 6]$

$\Rightarrow [a + 4 c b + 4 d 2 a + 5 c 2 b + 5 d 3 a + 6 c 3 b + 6 d] = [- 7 2 - 8 4 - 9 6]$ Equating the corresponding elements of the two matrices, we have

$a + 4 c = - 7, b + 4 d = 2, 2 a + 5 c = - 8, 2 b + 5 d = 4, 3 a + 6 c = - 9 3 b + 6 d = 6$

Now, $a + 4 c = - 7 \Rightarrow a = - 7 - 4 c$

$2 a + 5 c = - 8$

$\Rightarrow \Rightarrow \Rightarrow - 14 - 8 c + 5 c = - 8 - 3 c = 6 c = - 2$

Also; $3 (1) + 6 (- 2) = 3 - 12 = - 9$ Also, $b + 4 d = 2 \Rightarrow b = 2 - 4 d$ and $2 b + 5 d = 4 \Rightarrow 4 - 8 d + 5 d = 4$

$\Rightarrow ∴ - 3 d = 0 \Rightarrow d = 0 b = 2 - 4 (0) = 2$

Also; $3 (2) + 6 \times 0 = 6$ Thus, $a = 1, b = 2, c = - 2$ and $d = 0$ Hence, the required matrix $X$ is $[12 - 2 0]$ Choose the correct answer in the following questions from 9 to 11. 9. If $A = [α γ β - α]$ is such that $A^{2} = I$ , then (A) $1 + α^{2} + β γ = 0$ (B) $1 - α^{2} + β γ = 0$ (C) $1 - α^{2} - β γ = 0$ (D) $1 + α^{2} - β γ = 0$

Sol. (C) Given, $A^{2} = I$

$∴ \Rightarrow \Rightarrow A.A = I [α γ β - α] [α γ β - α] = [1001] [α^{2} + β γ α γ - γ α α β - α β γ β + α^{2}] = [1001]$

On comparing the corresponding elements, we have $α^{2} + β γ = 1$

$\Rightarrow \Rightarrow α^{2} + β γ - 1 = 0 1 - α^{2} - β γ = 0$

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

Sol. (B) Let A be a square matrix such that A is both symmetric and skew-symmetric matrix.

$\Rightarrow ∴ \Rightarrow \Rightarrow \Rightarrow A^{'} = A and A^{'} = A = - A A + A = O 2 A = O A = O$

If A is square matrix such that $A^{2} = A$ , then $(I + A)^{3} - 7 A$ is equal to (A) A (B) I - A (C) I (D) 3A

Sol. (C) Here, $A^{2} = A$

$(I + A)^{2} = (I + A) (I + A) = II + IA + AI + AA = I + 3 A$

$[∵ AI = A = IA, A^{2} = A]$

$∴ (I + A)^{3} = (I + A)^{2} \cdot (I + A) = (I + 3 A) \cdot (I + A) = II + IA + 3 (A I) + 3 (A A) = I + A + 3 A + 3 A = I + 7 A [∵ A^{2} = A] ∴ (I + A)^{3} - 7 A = I .$

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Maths chapter-wise, featuring exercise-wise answers, useful formulas, and clear step-by-step solutions for better understanding and practice.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

4.0Class 12 Maths Chapter 3 Matrices Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 3.1 Solutions	10 Questions	Types of matrices and basic matrix operations
Exercise 3.2 Solutions	22 Questions	Properties of matrices and identity matrix
Exercise 3.3 Solutions	12 Questions	Matrix multiplication, its properties and inverse of a matrix
Exercise 3.4 Solutions	1 Question	Transpose of a matrix and its properties
Miscellaneous Exercise Solutions	11 Questions	Mixed questions based on important matrix concept

5.0NCERT Solutions Class 12 Maths Chapter 3 Matrices: Key Features

The NCERT Solutions for Class 12 Maths Chapter 3 are prepared according to the latest NCERT syllabus followed by CBSE schools.
Step-by-step explanations make matrix operations easier to understand and apply while solving questions.
The exercises cover different types of matrices, their properties, and important questions based on the NCERT textbook.
Regular practice with NCERT questions can help students improve calculation accuracy and avoid common errors.
A clear understanding of matrices can support preparation for mathematics olympiads and other competitive examinations.
Strong basics in matrices can make it easier to understand determinants and other algebra topics in higher mathematics.

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NCERT Solutions Class 12 Maths Chapter 3 Matrices

1.0Key Concepts of Class 12 Maths Chapter 3 Matrices

2.0NCERT Class 12 Maths Chapter 3 Matrices : Detailed Solutions

EXERCISE - 3.1

EXERCISE - 3.2

EXERCISE - 3.3

EXERCISE - 3.4

MISCELLANEOUS EXERCISE

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 3 Matrices Exercise-wise Solutions

5.0NCERT Solutions Class 12 Maths Chapter 3 Matrices: Key Features

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