CBSE Class 12 Mathematics Matrices Worksheet Set 02

Question. \( A = [a_{ij}]_{m \times n} \) is a square matrix, if
(a) \( m < n \)
(b) \( m > n \)
(c) \( m = n \)
(d) None of the options
Answer: (c) \( m = n \)

Question. If \( A = [a_{ij}] \) is a square matrix of order 2 such that \( a_{ij} = \begin{cases} 1, & \text{when } i \neq j \\ 0, & \text{when } i = j \end{cases} \), then \( A^2 \) is
(a) \( \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \)
(b) \( \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \)
(c) \( \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \)
(d) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Answer: (d) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

Question. A matrix \( A = [a_{ij}]_{3 \times 3} \) is defined by \( a_{ij} = \begin{cases} 2i + 3j, & i < j \\ 5, & i = j \\ 3i - 2j, & i > j \end{cases} \). The number of elements in \( A \) which are more than 5, is
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (b) 4

Question. If product of rows and columns of matrix is 27, then the number of possible different ordered matrices are
(a) 3
(b) 5
(c) 6
(d) 4
Answer: (d) 4

Question. If \( A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix} \) and \( kA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix} \), then the value of \( k \), \( a \) and \( b \) respectively, are
(a) \( -6, -12, -18 \)
(b) \( -6, -4, -9 \)
(c) \( -6, 4, 9 \)
(d) \( -6, 12, 18 \)
Answer: (b) \( -6, -4, -9 \)

Question. If \( \begin{bmatrix} 2x + y & 4x \\ 5x - 7 & 4x \end{bmatrix} = \begin{bmatrix} 7 & 7y - 13 \\ y & x+6 \end{bmatrix} \), then
(a) \( x = 3, y = 1 \)
(b) \( x = 2, y = 3 \)
(c) \( x = 2, y = 4 \)
(d) \( x = 3, y = 3 \)
Answer: (b) \( x = 2, y = 3 \)

Question. If \( X, Y \) and \( XY \) are matrices of order \( 2 \times 3 \), \( m \times n \) and \( 2 \times 5 \) respectively, then number of elements in matrix \( Y \) is
(a) 6
(b) 10
(c) 15
(d) 35
Answer: (c) 15

Question. If \( A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1} \pi x & \tan^{-1} \frac{x}{\pi} \\ \sin^{-1} \frac{x}{\pi} & \cot^{-1} \pi x \end{bmatrix} \) and \( B = \frac{1}{\pi} \begin{bmatrix} -\cos^{-1} \pi x & \tan^{-1} \frac{x}{\pi} \\ \sin^{-1} \frac{x}{\pi} & -\tan^{-1} \pi x \end{bmatrix} \), then \( A - B \) is equal to
(a) \( I \)
(b) \( O \)
(c) \( 2I \)
(d) \( \frac{1}{2}I \)
Answer: (d) \( \frac{1}{2}I \)

Question. If \( A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \) and \( (3I + 4A)(3I - 4A) = x^2 I \), then the value(s) of \( x \) is/are
(a) \( \pm \sqrt{7} \)
(b) \( 0 \)
(c) \( \pm 5 \)
(d) \( 25 \)
Answer: (c) \( \pm 5 \)

Question. If \( A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \) 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) \( 1 + \alpha^2 - \beta\gamma = 0 \)
Answer: (c) \( 1 - \alpha^2 - \beta\gamma = 0 \)

Question. If \( A \) is a square matrix and \( A^2 = A \), then \( (I + A)^2 - 3A \) is equal to
(a) \( I \)
(b) \( A \)
(c) \( 2A \)
(d) \( 3I \)
Answer: (a) \( I \)

Question. If \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \), then \( A^{2023} \) is equal to
(a) \( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \)
(b) \( \begin{bmatrix} 0 & 2023 \\ 0 & 0 \end{bmatrix} \)
(c) \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
(d) \( \begin{bmatrix} 2023 & 0 \\ 0 & 2023 \end{bmatrix} \)
Answer: (c) \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

Assertion-Reason Based Questions

Question. Assertion (A): \( A = \text{diag}[3 \quad 5 \quad 2] \) is a scalar matrix of order \( 3 \times 3 \).
Reason (R): If a diagonal matrix has all non-zero elements equal, it is known as a scalar matrix.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Question. Assertion (A): \( A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) is an identity matrix.
Reason (R): \( A = [a_{ij}] \) is a scalar matrix, if \( a_{ij} = \begin{cases} k, & i = j \\ 0, & i \neq j \end{cases} \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (b) Both A and R are correct; R is not the correct explanation of A.

Question. Assertion (A): If \( \begin{bmatrix} xy & 4 \\ z+5 & x+y \end{bmatrix} = \begin{bmatrix} 4 & w \\ 0 & 4 \end{bmatrix} \), then \( x = 2, y = 2, z = -5 \) and \( w = 4 \).
Reason (R): Two matrices are equal, if their orders are same and their corresponding elements are same.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A): If \( A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} \), then \( 3A - B = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix} \).
Reason (R): If the matrices \( A \) and \( B \) of same order, say \( m \times n \), satisfy the commutative law, then \( A + B = B + A \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (b) Both A and R are correct; R is not the correct explanation of A.

Question. Assertion (A): The product of two matrices \( \begin{bmatrix} 1 & 3 & 4 \\ -1 & 2 & 0 \\ 1 & 3 & 2 \end{bmatrix} \) and \( \begin{bmatrix} -1 & 2 \\ 3 & 4 \\ -1 & 2 \end{bmatrix} \) is \( \begin{bmatrix} 4 & 22 \\ 7 & 6 \\ 6 & 18 \end{bmatrix} \).
Reason (R): The product of two matrices \( A \) and \( B \) is defined, if the number of columns of \( A \) is equal to the number of rows of \( B \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Case Based Questions - I

Two distributors of a chips company distribute two varieties of chips packets: Spicy Chips (SC) and Cheesy Chips (CC). At the beginning of a certain month, the number of chips packets available with the distributors is shown in matrix \( M \).
\[ M = \begin{bmatrix} 965 & 498 \\ 872 & 689 \end{bmatrix} \begin{matrix} \rightarrow \text{Distributor 1} \\ \rightarrow \text{Distributor 2} \end{matrix} \] with columns representing Spicy Chips (SC) and Cheesy Chips (CC) respectively.
The number of chips packets distributed during that month by them is shown in matrix \( N \).
\[ N = \begin{bmatrix} 956 & 399 \\ 650 & 511 \end{bmatrix} \begin{matrix} \rightarrow \text{Distributor 1} \\ \rightarrow \text{Distributor 2} \end{matrix} \] with columns representing SC and CC respectively.
Based on the above information, answer the following questions.

Question. Find the number of chips packets remaining with the distributors at the end of that month.
Answer: Number of chips packets remaining with the distributors at the end of month is given by: \[ M - N = \begin{bmatrix} 965 & 498 \\ 872 & 689 \end{bmatrix} - \begin{bmatrix} 956 & 399 \\ 650 & 511 \end{bmatrix} = \begin{bmatrix} 9 & 99 \\ 222 & 178 \end{bmatrix} \] Thus, Distributor 1 has 9 Spicy Chips and 99 Cheesy Chips remaining, and Distributor 2 has 222 Spicy Chips and 178 Cheesy Chips remaining. [6]

Question. A packet of Spicy Chips and Cheesy Chips costs ₹ 10 and ₹ 20, respectively. Find the total cost of the chips distributed by each distributor that month using matrix multiplication. Show your work and give your answer in the matrix form.
Answer: Let \( A \) denote the matrix showing cost of Spicy Chips and Cheesy Chips: \[ A = \begin{bmatrix} 10 \\ 20 \end{bmatrix} \] Cost of chips distributed by each distributor: \[ N \times A = \begin{bmatrix} 956 & 399 \\ 650 & 511 \end{bmatrix} \begin{bmatrix} 10 \\ 20 \end{bmatrix} = \begin{bmatrix} 9560 + 7980 \\ 6500 + 10220 \end{bmatrix} = \begin{bmatrix} 17540 \\ 16720 \end{bmatrix} \] Therefore, Distributor 1 distributed chips of cost ₹ 17,540 and Distributor 2 distributed chips of cost ₹ 16,720. [6]

Very Short Answer Type Questions

Question. If possible, then find the sum of the matrices \( A \) and \( B \), where \( A = \begin{bmatrix} \sqrt{3} & 1 \\ 2 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} x & y & z \\ a & b & c \end{bmatrix} \).
Answer: Here, order of matrix \( A = 2 \times 2 \) and order of matrix \( B = 2 \times 3 \).
Hence, the addition of given matrices is not possible because to add two matrices, the order of both the matrices must be same.

Question. Simplify \(\begin{bmatrix} \cos^2 x & \sin^2 x \\ \sin^2 x & \cos^2 x \end{bmatrix} + \begin{bmatrix} \sin^2 x & \cos^2 x \\ \cos^2 x & \sin^2 x \end{bmatrix}\).
Answer: We have, \(\begin{bmatrix} \cos^2 x & \sin^2 x \\ \sin^2 x & \cos^2 x \end{bmatrix} + \begin{bmatrix} \sin^2 x & \cos^2 x \\ \cos^2 x & \sin^2 x \end{bmatrix}\)
\(= \begin{bmatrix} \cos^2 x + \sin^2 x & \sin^2 x + \cos^2 x \\ \sin^2 x + \cos^2 x & \cos^2 x + \sin^2 x \end{bmatrix}\)
\(= \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \quad [\because \sin^2 x + \cos^2 x = 1]\)

Question. Find the value of \( y - x \) from the following equation:
\( 2 \begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \).
Answer: We have, \( 2 \begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\(\Rightarrow \begin{bmatrix} 2x & 10 \\ 14 & 2y-6 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\(\Rightarrow \begin{bmatrix} 2x+3 & 6 \\ 15 & 2y-4 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
On equating the corresponding elements of both sides, we get:
\( 2x+3 = 7 \text{ and } 2y-4 = 14 \)
\(\Rightarrow 2x = 4 \text{ and } 2y = 18 \)
\(\Rightarrow x = 2 \text{ and } y = 9 \)
Now, \( y - x = 9 - 2 = 7 \).

Question. Find non-zero values of \( x \) satisfying the matrix equation \( x \begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} + 2 \begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix} = 2 \begin{bmatrix} x^2 + 8 & 24 \\ 10 & 6x \end{bmatrix} \).
Answer: We have, \( x \begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} + 2 \begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix} = 2 \begin{bmatrix} x^2+8 & 24 \\ 10 & 6x \end{bmatrix} \)
\(\Rightarrow \begin{bmatrix} 2x^2 & 2x \\ 3x & x^2 \end{bmatrix} + \begin{bmatrix} 16 & 10x \\ 8 & 8x \end{bmatrix} = \begin{bmatrix} 2x^2+16 & 48 \\ 20 & 12x \end{bmatrix} \)
\(\Rightarrow \begin{bmatrix} 2x^2+16 & 12x \\ 3x+8 & x^2+8x \end{bmatrix} = \begin{bmatrix} 2x^2+16 & 48 \\ 20 & 12x \end{bmatrix} \)
On equating the corresponding elements of both sides, we get:
\( 3x+8 = 20 \Rightarrow 3x = 12 \Rightarrow x = 4 \)
\( 12x = 48 \Rightarrow x = 4 \)
\( x^2+8x = 12x \Rightarrow x^2-4x = 0 \Rightarrow x(x-4) = 0 \Rightarrow x = 4 \text{ (since } x \neq 0\text{)} \)
Thus, the non-zero value of \( x \) is \( 4 \).

Question. If \( A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \), then find \( A - B \).
Answer: Given, \( A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \)
Now, \( A-B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix} \).

Question. If \( \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} = A + \begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix} \), then find the matrix \( A \).
Answer: We have, \( \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} = A + \begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix} \)
\(\Rightarrow A = \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix} = \begin{bmatrix} 9-1 & -1-2 & 4-(-1) \\ -2-0 & 1-4 & 3-9 \end{bmatrix} = \begin{bmatrix} 8 & -3 & 5 \\ -2 & -3 & -6 \end{bmatrix} \).

Question. If \( x \begin{bmatrix} 1 \\ 2 \end{bmatrix} + y \begin{bmatrix} 2 \\ 5 \end{bmatrix} = \begin{bmatrix} 4 \\ 9 \end{bmatrix} \), then find the value of \( x \).
Answer: We have, \( x\begin{bmatrix}1\\2\end{bmatrix} + y\begin{bmatrix}2\\5\end{bmatrix} = \begin{bmatrix}4\\9\end{bmatrix} \)
\(\Rightarrow \begin{bmatrix} x \\ 2x \end{bmatrix} + \begin{bmatrix} 2y \\ 5y \end{bmatrix} = \begin{bmatrix} 4 \\ 9 \end{bmatrix} \)
\(\Rightarrow \begin{bmatrix} x+2y \\ 2x+5y \end{bmatrix} = \begin{bmatrix} 4 \\ 9 \end{bmatrix} \)
On equating the corresponding elements, we get:
\( x+2y = 4 \) ...(i)
\( 2x+5y = 9 \) ...(ii)
On solving Eqs. (i) and (ii), we get:
\( x = 2, y = 1 \).

Question. If matrices \( A \) and \( B \) are of order \( 3 \times n \) and \( m \times 5 \) respectively, then find the order of matrix \( C = 5A + 3B \).
Answer: Given, order of \( A = 3 \times n \) and order of \( B = m \times 5 \).
\( C = 5A + 3B \)
Two matrices are additive, if their orders are same.
\( \therefore 3 \times n = m \times 5 \Rightarrow m = 3 \text{ and } n = 5 \)
\( \therefore \) Order of \( C = 3 \times 5 \).

Question. Let \( A = [a_{ij}]_{n \times n} \) be a diagonal matrix whose diagonal elements are different and \( B = [b_{ij}]_{n \times n} \) is some another matrix. If \( AB = [c_{ij}]_{n \times n} \), then find \( c_{ij} \).
Answer: Here, \( A = [a_{ij}]_{n \times n} \) where
\( a_{ij} = \begin{cases} 0, & \text{if } i \neq j \\ a_{ii}, & \text{if } i = j \end{cases} \)
and \( B = [b_{ij}]_{n \times n} \).
So, \( c_{ij} = a_{ii} b_{ij} \).

Question. Suppose \( A = \begin{bmatrix} 5 & 4 \\ 2 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 5 & 1 \\ 6 & 8 & 4 \end{bmatrix} \), then find \( AB \) and \( BA \), if they exist.
Answer: Here, number of columns of \( A = 2 \) = number of rows of \( B \).
And number of columns of \( B = 3 \neq \) number of rows of \( A = 2 \).
So, product \( AB \) of matrices \( A \) and \( B \) is possible but product \( BA \) is not possible.
\( AB = \begin{bmatrix} 5 & 4 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 3 & 5 & 1 \\ 6 & 8 & 4 \end{bmatrix} = \begin{bmatrix} 15+24 & 25+32 & 5+16 \\ 6+18 & 10+24 & 2+12 \end{bmatrix} = \begin{bmatrix} 39 & 57 & 21 \\ 24 & 34 & 14 \end{bmatrix} \)
\( BA \) does not exist.

Short Answer Type Questions

Question. If \(\begin{bmatrix} x + 3 & z + 4 & 2y - 7 \\ -6 & a - 1 & 0 \\ b - 3 & -21 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 6 & 3y - 2 \\ -6 & -3 & 2c + 2 \\ 2b + 4 & -21 & 0 \end{bmatrix}\), then find the values of \( a, b, c \) and \( z \).
Answer: On comparing the corresponding elements of both the matrices, we get:
\( x+3 = 0 \Rightarrow x = -3 \)
\( z+4 = 6 \Rightarrow z = 2 \)
\( 2y-7 = 3y-2 \Rightarrow y = -5 \)
\( a-1 = -3 \Rightarrow a = -2 \)
\( 2c+2 = 0 \Rightarrow 2c = -2 \Rightarrow c = -1 \)
\( b-3 = 2b+4 \Rightarrow b = -7 \)
Thus, \( a = -2, b = -7, c = -1 \) and \( z = 2 \).

Question. Assume \( Y \), \( W \) and \( P \) are the matrices of orders \( 3 \times k \), \( n \times 3 \) and \( p \times k \). Find the restrictions on \( n, k \) and \( p \), so that \( PY + WY \) will be defined.
Answer: Given, matrices \( Y, W \) and \( P \) are the matrices of orders \( 3 \times k, n \times 3 \) and \( p \times k \).
Order of \( P = p \times k \) and order of \( Y = 3 \times k \).
\( PY \) is possible, if \( k = 3 \).
\(\therefore\) Order of \( PY = p \times 3 \).
Order of \( W = n \times 3 \) and order of \( Y = 3 \times k \).
\(\therefore\) Order of \( WY = n \times 3 \) (since \( k = 3 \)).
Now, sum of \( PY \) and \( WY \) is possible only if both of them are of same order.
\(\therefore p \times 3 = n \times 3 \Rightarrow p = n \).
Hence, \( PY + WY \) is defined if \( p = n \) and \( k = 3 \).

Question. If \( A = \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix} \), then verify that \( (BA)^2 \neq B^2 A^2 \).
Answer: Now, \( BA = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 6+1+4 & -8+1+0 \\ 3+2+8 & -4+2+0 \end{bmatrix} = \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix} \)
and \( A^2 = A \cdot A = \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix} \).
Here, number of columns of first matrix i.e., 2 is not equal to the number of rows of second matrix i.e., 3.
So, \( A^2 \) is not possible.
Similarly, \( B^2 \) is also not possible.
Hence, \( (BA)^2 \neq B^2 A^2 \).

Question. If \( A = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \), then show that \( (A+B)(A-B) \neq A^2 - B^2 \).
Answer: Now, \( A^2 = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 0+1 & 0+1 \\ 0+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \) ...(i)
\( B^2 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0-1 & 0+0 \\ 0+0 & -1+0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \) ...(ii)
\( A+B = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0+0 & 1-1 \\ 1+1 & 1+0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 2 & 1 \end{bmatrix} \) ...(iii)
\( A-B = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0-0 & 1-(-1) \\ 1-1 & 1-0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 0 & 1 \end{bmatrix} \) ...(iv)
From Eqs. (iii) and (iv), we get:
\( (A+B)(A-B) = \begin{bmatrix} 0 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0+0 & 0+0 \\ 0+0 & 4+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix} \) ...(v)
From Eqs. (i) and (ii), we get:
\( A^2 - B^2 = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \) ...(vi)
From Eqs. (v) and (vi), we get:
\( (A+B)(A-B) \neq A^2 - B^2 \).
Hence proved.

Question. Solve the matrix equation: \( \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = 0 \).
Answer: We have, \( \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = 0 \)
\(\Rightarrow \begin{bmatrix} x-0-2 & 0-10-0 & 2x-5-3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = 0 \)
\(\Rightarrow \begin{bmatrix} x-2 & -10 & 2x-8 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = 0 \)
\(\Rightarrow [x^2 - 2x - 40 + 2x - 8] = [0] \)
On comparing corresponding elements, we get:
\( x^2 - 48 = 0 \)
\(\Rightarrow x^2 = 48 \)
\(\Rightarrow x = \pm 4\sqrt{3} \).

Question. If \( [2 \quad 1 \quad 3] \begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = A \), then find the value of \( A \).
Answer: We have, \( A = [2 \quad 1 \quad 3] \begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \)
\( = [-2-1+0 \quad 0+1+3 \quad -2+0+3] \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \)
\( = [-3 \quad 4 \quad 1] \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = [-3 + 0 - 1] = [-4] \).

Question. If \( A = \begin{bmatrix} 4 & 0 \\ 2k & 5k \end{bmatrix} \) and \( B = \begin{bmatrix} k & 0 \\ 3 & -1 \end{bmatrix} \), such that \( AB = BA \). Then, show that \( 2k^2 + 17k - 12 = 0 \).
Answer: Given, \( A = \begin{bmatrix} 4 & 0 \\ 2k & 5k \end{bmatrix} \) and \( B = \begin{bmatrix} k & 0 \\ 3 & -1 \end{bmatrix} \)
Also given, \( AB = BA \)
\( \Rightarrow \begin{bmatrix} 4 & 0 \\ 2k & 5k \end{bmatrix} \begin{bmatrix} k & 0 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} k & 0 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 0 \\ 2k & 5k \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 4k + 0 & 0 + 0 \\ 2k^2 + 15k & 0 - 5k \end{bmatrix} = \begin{bmatrix} 4k + 0 & 0 + 0 \\ 12 - 2k & 0 - 5k \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 4k & 0 \\ 2k^2 + 15k & -5k \end{bmatrix} = \begin{bmatrix} 4k & 0 \\ 12 - 2k & -5k \end{bmatrix} \)
On equating the corresponding elements of both the matrices, we get
\( 2k^2 + 15k = 12 - 2k \)
\( \Rightarrow 2k^2 + 15k + 2k - 12 = 0 \)
\( \Rightarrow 2k^2 + 17k - 12 = 0 \)
Hence proved.

Question. Find the matrix \( A \) satisfying the matrix equation \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Answer: Let \( P = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \) and \( Q = \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} \).
The given equation is \( P A Q = I \), where \( I \) is the identity matrix of order 2.
Since \( |P| = 2(2) - 1(3) = 4 - 3 = 1 \neq 0 \), \( P^{-1} \) exists and is given by:
\( P^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \)
Since \( |Q| = -3(-3) - 2(5) = 9 - 10 = -1 \neq 0 \), \( Q^{-1} \) exists and is given by:
\( Q^{-1} = \frac{1}{-1} \begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} \)
Pre-multiplying by \( P^{-1} \) and post-multiplying by \( Q^{-1} \) on both sides of \( P A Q = I \), we get:
\( A = P^{-1} I Q^{-1} = P^{-1} Q^{-1} \)
\( A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} \)
\( A = \begin{bmatrix} 2(3) + (-1)(5) & 2(2) + (-1)(3) \\ -3(3) + 2(5) & -3(2) + 2(3) \end{bmatrix} \)
\( A = \begin{bmatrix} 6 - 5 & 4 - 3 \\ -9 + 10 & -6 + 6 \end{bmatrix} \)
\( A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \).

Question. If \( A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \), then find \( A^2 - 5A + 6I \).
Answer: Given, \( A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \)
Now, \( A^2 - 5A + 6I \)
\( = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} - 5 \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} + 6 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1+0 & 1-3+0 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} \)
\( = \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+6 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix} \).

Question. For the matrix \( A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \), find \( a \) and \( b \) such that \( A^2 + aI = bA \), where \( I \) is a \( 2 \times 2 \) identity matrix.
Answer: Let \( A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \)
Now, \( A^2 + aI = bA \)
\( \Rightarrow \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} + a \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = b \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} + \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} = \begin{bmatrix} 3b & b \\ 7b & 5b \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 16+a & 8 \\ 56 & 32+a \end{bmatrix} = \begin{bmatrix} 3b & b \\ 7b & 5b \end{bmatrix} \)
On comparing the corresponding elements of both matrices, we get
\( b = 8 \)
and \( 16+a = 3b \)
\( \Rightarrow a = 3 \times 8 - 16 \)
\( \Rightarrow a = 8 \)
Hence, \( a = 8 \) and \( b = 8 \).

Question. If \( A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \), then prove that \( A^3 - 4A^2 + A = O \).
Answer: Given, \( A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \)
\( A^2 = A \cdot A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \)
\( A^3 = A^2 \cdot A = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 26 & 45 \\ 15 & 26 \end{bmatrix} \)
Now, \( \text{LHS} = A^3 - 4A^2 + A \)
\( = \begin{bmatrix} 26 & 45 \\ 15 & 26 \end{bmatrix} - 4 \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} + \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 26 - 28 + 2 & 45 - 48 + 3 \\ 15 - 16 + 1 & 26 - 28 + 2 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O = \text{RHS} \)
Hence proved.

Long Answer Type Questions

Question. Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices \( A \) and \( B \).
September sales (in rupees):
\( A = \begin{bmatrix} 10000 & 20000 & 30000 \\ 50000 & 30000 & 10000 \end{bmatrix} \begin{matrix} \text{Ramkishan} \\ \text{Gurcharan Singh} \end{matrix} \)
October sales (in rupees):
\( B = \begin{bmatrix} 5000 & 10000 & 6000 \\ 20000 & 10000 & 10000 \end{bmatrix} \begin{matrix} \text{Ramkishan} \\ \text{Gurcharan Singh} \end{matrix} \)
(i) Find the combined sales in September and October for each farmer in each variety.
(ii) Find the decrease in sales from September to October.
(iii) If both farmers receive 2% profit on gross sales, then compute the profit for each farmer and for each variety sold in October.
Answer: (i) Combined sales in September and October for each farmer in each variety is given by \( A + B \):
\( A + B = \begin{bmatrix} 10000+5000 & 20000+10000 & 30000+6000 \\ 50000+20000 & 30000+10000 & 10000+10000 \end{bmatrix} \)
\( = \begin{bmatrix} 15000 & 30000 & 36000 \\ 70000 & 40000 & 20000 \end{bmatrix} \begin{matrix} \text{Ramkishan} \\ \text{Gurcharan Singh} \end{matrix} \)
(ii) Decrease in sales from September to October is given by \( A - B \):
\( A - B = \begin{bmatrix} 10000-5000 & 20000-10000 & 30000-6000 \\ 50000-20000 & 30000-10000 & 10000-10000 \end{bmatrix} \)
\( = \begin{bmatrix} 5000 & 10000 & 24000 \\ 30000 & 20000 & 0 \end{bmatrix} \begin{matrix} \text{Ramkishan} \\ \text{Gurcharan Singh} \end{matrix} \)

(iii) 2% profit of October sales \( B \) is given by:
\( 2\% \text{ of } B = \frac{2}{100} \times B = 0.02 \times B \)
\( = 0.02 \begin{bmatrix} 5000 & 10000 & 6000 \\ 20000 & 10000 & 10000 \end{bmatrix} \)
\( = \begin{bmatrix} 100 & 200 & 120 \\ 400 & 200 & 200 \end{bmatrix} \begin{matrix} \text{Ramkishan} \\ \text{Gurcharan Singh} \end{matrix} \)
Thus, in October, Ramkishan receives Rs 100, Rs 200, and Rs 120 as profit in the sale of each variety of rice, respectively, and Gurcharan Singh receives profit of Rs 400, Rs 200, and Rs 200 in the sale of each variety of rice, respectively.

Question. Find matrix \( A \), if \( \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix} A = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix} \).
Answer: Let the order of \( A \) is \( m \times n \).
Since the pre-multiplying matrix is of order \( 3 \times 2 \) and the resulting matrix is of order \( 3 \times 3 \), we must have \( m = 2 \) and \( n = 3 \).
So, order of matrix \( A \) is \( 2 \times 3 \).
Let \( A = \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix} \).
Now, \( \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix} = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2a-d & 2b-e & 2c-f \\ a & b & c \\ -3a+4d & -3b+4e & -3c+4f \end{bmatrix} = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix} \)
By equating the corresponding elements, we directly get:
\( a = 1, \quad b = -2, \quad c = -5 \)
Also, \( 2a-d = -1 \Rightarrow 2(1) - d = -1 \Rightarrow d = 3 \)
\( 2b-e = -8 \Rightarrow 2(-2) - e = -8 \Rightarrow e = 4 \)
\( 2c-f = -10 \Rightarrow 2(-5) - f = -10 \Rightarrow f = 0 \)
Thus, \( A = \begin{bmatrix} 1 & -2 & -5 \\ 3 & 4 & 0 \end{bmatrix} \).

Question. If \( A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \), then prove that \( A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}, n \in \mathbb{N} \).
Answer: We shall prove the result using the principle of mathematical induction.
Let \( P(n) \): If \( A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \), then \( A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix} \).
For \( n=1 \):
\( A^1 = \begin{bmatrix} \cos 1\theta & \sin 1\theta \\ -\sin 1\theta & \cos 1\theta \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \), which is true.
Therefore, the result is true for \( n=1 \).
Let the result be true for \( n=k \).
So, \( P(k) \): \( A^k = \begin{bmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{bmatrix} \).
Now, we prove that the result holds for \( n = k + 1 \).
\( A^{k+1} = A \cdot A^k = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{bmatrix} \)
\( = \begin{bmatrix} \cos\theta \cos k\theta - \sin\theta \sin k\theta & \cos\theta \sin k\theta + \sin\theta \cos k\theta \\ -\sin\theta \cos k\theta - \cos\theta \sin k\theta & -\sin\theta \sin k\theta + \cos\theta \cos k\theta \end{bmatrix} \)
\( = \begin{bmatrix} \cos(\theta + k\theta) & \sin(\theta + k\theta) \\ -\sin(\theta + k\theta) & \cos(\theta + k\theta) \end{bmatrix} \)
\( = \begin{bmatrix} \cos(k+1)\theta & \sin(k+1)\theta \\ -\sin(k+1)\theta & \cos(k+1)\theta \end{bmatrix} \).
Therefore, the result is true for \( n=k+1 \).
Thus, by principle of mathematical induction, \( A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix} \) holds for all natural numbers.

Question. If \( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \), then prove that \( A \) is a root of the polynomial \( f(x) = x^3 - 6x^2 + 7x + 2 \).
Answer: To prove that \( A \) is a root of the polynomial \( f(x) = x^3 - 6x^2 + 7x + 2 \), we must show that \( f(A) = O \), i.e., \( A^3 - 6A^2 + 7A + 2I = O \).
Given, \( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \).
Now, \( A^2 = A \cdot A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \).
Now, \( A^3 = A^2 \cdot A = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39 \end{bmatrix} = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} \).
Now, \( f(A) = A^3 - 6A^2 + 7A + 2I \)
\( = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - 6 \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 21-30+7+2 & 0-0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0-0+0+0 & 55-78+21+2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O \).
Hence proved.