CBSE Class 12 Mathematics Matrices Worksheet Set 03

Question. If \( A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & 0 \\ -1 & 1 \end{bmatrix} \), then the value of \( x \) for which \( A^2 = B \) is
(a) \( -2 \)
(b) 2
(c) 2 or -2
(d) 4
Answer: (a) \( -2 \)

Question. Find the matrix \( A^2 \), where \( A = [a_{ij}] \) is a \( 2 \times 2 \) matrix whose elements are given by \( a_{ij} = \text{maximum }(i,j) - \text{minimum }(i,j) \)
(a) \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
(b) \( \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)
(c) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
(d) \( \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \)
Answer: (b) \( \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)

Question. Given that \( A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \) and \( A^2 = 3I \), then
(a) \( 1 + \alpha^2 + \beta\gamma = 0 \)
(b) \( 1 - \alpha^2 - \beta\gamma = 0 \)
(c) \( 3 - \alpha^2 - \beta\gamma = 0 \)
(d) \( 3 + \alpha^2 + \beta\gamma = 0 \)
Answer: (c) \( 3 - \alpha^2 - \beta\gamma = 0 \)

Question. If \( \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix} \), then the value of \( (2x + y - z) \) is
(a) 1
(b) 2
(c) 3
(d) 5
Answer: (d) 5

Question. If \( \begin{bmatrix} 4 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 1 & 3 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = O \), then \( x + 3y - 3z \) is
(a) 1
(b) 3
(c) 4
(d) 0
Answer: (d) 0

Question. For what value of \( x \in \left[0, \frac{\pi}{2}\right] \), is \( A + A^T = \sqrt{3}I \), where \( A = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} \)?
(a) \( \frac{\pi}{3} \)
(b) \( \frac{\pi}{6} \)
(c) \( 0 \)
(d) \( \frac{\pi}{2} \)
Answer: (b) \( \frac{\pi}{6} \)

Question. If \( A \) and \( B \) are symmetric matrices, then \( (BA)^T \) is
(a) \( AB \)
(b) \( BA \)
(c) \( I \)
(d) Null matrix
Answer: (a) \( AB \)

Question. If \( A = [a_{ij}] \) is a skew-symmetric matrix of order \( n \), then
(a) \( a_{ij} = \frac{1}{a_{ji}}, \forall i,j \)
(b) \( a_{ij} \neq 0, \forall i,j \)
(c) \( a_{ij} = 0 \), where \( i = j \)
(d) \( a_{ij} \neq 0 \), where \( i = j \)
Answer: (c) \( a_{ij} = 0 \), where \( i = j \)

Question. If a matrix \( A \) is both symmetric and skew-symmetric, then \( A \) is necessarily a
(a) diagonal matrix
(b) zero square matrix
(c) square matrix
(d) identity matrix
Answer: (b) zero square matrix

Question. If \( A = \begin{bmatrix} 0 & x & y \\ 2 & 0 & 5 \\ 6 & -5 & z \end{bmatrix} \) is skew-symmetric matrix, then the value of \( x + y^2 + z \) is
(a) -34
(b) 38
(c) 34
(d) 35
Answer: (c) 34

Question. A and B are skew-symmetric matrices of same order. \( AB \) is symmetric, if
(a) \( AB = 0 \)
(b) \( AB = -BA \)
(c) \( AB = BA \)
(d) \( BA = 0 \)
Answer: (c) \( AB = BA \)

Question. If for a square matrix \( A \), \( A^2 - 3A + I = O \) and \( A^{-1} = xA + yI \), then the value of \( x + y \) is
(a) -2
(b) 2
(c) 3
(d) -3
Answer: (b) 2

Question. If for a square matrix \( A \), \( A^2 - A + I = O \), then \( A^{-1} \) is equal to
(a) \( A \)
(b) \( A + I \)
(c) \( I - A \)
(d) \( A - I \)
Answer: (c) \( I - A \)

Assertion-Reason Based Questions

Question. Assertion (A) If \( A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & 4 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & 5 \\ 4 & -2 \\ 2 & 1 \end{bmatrix} \) and \( AB \) and \( BA \) both are defined.
Reason (R) For two matrices \( A \) and \( B \), product \( AB \) is defined, if number of columns in \( A \) is equal to the number of rows in \( B \).
Answer: Assertion and Reason both are correct and Reason is the correct explanation of Assertion. Order of \( A = 2 \times 3 \) and order of \( B = 3 \times 2 \). For \( AB \), the number of columns of \( A = \text{number of rows of } B = 3 \). So, \( AB \) is defined. For \( BA \), the number of columns of \( B = \text{number of rows of } A = 2 \). So, \( BA \) is defined. [5]

Question. Let \( A, B \) and \( C \) be three square matrices of same order. Now, consider the following statements.
Assertion (A) If \( A = B \), then \( AC = BC \).
Reason (R) If \( AC = BC \), then \( A = B \).
Answer: Assertion is correct and Reason is incorrect. Clearly, if \( A = B \), then \( AC = BC \). But if \( AC = BC \) it may be \( A \neq B \). For example, let \( A = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 4 \\ 0 & 0 \end{bmatrix} \) and \( C = \begin{bmatrix} 0 & 7 \\ 0 & 0 \end{bmatrix} \). Here, \( AC = BC = \begin{bmatrix} 0 & 7 \\ 0 & 0 \end{bmatrix} \), but \( A \neq B \). [5]

Question. Assertion (A) If \( A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), then value of \( k \) such that \( A^2 = kA - 2I \) is \( -1 \).
Reason (R) If \( A \) and \( B \) are square matrices of the same order, then \( (A + B)(A + B) = A^2 + AB + BA + B^2 \).
Answer: Assertion is incorrect but Reason is correct.
Since \( A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \).
Using \( A^2 = kA - 2I \):
\( \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = k \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix} \).
On comparing, \( 1 = 3k - 2 \implies k = 1 \). So, Assertion is incorrect.
Now, by distributive law, \( (A + B)(A + B) = A^2 + AB + BA + B^2 \). Reason is correct. [5]

Question. Assertion (A) Matrix \( A = \begin{bmatrix} 1 & 2 & 4 \\ 2 & 3 & -1 \\ 4 & -1 & 5 \end{bmatrix} \) is symmetric matrix.
Reason (R) Matrix \( B = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix} \) is a skew-symmetric matrix.
Answer: Assertion and Reason both are correct but Reason is not the correct explanation of Assertion. Since \( A' = \begin{bmatrix} 1 & 2 & 4 \\ 2 & 3 & -1 \\ 4 & -1 & 5 \end{bmatrix} = A \), Assertion is correct. Since \( B' = \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix} = -B \), Reason is correct. [6]

Question. Assertion (A) For any square matrix \( B \) with real number entries, \( B + B' \) is a skew-symmetric matrix and \( B - B' \) is a symmetric matrix.
Reason (R) A square matrix \( B \) can be expressed as the sum of a symmetric matrix and skew-symmetric matrix.
Answer: Assertion is incorrect but Reason is correct. For any square matrix \( B \) with real number entries, \( B + B' \) is a symmetric matrix and \( B - B' \) is a skew-symmetric matrix. Reason is correct. [6]

Case Based Questions - I

A company wanted to outsource the creation of its video and picture content for social media. They were approached by two firms offering the following rates per post (in ₹).
Rates matrix \( A \):
\[ A = \begin{bmatrix} 30 & 10 \\ 25 & 15 \end{bmatrix} \begin{matrix} \text{Firm 1} \\ \text{Firm 2} \end{matrix} \] with columns representing Video and Picture rates.
They gave the contract to both the firms for an equal number of posts. After 3 months, each firm created the following number of posts.
Matrix \( B \):
\[ B = \begin{bmatrix} 70 \\ 100 \end{bmatrix} \begin{matrix} \text{Video} \\ \text{Picture} \end{matrix} \]
Each firm's posts created the following number of sales for the company.
Matrix \( C \):
\[ C = \begin{bmatrix} 500 & 200 \\ 400 & 300 \end{bmatrix} \begin{matrix} \text{Firm 1} \\ \text{Firm 2} \end{matrix} \] with columns representing Video and Picture sales.
Based on the above information, answer the following questions.

Question. Which firm cost more to the company?
Answer: Cost of each firm = \( A \times B \): \[ A \times B = \begin{bmatrix} 30 & 10 \\ 25 & 15 \end{bmatrix} \begin{bmatrix} 70 \\ 100 \end{bmatrix} = \begin{bmatrix} 2100 + 1000 \\ 1750 + 1500 \end{bmatrix} = \begin{bmatrix} 3100 \\ 3250 \end{bmatrix} \] Cost for Firm 1 = ₹ 3100 and cost for Firm 2 = ₹ 3250. Therefore, Firm 2 cost more to the company. [6]

Question. If each sale was worth ₹ 300, which firm was more profitable for the company? Show your work.
Answer: Total revenue generated by each firm = \( 300 \times C \): \[ 300 \times \begin{bmatrix} 500 & 200 \\ 400 & 300 \end{bmatrix} = \begin{bmatrix} 150000 & 60000 \\ 120000 & 90000 \end{bmatrix} \] Revenue generated by Firm 1 = \( 150000 + 60000 = \text{₹ } 210000 \).
Revenue generated by Firm 2 = \( 120000 + 90000 = \text{₹ } 210000 \).
Profit generated by Firm 1 = \( 210000 - 3100 = \text{₹ } 206900 \).
Profit generated by Firm 2 = \( 210000 - 3250 = \text{₹ } 206750 \).
Therefore, Firm 1 was more profitable for the company. [6]

Very Short Answer Type Questions

Question. Find \( x \) from the matrix equation \(\begin{bmatrix} 1 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ 2 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}\).
Answer: We have, \(\begin{bmatrix} 1 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ 2 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} x + 6 \\ 4x + 10 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}\)
On equating the corresponding elements of both sides, we get:
\( x+6 = 5 \Rightarrow x = -1 \).

Question. If \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ -1 \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}\), find \( x+y+z \).
Answer: We have, \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ -1 \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \Rightarrow \begin{bmatrix} x \\ -y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}\)
On equating the corresponding elements of both the matrices, we get:
\( x = 1, y = -2 \text{ and } z = 1 \)
\(\therefore x+y+z = 1 - 2 + 1 = 0 \).

Question. If matrix \( A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) and \( A^2 = kA \), then write the value of \( k \).
Answer: Given, \( A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) and \( A^2 = kA \)
\(\therefore \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = k \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\(\Rightarrow \begin{bmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{bmatrix} = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \)
\(\Rightarrow \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \)
On equating the corresponding elements of both sides, we get:
\( k=2 \).

Question. If \( A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \) and \( A = B^2 \), then find the value of \( x \).
Answer: We have, \( B = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \)
\(\therefore B^2 = B \cdot B = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} x^2+0 & 0+0 \\ x+1 & 0+1 \end{bmatrix} = \begin{bmatrix} x^2 & 0 \\ x+1 & 1 \end{bmatrix} \)
Since, \( A = B^2 \Rightarrow \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} x^2 & 0 \\ x+1 & 1 \end{bmatrix} \)
By equality of matrices,
\( x^2 = 1 \text{ and } x+1 = 2 \)
\(\Rightarrow x = \pm 1 \text{ and } x = 1 \Rightarrow x = 1 \).

Question. If \( A = \begin{bmatrix} 2 & -3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix} \), \( X = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \) and \( Y = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} \), then find the value of \( AB - XY \).
Answer: Given, \( A = \begin{bmatrix} 2 & -3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix} \), \( X = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \) and \( Y = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} \).
Now, \( AB = \begin{bmatrix} 2 & -3 & 4 \end{bmatrix} \begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix} = [6 - 6 + 8] = [8] \)
and \( XY = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} = [2+6+12] = [20] \)
\(\therefore AB - XY = [8] - [20] = [-12] \).

Question. Find the product of \( A = \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \) and \( B = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \).
Answer: \( AB = \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix} = \begin{bmatrix} a^2+b^2 & -ab+ba \\ -ba+ab & b^2+a^2 \end{bmatrix} = \begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix} \).

Question. If \( A \) is a \( 3 \times 4 \) matrix and \( B \) is a matrix such that \( A'B \) and \( AB' \) are both defined, then what is the order of matrix \( B \)?
Answer: Given, order of \( A \) is \( 3 \times 4 \).
\(\therefore\) Order of \( A' \) is \( 4 \times 3 \).
Let order of \( B \) be \( m \times n \).
\(\therefore\) Order of \( B' \) is \( n \times m \).
Since, \( A'B \) is defined:
\(\therefore\) Number of columns of \( A' \) = Number of rows of \( B \)
\(\Rightarrow m = 3 \)
Since, \( AB' \) is defined:
\(\dots\) Number of columns of \( A \) = Number of rows of \( B' \)
\(\Rightarrow n = 4 \)
\(\therefore\) Order of the matrix \( B \) is \( 3 \times 4 \).

Question. If a matrix \( A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \), then find the matrix \( AA' \) (where \( A' \) is the transpose of \( A \)).
Answer: Given, \( A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \)
\(\therefore A' = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
Now, \( AA' = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = [1+4+9] = [14] \).

Question. For what value of \( x \), is the matrix \( A = \begin{bmatrix} 0 & 1 & -4 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{bmatrix} \) a skew-symmetric matrix?
Answer: If matrix \( A \) is skew-symmetric, then \( A' = -A \)
\(\Rightarrow \begin{bmatrix} 0 & -1 & x \\ 1 & 0 & -3 \\ -4 & 3 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & 1 & -4 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} 0 & -1 & x \\ 1 & 0 & -3 \\ -4 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 4 \\ 1 & 0 & -3 \\ -x & 3 & 0 \end{bmatrix}\)
On comparing the corresponding elements of both sides, we get:
\( x=4 \).

Question. If \( A = \begin{bmatrix} a & c & -1 \\ b & 0 & 5 \\ 1 & -5 & 0 \end{bmatrix} \) is a skew-symmetric matrix, then find the value of \( 2a - (b+c) \).
Answer: Since, \( A \) is a skew-symmetric matrix.
\(\therefore A^T = -A \)
\(\Rightarrow \begin{bmatrix} a & b & 1 \\ c & 0 & -5 \\ -1 & 5 & 0 \end{bmatrix} = -\begin{bmatrix} a & c & -1 \\ b & 0 & 5 \\ 1 & -5 & 0 \end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} a & b & 1 \\ c & 0 & -5 \\ -1 & 5 & 0 \end{bmatrix} = \begin{bmatrix} -a & -c & 1 \\ -b & 0 & -5 \\ -1 & 5 & 0 \end{bmatrix}\)
On comparing the corresponding elements of both sides, we get:
\( a = -a \Rightarrow 2a = 0 \Rightarrow a = 0 \)
and \( b = -c \Rightarrow b+c = 0 \)
Thus, \( 2a - (b+c) = 2(0) - 0 = 0 \).

Short Answer Type Questions

Question. If \( A = \begin{bmatrix} 1 & 0 \\ -1 & 5 \end{bmatrix} \), then find the value of \( k \) if \( A^2 = 6A + kI_2 \), where \( I_2 \) is an identity matrix.
Answer: Given, \( A = \begin{bmatrix} 1 & 0 \\ -1 & 5 \end{bmatrix} \) ... (i)
and \( A^2 - 6A - kI_2 = O \) ... (ii)
Now, \( A^2 = A \cdot A = \begin{bmatrix} 1 & 0 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 5 \end{bmatrix} = \begin{bmatrix} 1+0 & 0+0 \\ -1-5 & 0+25 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -6 & 25 \end{bmatrix} \) ... (iii)
From Eq. (ii), we get:
\( A^2 - 6A + kI_2 = O \)
\( \Rightarrow \begin{bmatrix} 1 & 0 \\ -6 & 25 \end{bmatrix} - 6 \begin{bmatrix} 1 & 0 \\ -1 & 5 \end{bmatrix} - k \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 1 & 0 \\ -6 & 25 \end{bmatrix} - \begin{bmatrix} 6 & 0 \\ -6 & 30 \end{bmatrix} - \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 1-6 & 0-0 \\ -6+6 & 25-30 \end{bmatrix} - \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} -5 & 0 \\ 0 & -5 \end{bmatrix} - \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} -5-k & 0 \\ 0 & -5-k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
On equating the corresponding elements of both sides, we get:
\( -5-k = 0 \Rightarrow k = -5 \).

Question. If \( A = \text{diag } [a, b, c] \), then show that \( A^n = \text{diag } [a^n, b^n, c^n], \forall n \in \mathbb{N} \).
Answer: Given, \( A = \text{diag } [a, b, c] = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \)
\( A^2 = A \cdot A = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} = \begin{bmatrix} a^2 & 0 & 0 \\ 0 & b^2 & 0 \\ 0 & 0 & c^2 \end{bmatrix} = \text{diag }[a^2, b^2, c^2] \)
Similarly, \( A^3 = \text{diag }[a^3, b^3, c^3] \) and \( A^4 = \text{diag }[a^4, b^4, c^4] \)
Therefore, \( A^n = \text{diag }[a^n, b^n, c^n] \)
Hence proved.

Question. In a legislative assembly election, a political group hired a public relations firm to promote its candidate in three ways: telephone, house calls and letters. The cost per contact (in paise) is given in matrix \( A \) as:
\( A = \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix} \begin{matrix} \text{Telephone} \\ \text{House call} \\ \text{Letter} \end{matrix} \)
The number of contacts of each type made in two cities \( X \) and \( Y \) is given by matrix \( B \) as:
\( B = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{matrix} \rightarrow X \\ \rightarrow Y \end{matrix} \)
Find the total amount spent by the group in the two cities \( X \) and \( Y \).
Answer: Total amount spent is given by the matrix product \( BA \):
\( BA = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix} \)
\( = \begin{bmatrix} 1000 \times 40 + 500 \times 100 + 5000 \times 50 \\ 3000 \times 40 + 1000 \times 100 + 10000 \times 50 \end{bmatrix} \)
\( = \begin{bmatrix} 40000 + 50000 + 250000 \\ 120000 + 100000 + 500000 \end{bmatrix} \)
\( = \begin{bmatrix} 340000 \\ 720000 \end{bmatrix} \)
So, the total amount spent by the group in the two cities is \( 340000 \) paise and \( 720000 \) paise, i.e., \( \text{Rs } 3400 \) and \( \text{Rs } 7200 \), respectively.

Question. If \( A \) and \( B \) are symmetric matrices of the same order, then show that \( AB \) is symmetric if and only if \( A \) and \( B \) commute, that is \( AB = BA \).
Answer: Since, \( A \) and \( B \) both are symmetric matrices, therefore \( A' = A \) and \( B' = B \).
Let \( AB \) be symmetric, then \( (AB)' = AB \).
But \( (AB)' = B'A' = BA \) [since \( B' = B \) and \( A' = A \)]
Therefore, \( BA = AB \).
Conversely, if \( AB = BA \), then we shall show that \( AB \) is symmetric.
\( (AB)' = B'A' = BA = AB \).
Hence proved.

Question. If \( A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix} \), then verify that:
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = A' - B' \)
Answer: We have:
\( A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix} \) and \( B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix} \)
Now, \( A+B = \begin{bmatrix} -1-4 & 2+1 & 3-5 \\ 5+1 & 7+2 & 9+0 \\ -2+1 & 1+3 & 1+1 \end{bmatrix} = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix} \)
Therefore, \( \text{LHS} = (A+B)' = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix} \) ... (i)
Now, \( \text{RHS} = A' + B' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix} + \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix} \) ... (ii)
From Eqs. (i) and (ii), we get \( (A+B)' = A' + B' \). Hence proved.

(ii) \( A-B = \begin{bmatrix} -1-(-4) & 2-1 & 3-(-5) \\ 5-1 & 7-2 & 9-0 \\ -2-1 & 1-3 & 1-1 \end{bmatrix} = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix} \)
Therefore, \( \text{LHS} = (A-B)' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix} \) ... (iii)
Now, \( \text{RHS} = A' - B' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix} - \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix} \) ... (iv)
From Eqs. (iii) and (iv), we get \( (A-B)' = A' - B' \). Hence proved.

Question. Find \( \frac{1}{2}(A+A') \) and \( \frac{1}{2}(A-A') \), where \( A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \).
Answer: Given, \( A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \). Then \( A' = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} \).
Now, \( \frac{1}{2}(A+A') = \frac{1}{2} \left( \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} + \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} \right) \)
\( = \frac{1}{2} \begin{bmatrix} 0+0 & a-a & b-b \\ -a+a & 0+0 & c-c \\ -b+b & -c+c & 0+0 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \).

And, \( \frac{1}{2}(A-A') = \frac{1}{2} \left( \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} - \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} \right) \)
\( = \frac{1}{2} \begin{bmatrix} 0-0 & a-(-a) & b-(-b) \\ -a-a & 0-0 & c-(-c) \\ -b-b & -c-c & 0-0 \end{bmatrix} \)
\( = \frac{1}{2} \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix} = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} \).

Question. If \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \), then show that \( (A-A') \) is a skew-symmetric matrix, where \( A' \) is the transpose of matrix \( A \).
Answer: Given, \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \), so \( A' = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \).
Now, \( A - A' = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} = \begin{bmatrix} 3-3 & -4-1 \\ 1-(-4) & -1-(-1) \end{bmatrix} = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} \) ... (i)
Now, let \( C = A-A' = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} \).
Then, \( C' = \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} = -C \).
Hence, \( (A-A') \) is a skew-symmetric matrix.

Question. Show that the matrix \( B^T AB \) is skew-symmetric, if \( A \) is skew-symmetric.
Answer: Since, \( A \) is a skew-symmetric matrix, therefore \( A^T = -A \).
Now, \( (B^T AB)^T = (AB)^T (B^T)^T \)
\( = (B^T A^T) B \)
\( = B^T (-A) B \)
\( = -B^T AB \).
Hence, \( B^T AB \) is a skew-symmetric matrix.

Question. Give one example of skew-symmetric matrix of order 2 and order 3.
Answer: We know that the diagonal elements of a skew-symmetric matrix are zero and \( a_{ij} = -a_{ji} \).
Therefore, skew-symmetric matrices of order 2 and 3 are:
\( \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 2 & 3 & 0 \end{bmatrix} \circ \), respectively.

Question. If \( A \) and \( B \) are symmetric matrices, then prove that \( BA - 2AB \) is neither a symmetric matrix nor skew-symmetric matrix.
Answer: Given, \( A \) and \( B \) are symmetric matrices.
\( \therefore A' = A \) and \( B' = B \).
Now, \( (BA - 2AB)' = (BA)' - (2AB)' \)
\( = (A'B') - 2(AB)' \) [since \( (kA)' = kA' \) and \( (XY)' = Y'X' \)]
\( = A'B' - 2B'A' \)
\( = AB - 2BA \) [since \( A' = A \) and \( B' = B \)]
Which is neither equal to \( (BA - 2AB) \) nor equal to \( -(BA - 2AB) \).
Hence, \( BA - 2AB \) is neither a symmetric matrix nor skew-symmetric matrix.

Question. If \( A = \begin{bmatrix} 5 & 2 \\ 3 & -6 \end{bmatrix} \), then prove that \( A \) can be expressed as sum of symmetric and skew-symmetric matrices.
Answer: Let \( A = \begin{bmatrix} 5 & 2 \\ 3 & -6 \end{bmatrix} \). Then \( A' = \begin{bmatrix} 5 & 3 \\ 2 & -6 \end{bmatrix} \).
Now, \( A + A' = \begin{bmatrix} 5 & 2 \\ 3 & -6 \end{bmatrix} + \begin{bmatrix} 5 & 3 \\ 2 & -6 \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 5 & -12 \end{bmatrix} \)
\( \Rightarrow \frac{1}{2}(A+A') = \frac{1}{2} \begin{bmatrix} 10 & 5 \\ 5 & -12 \end{bmatrix} \circ \), which is symmetric. ... (i)
Also, \( A - A' = \begin{bmatrix} 5 & 2 \\ 3 & -6 \end{bmatrix} - \begin{bmatrix} 5 & 3 \\ 2 & -6 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \)
\( \Rightarrow \frac{1}{2}(A-A') = \frac{1}{2} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \), which is skew-symmetric. ... (ii)
On adding Eqs. (i) and (ii), we get:
\( \frac{1}{2}(A+A') + \frac{1}{2}(A-A') = \frac{1}{2} \begin{bmatrix} 10 & 5 \\ 5 & -12 \end{bmatrix} + \frac{1}{2} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \)
\( = \frac{1}{2} \begin{bmatrix} 10 & 4 \\ 6 & -12 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 3 & -6 \end{bmatrix} = A \).
Hence proved.

Long Answer Type Questions

Question. Let \( A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \) and \( f(x) = x^2 - 4x + 7 \). Show that \( f(A) = O \). Use this result to find \( A^5 \).
Answer: Given, \( A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \) and \( f(x) = x^2 - 4x + 7 \).
Then, \( f(A) = A^2 - 4A + 7I \)
\( A^2 = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4-3 & 6+6 \\ -2-2 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} \)
Now, \( A^2 - 4A + 7I = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - 4 \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \)
\( = \begin{bmatrix} 1-8+7 & 12-12+0 \\ -4+4+0 & 1-8+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).
Hence proved.
We have \( A^2 - 4A + 7I = O \Rightarrow A^2 = 4A - 7I \)
Thus, \( A^3 = A \cdot A^2 = A(4A - 7I) = 4A^2 - 7A \)
\( = 4(4A - 7I) - 7A = 16A - 28I - 7A = 9A - 28I \)
And, \( A^5 = A^3 \cdot A^2 = (9A - 28I)(4A - 7I) \)
\( = 36A^2 - 63A - 112A + 196I^2 \)
\( = 36(4A - 7I) - 175A + 196I \)
\( = 144A - 252I - 175A + 196I = -31A - 56I \)
\( = -31 \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} - 56 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} -62 & -93 \\ 31 & -62 \end{bmatrix} - \begin{bmatrix} 56 & 0 \\ 0 & 56 \end{bmatrix} \)
\( = \begin{bmatrix} -62-56 & -93-0 \\ 31-0 & -62-56 \end{bmatrix} = \begin{bmatrix} -118 & -93 \\ 31 & -118 \end{bmatrix} \).

Question. A manufacturer sells the products \( x \), \( y \) and \( z \) in two markets. Annual sales are indicated below:

(i) If unit sale prices of \( x, y \) and \( z \) are Rs 2.50, Rs 1.50 and Rs 1.00 respectively, then find the total revenue in each market with the help of matrix algebra.
(ii) If the unit cost of the above three commodities are Rs 2.00, Rs 1.00 and 50 paise, respectively. Find the gross profit. [Hint Profit = Revenue - Cost]
Answer: (i) Let the sales of products \( x, y \) and \( z \) per market be denoted by matrix \( A \):
\( A = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{matrix} \text{Market I} \\ \text{Market II} \end{matrix} \)
And the unit sale price of products \( x, y \) and \( z \) be denoted by column matrix \( B \):
\( B = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} \)
Now, total revenue = Total sales \(\times\) Unit sales price = \( AB \)
\( = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} \)
\( = \begin{bmatrix} 10000 \times 2.50 + 2000 \times 1.50 + 18000 \times 1.00 \\ 6000 \times 2.50 + 20000 \times 1.50 + 8000 \times 1.00 \end{bmatrix} \)
\( = \begin{bmatrix} 25000 + 3000 + 18000 \\ 15000 + 30000 + 8000 \end{bmatrix} = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix} \) ... (i)
Hence, total revenue of Market I = Rs 46000 and total revenue of Market II = Rs 53000.

(ii) Let the unit cost price of products \( x, y \) and \( z \) be denoted by matrix \( C \):
\( C = \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} \)
Now, total cost = Total sales \(\times\) Unit cost price = \( AC \)
\( = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} \)
\( = \begin{bmatrix} 10000 \times 2.00 + 2000 \times 1.00 + 18000 \times 0.50 \\ 6000 \times 2.00 + 20000 \times 1.00 + 8000 \times 0.50 \end{bmatrix} \)
\( = \begin{bmatrix} 20000 + 2000 + 9000 \\ 12000 + 20000 + 4000 \end{bmatrix} = \begin{bmatrix} 31000 \\ 36000 \end{bmatrix} \) ... (ii)
Now, Profit = Revenue - Cost
\( = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix} - \begin{bmatrix} 31000 \\ 36000 \end{bmatrix} = \begin{bmatrix} 15000 \\ 17000 \end{bmatrix} \)
Therefore, total profit of Market I = Rs 15000 and total profit of Market II = Rs 17000.
Hence, gross profit = Rs 15000 + Rs 17000 = Rs 32000.

Question. A trust fund has Rs 30000 that must be invested in two different types of bond. The first bond pays 5% interest per year and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30000 among the two type of bonds. If the trust fund must obtain an annual total interest of:
(a) Rs 1800
(b) Rs 2000
Answer: Let the amount invested in first bond = Rs \( x \).
Therefore, the amount invested in second bond = Rs \( (30000 - x) \).
(a) If the total interest is Rs 1800, then:
Interest = Investment in bonds \(\times\) Annual rate of interest
\( \Rightarrow [1800] = \begin{bmatrix} x & 30000-x \end{bmatrix} \begin{bmatrix} 5\% \\ 7\% \end{bmatrix} \)
\( \Rightarrow [1800] = \begin{bmatrix} x \times 5\% + (30000 - x) \times 7\% \end{bmatrix} \)
On comparing the corresponding elements, we get:
\( \frac{5x}{100} + \frac{7}{100}(30000-x) = 1800 \)
\( \Rightarrow 5x + 210000 - 7x = 180000 \)
\( \Rightarrow -2x = -30000 \Rightarrow x = 15000 \).
Therefore, the amount invested in first bond is Rs 15000 and second bond is Rs 15000.
(b) If the total interest is Rs 2000, then:
Interest = Investment in bonds \(\times\) Annual rate of interest
\( \Rightarrow [2000] = \begin{bmatrix} x & 30000-x \end{bmatrix} \begin{bmatrix} 5\% \\ 7\% \end{bmatrix} \)
\( \Rightarrow [2000] = \begin{bmatrix} x \times 5\% + (30000 - x) \times 7\% \end{bmatrix} \)
On comparing the corresponding elements, we get:
\( \frac{5x}{100} + \frac{7}{100}(30000-x) = 2000 \)
\( \Rightarrow 5x + 210000 - 7x = 200000 \)
\( \Rightarrow -2x = -10000 \)
\( \Rightarrow x = 5000 \).
Therefore, the amount invested in first bond is Rs 5000 and second bond is Rs 25000.