CBSE Class 12 Mathematics Matrices VBQs Set 02

Fill in the Blanks

Question. A matrix which is not a square matrix is called a ____________ matrix.
Answer: rectangular

Question. If A and B are square matrix of the same order then \( (AB)' = \) ____________ .
Answer: \( B' A' \)

Question. If A is a skew-symmetric matrix and \( n \in N \) such that \( (A^n)^T = \lambda A^n \) then \( \lambda = \) ____________ .
Answer: \( (-1)^n \)

Question. Given a skew-symmetric matrix \( A = \begin{bmatrix} 0 & a & 1 \\ -1 & b & 1 \\ -1 & c & 0 \end{bmatrix} \), then value of \( (a + b + c)^2 \) is ____________ .
Answer: 0
Solution: We have \( A = -A^T \) (Since Matrix A is skew symmetric)
\( \Rightarrow \begin{bmatrix} 0 & a & 1 \\ -1 & b & 1 \\ -1 & c & 0 \end{bmatrix} = -\begin{bmatrix} 0 & -1 & -1 \\ a & b & c \\ 1 & 1 & 0 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 0 & a & 1 \\ -1 & b & 1 \\ -1 & c & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 1 \\ -a & -b & -c \\ -1 & -1 & 0 \end{bmatrix} \)
\( \Rightarrow a = 1, b = -b \Rightarrow 2b = 0 \Rightarrow b = 0 \) and \( c = -1 \)
\( \therefore (a + b + c)^2 = (1 + 0 - 1)^2 = 0 \)

Question. If \( A + B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \) and \( A - 2B = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix} \), then \( A = \) ____________ .
Answer: \( \begin{bmatrix} 1/3 & 1/3 \\ 2/3 & 1/3 \end{bmatrix} \)
Solution: We have,
\( 2(A + B) + (A - 2B) = 2 \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix} \)
\( \Rightarrow 3A = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} \Rightarrow A = \begin{bmatrix} 1/3 & 1/3 \\ 2/3 & 1/3 \end{bmatrix} \)

Very Short Answer Questions

Question. If \( \begin{bmatrix} x - y & z \\ 2x - y & w \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ 0 & 5 \end{bmatrix} \), find the value of \( x + y \).
Answer: Given \( \begin{bmatrix} x - y & z \\ 2x - y & w \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ 0 & 5 \end{bmatrix} \)
Equating, we get
\( x - y = -1 \) ...(i)
\( 2x - y = 0 \) ...(ii)
\( z = 4, w = 5 \)
(ii) - (i) \( \Rightarrow 2x - y - x + y = 0 + 1 \Rightarrow x = 1 \)
Putting it in eq. (i), we get \( 1 - y = -1 \Rightarrow y = 2 \)
\( \therefore x + y = 1 + 2 = 3 \)

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: Given \( \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} = A + \begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix} \)
\( \Rightarrow A = \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix} = \begin{bmatrix} 8 & -3 & 5 \\ -2 & -3 & -6 \end{bmatrix} \)

Question. Write the element \( a_{23} \) of a \( 3 \times 3 \) matrix \( A = (a_{ij}) \) whose elements \( a_{ij} \) are given by \( a_{ij} = \frac{|i - j|}{2} \).
Answer: \( a_{23} = \frac{|2 - 3|}{2} = \frac{|-1|}{2} = \frac{1}{2} \)

Question. If matrix \( \begin{bmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{bmatrix} \) is a skew-symmetric matrix, then find the values of \( a, b \) and \( c \).
Answer: Let \( A = \begin{bmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{bmatrix} \)
Since A is skew-symmetric matrix \( \therefore A^T = -A \)
\( \Rightarrow \begin{bmatrix} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -a & -3 \\ -2 & -b & 1 \\ -c & -1 & 0 \end{bmatrix} \)
By equating corresponding elements, we get
\( a = -2, c = -3 \) and \( b = -b \Rightarrow 2b = 0 \Rightarrow b = 0 \)
\( \therefore a = -2, b = 0 \) and \( c = -3 \)

Question. If \( A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \), then for what value of \( \alpha \), A is an identity matrix.
Answer: If A is identity matrix, then \( A = I_2 \)
\( \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
On equating corresponding elements, we get
\( \Rightarrow \cos \alpha = 1, \sin \alpha = 0 \Rightarrow \alpha = 0 \)

Question. If \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \), then find the value of \( k \).
Answer: Given: \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} (1)(3) + (2)(2) & (1)(1) + (2)(5) \\ (3)(3) + (4)(2) & (3)(1) + (4)(5) \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 7 & 11 \\ 17 & 23 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \)
Equating the corresponding elements, we get \( k = 17 \)

Question. If the matrix \( A = \begin{bmatrix} 0 & a & -3 \\ 2 & 0 & -1 \\ b & 1 & 0 \end{bmatrix} \) is skew symmetric, find the values of \( 'a' \) and \( 'b' \).
Answer: Given \( A = \begin{bmatrix} 0 & a & -3 \\ 2 & 0 & -1 \\ b & 1 & 0 \end{bmatrix} \)
For skew symmetric matrix \( A^T = -A \)
\( \begin{bmatrix} 0 & 2 & b \\ a & 0 & 1 \\ -3 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -a & 3 \\ -2 & 0 & 1 \\ -b & -1 & 0 \end{bmatrix} \)
On comparing both sides, we get \( a = -2 \) and \( -b = -3 \Rightarrow b = 3 \)

Question. Simplify: \( \cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} \)
Answer: Given: \( \cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2 \theta & \sin \theta \cos \theta \\ -\sin \theta \cos \theta & \cos^2 \theta \end{bmatrix} + \begin{bmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix} \)
\( = \begin{bmatrix} \sin^2 \theta + \cos^2 \theta & 0 \\ 0 & \sin^2 \theta + \cos^2 \theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

Question. If A is a \( 3 \times 3 \) matrix, whose elements are given by \( a_{ij} = \frac{1}{3} |-3i + j| \), then write the value of \( a_{23} \).
Answer: \( a_{23} = \frac{1}{3} |-3 \times 2 + 3| = \frac{1}{3} |-6 + 3| = \frac{1}{3} \times 3 = 1 \)

Question. If \( A^T = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} \), then find \( A^T - B^T \).
Answer: Given: \( B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} \Rightarrow B^T = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} \)
Now \( A^T - B^T = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix} \)

Question. For what value of \( x \), is the matrix \( A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{bmatrix} \) a skew-symmetric matrix?
Answer: A will be skew symmetric matrix if \( A = -A^T \).
\( \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & -1 & x \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -x \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix} \)
Equating the corresponding elements, we get \( x = 2 \).

Question. If \( A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \), find \( AA^T \).
Answer: We have, \( AA^T = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & \cos \theta \sin \theta - \sin \theta \cos \theta \\ \sin \theta \cos \theta - \cos \theta \sin \theta & \sin^2 \theta + \cos^2 \theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_2 \)

Question. Let A and B are matrices of order \( 3 \times 2 \) and \( 2 \times 4 \) respectively. Write the order of matrix (AB).
Answer: Order of \( AB = [a_{ij}]_{3 \times 2} [b_{ij}]_{2 \times 4} = [c_{ij}]_{3 \times 4} \) i.e., order of \( AB \) is \( 3 \times 4 \).

Question. Construct a \( 2 \times 2 \) matrix \( A = [a_{ij}] \) whole elements are given by \( a_{ij} = |(i)^2 - j| \).
Answer: We have, elements of the Matrix are given by \( a_{ij} = |(i)^2 - j| \)
\( \therefore a_{11} = |(1)^2 - 1| = 0, a_{21} = |(2)^2 - 1| = 3 \)
\( a_{12} = |(1)^2 - 2| = 1, a_{22} = |(2)^2 - 2| = 2 \)
\( \therefore \) Matrix \( A = \begin{bmatrix} 0 & 1 \\ 3 & 2 \end{bmatrix} \)

Short Answer Questions-

Question. Find the value of \( x + y \) 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: Given, \( 2 \begin{bmatrix} x & 5 \\ 7 & y - 3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2x & 10 \\ 14 & 2y - 6 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2x + 3 & 6 \\ 15 & 2y - 4 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
Equating the corresponding elements, we get
\( 2x + 3 = 7 \text{ and } 2y - 4 = 14 \)
\( \Rightarrow 2x = 4 \Rightarrow x = 2 \) and \( 2y = 18 \Rightarrow y = 9 \)
\( \therefore x + y = 2 + 9 = 11 \)

Question. If matrix \( A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \) and \( A^2 = kA \), then write the value of \( k \). 
Answer: Given: \( A^2 = kA \)
\( \Rightarrow \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = k \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = k \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\( \Rightarrow 2 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = k \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\( \Rightarrow k = 2 \)

Question. If \( A = \begin{bmatrix} -3 & 2 \\ 1 & -1 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), find scalar \( k \) so that \( A^2 + I = kA \). 
Answer: We have,
\( A^2 = A \times A = \begin{bmatrix} -3 & 2 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -3 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 11 & -8 \\ -4 & 3 \end{bmatrix} \)
and \( kA = k \begin{bmatrix} -3 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} -3k & 2k \\ k & -k \end{bmatrix} \)
Given, \( A^2 + I = kA \Rightarrow \begin{bmatrix} 11 & -8 \\ -4 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -3k & 2k \\ k & -k \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 12 & -8 \\ -4 & 4 \end{bmatrix} = \begin{bmatrix} -3k & 2k \\ k & -k \end{bmatrix} \)
\( \Rightarrow k = -4 \)

Question. Show that \( A' A \) and \( A A' \) are both symmetric matrices for any matrix \( A \). 
Answer: Let \( P = A'A \)
\( P' = (A'A)' \)
\( = A'(A')' = A'A = P \) [using \( (AB)' = B'A' \)]
So, \( A'A \) is symmetric matrix for any matrix \( A \).
Similarly, let \( Q = AA' \)
\( Q' = (AA')' = (A')'(A)' \)
\( = A(A')' = Q \)
So, \( AA' \) is symmetric matrix for any matrix \( A \).

Question. Matrix \( A = \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix} \) is given to be symmetric, find values of \( a \) and \( b \). [CBSE Delhi 2016]
Answer: We have \( A = \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix} \)
\(\because\) \( A \) is symmetric matrix.
\( \Rightarrow A^T = A \Rightarrow \begin{bmatrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix} \)
Equating the corresponding elements, we get
\( 2b = 3 \) and \( 3a = -2 \Rightarrow b = \frac{3}{2} \) and \( a = -\frac{2}{3} \)

Question. Show that all the diagonal elements of a skew symmetric matrix are zero. 
Answer: Since, \( A = [a_{ij}] \) is skew symmetric matrix
\( \therefore A^T = -A \)
\( [a_{ij}]^T = -[a_{ij}] \Rightarrow [a_{ji}] = [-a_{ij}] \)
For diagonal elements \( i = j \)
\( \Rightarrow a_{ii} = -a_{ii} \Rightarrow 2a_{ii} = 0 \Rightarrow a_{ii} = 0 \forall i \)
Hence, diagonal elements of skew symmetric matrix are zero.

Question. If \( A = \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix} \) and \( BA = (b_{ij}) \), find \( b_{21} + b_{32} \).
Answer: We have, \( A = \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix} \)
\( \therefore BA = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}_{3 \times 2} \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix}_{2 \times 3} \)
\( [b_{ij}] = \begin{bmatrix} 2-12 & -4+6 & 6+15 \\ 4-20 & -8+10 & 12+25 \\ 2-4 & -4+2 & 6+5 \end{bmatrix}_{3 \times 3} \Rightarrow [b_{ij}] = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}_{3 \times 3} \)
Now, \( b_{21} = -16 \); \( b_{32} = -2 \)
\( \therefore b_{21} + b_{32} = -16 - 2 = -18 \)

Question. Find the value of \( (x - y) \) from the matrix 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: \( \begin{bmatrix} 2x & 10 \\ 14 & 2y-6 \end{bmatrix} + \begin{bmatrix} -3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2x-3 & 10-4 \\ 14+1 & 2y-6+2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2x-3 & 6 \\ 15 & 2y-4 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
We know that two matrices of same order are equal if the corresponding entries are equal.
i.e., \( 2x-3 = 7 \Rightarrow 2x = 10 \Rightarrow x = 5 \)
and \( 2y-4 = 14 \Rightarrow 2y = 18 \Rightarrow y = 9 \)
\( \therefore x - y = 5 - 9 = -4 \)

Question. If \( A \) and \( B \) are symmetric matrices, such that \( AB \) and \( BA \) are both defined, then prove that \( AB - BA \) is a skew-symmetric matrix. 
Answer: We have \( A^T = A \) and \( B^T = B \) and \( AB \) and \( BA \) are both defined.
Now \( (AB - BA)^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T \) (\(\because (AB)^T = B^T A^T\))
\( = BA - AB = -(AB - BA) \)
\( \Rightarrow AB - BA \) is a skew-symmetric matrix. Hence proved.

Question. For the matrix \( A = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \), find \( A + A^T \) and verify it is a symmetric matrix. 
Answer: We have \( A = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \), \( A^T = \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix} \)
\( \Rightarrow A + A^T = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} + \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 8 & 14 \end{bmatrix} \)
\( \Rightarrow (A + A^T)^T = \begin{bmatrix} 4 & 8 \\ 8 & 14 \end{bmatrix}^T = \begin{bmatrix} 4 & 8 \\ 8 & 14 \end{bmatrix} = A + A^T \)
Hence, \( A + A^T \) is a symmetric matrix.

Short Answer Questions-II

Question. For the following matrices \( A \) and \( B \), verify that \( (AB)' = B'A' \). 
\( A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \), \( B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} \)
Answer: Given: \( A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \), \( B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} \)
\( AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix} \)
\( (AB)' = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \)
\( B' A' = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}' \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \)
\( \therefore (AB)' = B'A' \).

Question. If \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \) and \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \) and \( (A + B)^2 = A^2 + B^2 \), then find the values of \( a \) and \( b \).
Answer: Here, \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \) and \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \)
\( \therefore A + B = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} = \begin{bmatrix} 1+a & 0 \\ 2+b & -2 \end{bmatrix} \)
\( \Rightarrow (A + B)^2 = \begin{bmatrix} 1+a & 0 \\ 2+b & -2 \end{bmatrix} \begin{bmatrix} 1+a & 0 \\ 2+b & -2 \end{bmatrix} = \begin{bmatrix} (1+a)^2 & 0 \\ (2+b)(1+a) - 2(2+b) & 4 \end{bmatrix} = \begin{bmatrix} a^2+2a+1 & 0 \\ 2a-b+ab-2 & 4 \end{bmatrix} \)
Again \( A^2 + B^2 = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \)
\( = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} + \begin{bmatrix} a^2+b & a-1 \\ ab-b & b+1 \end{bmatrix} = \begin{bmatrix} a^2+b-1 & a-1 \\ ab-b & b \end{bmatrix} \)
Given, \( (A + B)^2 = A^2 + B^2 \)
\( \begin{bmatrix} a^2+2a+1 & 0 \\ 2a-b+ab-2 & 4 \end{bmatrix} = \begin{bmatrix} a^2+b-1 & a-1 \\ ab-b & b \end{bmatrix} \)
Equating the corresponding elements, we get
\( a^2 + 2a + 1 = a^2 + b - 1 \Rightarrow 2a - b = -2 \) ...(i)
\( a - 1 = 0 \Rightarrow a = 1 \) ...(ii)
\( 2a - b + ab - 2 = ab - b \Rightarrow 2a - 2 = 0 \) ...(iii)
\( b = 4 \) ...(iv)
\( a = 1, b = 4 \) satisfy all four equations (i), (ii), (iii) and (iv)
Hence, \( a = 1, b = 4 \).

Question. Let \( A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} \) and \( C = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \). Find a matrix \( D \) such that \( CD - AB = O \).
Answer: Since \( A, B, C \) are all square matrices of order 2, and \( CD - AB \) is well defined, \( D \) must be a square matrix of order 2.
Let \( D = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \). Then \( CD - AB = O \) gives
\( \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} = O \)
\( \begin{bmatrix} 2a+5c & 2b+5d \\ 3a+8c & 3b+8d \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
or \( \begin{bmatrix} 2a+5c-3 & 2b+5d \\ 3a+8c-43 & 3b+8d-22 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
By equating the corresponding elements of matrices, we get
\( 2a + 5c - 3 = 0 \) ...(i)
\( 3a + 8c - 43 = 0 \) ...(ii)
\( 2b + 5d = 0 \) ...(iii)
and \( 3b + 8d - 22 = 0 \) ...(iv)
Solving (i) and (ii), we get \( a = -191, c = 77 \) and solving (iii) and (iv), we get \( b = -110, d = 44 \).
Therefore \( D = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix} \).

Question. Express the following matrix as the sum of a symmetric and skew symmetric matrix, and verify your result. 
\( \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \)
Answer: Let \( A = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \)
\( A \) can be expressed as
\( A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') \) ...(i) [\(\because \frac{1}{2}(A + A') + \frac{1}{2}(A - A') = \frac{2A}{2} = A\)]
where, \( A + A' \) and \( A - A' \) are symmetric and skew symmetric matrices respectively.
Now, \( A + A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} \)
\( A - A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} = \begin{bmatrix} 0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0 \end{bmatrix} \)
Putting these values in (i), we get
\( A = \frac{1}{2} \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} + \frac{1}{2} \begin{bmatrix} 0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & -5/2 & -3/2 \\ 5/2 & 0 & -3 \\ 3/2 & 3 & 0 \end{bmatrix} \)
Verification:
\( \begin{bmatrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & -5/2 & -3/2 \\ 5/2 & 0 & -3 \\ 3/2 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 3+0 & \frac{1}{2}-\frac{5}{2} & -\frac{5}{2}-\frac{3}{2} \\ \frac{1}{2}+\frac{5}{2} & -2+0 & -2-3 \\ -\frac{5}{2}+\frac{3}{2} & -2+3 & 2+0 \end{bmatrix} = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} = A \)

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: We have, \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Let \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)
\( \therefore \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} 2a+c & 2b+d \\ 3a+2c & 3b+2d \end{bmatrix} \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \Rightarrow \begin{bmatrix} -6a-3c+10b+5d & 4a+2c-6b-3d \\ -9a-6c+15b+10d & 6a+4c-9b-6d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Equating the elements, we get:
\( -6a - 3c + 10b + 5d = 1 \) ...(i)
\( 4a + 2c - 6b - 3d = 0 \) ...(ii)
\( -9a - 6c + 15b + 10d = 0 \) ...(iii)
\( 6a + 4c - 9b - 6d = 1 \) ...(iv)
On adding equations (i) and (iv), we get \( c + b = 2-d+d \Rightarrow c + b = 2 \) ...(v)
On adding equations (ii) and (iii), we get \( -5a - 4c + 9b + 7d = 0 \) ...(vi)
On adding equations (vi) and (iv), we get \( a + 0 + 0 + d = 1 \Rightarrow d = 1 - a \) ...(vii)
From equations (v) and (vii), \( c + b = 2, d = 1-a \Rightarrow a + b + c = 3 \) ...(viii) \( \Rightarrow a = 3 - b - c \)
Now, using the values of \( a \) and \( d \) in equation (iii), we get \( -9(3 - b - c) - 6c + 15b + 10( -2 + b + c) = 0 \)
\( \Rightarrow -27 + 9b + 9c - 6c + 15b - 20 + 10b + 10c = 0 \Rightarrow 34b + 13c = 47 \) ...(ix)
Now, using the values of \( a \) and \( d \) in equation (ii), we get \( 4(3 - b - c) + 2c - 6b - 3(b + c - 2) = 0 \)
\( \Rightarrow 12 - 4b - 4c + 2c - 6b - 3b - 3c + 6 = 0 \Rightarrow -13b - 5c = -18 \) ...(x)
On multiplying equation (ix) by 5 and equation (x) by 13, then adding, we get \( b = 1 \)
\( \Rightarrow -13 \times 1 - 5c = -18 \Rightarrow c = 1 \)
\( \therefore a = 3 - 1 - 1 = 1 \) and \( d = 1 - 1 = 0 \)
\( \therefore A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \)

Question. If \( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \) and \( I \) is the identity matrix of order 2, then show that \( A^2 = 4A - 3I \). Hence find \( A^{-1} \). 
Answer: Here, \( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \)
\( \therefore A^2 = A.A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4+1 & -2-2 \\ -2-2 & 1+4 \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} \) ...(i)
Also, \( 4A - 3I = 4\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} \) ...(ii)
From (i) and (ii), we get \( A^2 = 4A - 3I \)
Pre-multiplying both sides by \( A^{-1} \)
\( A^{-1} . A^2 = A^{-1} . (4A - 3I) \Rightarrow (A^{-1} . A) . A = 4 A^{-1} . A - 3 A^{-1} . I \)
\( \Rightarrow IA = 4I - 3A^{-1} \Rightarrow A = 4I - 3A^{-1} \) [\(\because AA^{-1} = I, A^{-1} I = A^{-1}\)]
\( \Rightarrow 3A^{-1} = 4I - A \)
\( \Rightarrow A^{-1} = \frac{1}{3} \left( 4 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \right) = \frac{1}{3} \left( \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \right) = \frac{1}{3} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2/3 & 1/3 \\ 1/3 & 2/3 \end{bmatrix} \)

Question. Let \( A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \). Then show that \( A^2 - 4A + 7I = 0 \). Using this result calculate \( A^5 \). 
Answer: Here, \( A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \)
\( \Rightarrow A^2 = A \times A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} \)
Now, \( A^2 - 4A + 7I = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - 4 \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \text{ (zero matrix)} \)
\( \Rightarrow A^2 - 4A + 7I = 0 \Rightarrow A^2 = 4A - 7I \)
\( \Rightarrow A.A^2 = 4A.A - 7A.I \) [Pre multiplying by \( A \)]
\( \Rightarrow A^3 = 4A^2 - 7A \) [\( AI = A \)]
\( \Rightarrow A^3 = 4(4A - 7I) - 7A \) [Putting the value of \( A^2 \)]
\( \Rightarrow A^3 = 16A - 28I - 7A \Rightarrow A^3 = 9A - 28I \)
\( \Rightarrow A.A^3 = 9A.A - 28A.I \) [Pre multiplying by \( A \)]
\( \Rightarrow A^4 = 9A^2 - 28A \)
\( \Rightarrow A^4 = 9(4A - 7I) - 28A = 8A - 63I \)
\( \Rightarrow A.A^4 = 8A^2 - 63A \)
\( \Rightarrow A^5 = 8(4A - 7I) - 63A = -31A - 56I \)
\( = -31 \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} - 56 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -118 & -93 \\ 31 & -118 \end{bmatrix} \)

Question. Prove that every square matrix can be uniquely expressed as the sum of a symmetric and skew-symmetric matrix.
Answer: Let \( A \) be any square matrix. Then,
\( A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T) = P + Q \), (say),
where, \( P = \frac{1}{2}(A + A^T), Q = \frac{1}{2}(A - A^T) \)
Now, \( P^T = \left( \frac{1}{2}(A + A^T) \right)^T = \frac{1}{2}(A^T + (A^T)^T) = \frac{1}{2}(A^T + A) = P \)
\( \therefore P \) is symmetric matrix.
Also, \( Q^T = \left( \frac{1}{2}(A - A^T) \right)^T = \frac{1}{2}(A^T - (A^T)^T) = \frac{1}{2}(A^T - A) = - \frac{1}{2}(A - A^T) = -Q \)
\( \therefore Q \) is skew-symmetric matrix.
Thus, \( A = P + Q \), where \( P \) is a symmetric matrix and \( Q \) is a skew-symmetric matrix. Hence, \( A \) is expressible as the sum of a symmetric and a skew-symmetric matrix.
Uniqueness: If possible, let \( A = R + S \), where \( R \) is symmetric and \( S \) is skew-symmetric, then,
\( A^T = (R + S)^T = R^T + S^T \)
\( \Rightarrow A^T = R - S \) [\(\because R^T = R \text{ and } S^T = -S\)]
Now, \( A = R + S \) and \( A^T = R - S \)
\( \Rightarrow R = \frac{1}{2}(A + A^T) = P, S = \frac{1}{2}(A - A^T) = Q \)
Hence, \( A \) is uniquely expressible as the sum of a symmetric and a skew-symmetric matrix.

Question. Choose and write the correct option in each of the following questions.
If \( A = \begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix} \), then \( A^n \) is equal to
(a) \( \begin{bmatrix} a^n & 0 & 0 \\ 0 & a^n & 0 \\ 0 & 0 & a^n \end{bmatrix} \)
(b) \( \begin{bmatrix} a^n & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix} \)
(c) \( \begin{bmatrix} a & 0 & 0 \\ 0 & a^n & 0 \\ 0 & 0 & a \end{bmatrix} \)
(d) \( \begin{bmatrix} na & 0 & 0 \\ 0 & na & 0 \\ 0 & 0 & na \end{bmatrix} \)
Answer: (a)

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

Question. The number of all possible matrices of order \( 3 \times 3 \) with each entry 0 or 1 is 
(a) 27
(b) 18
(c) 81
(d) 512
Answer: (d)

Question. The \( A = \begin{bmatrix} 2 & -3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix} \), \( X = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \) and \( y = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} \), then \( AB + XY \) equals
(a) [28]
(b) [24]
(c) 28
(d) 24
Answer: (c)

Question. The matrix \( A = \begin{bmatrix} 0 & 0 & 5 \\ 0 & 5 & 0 \\ 5 & 0 & 0 \end{bmatrix} \) 
(a) scalar matrix
(b) diagonal matrix
(c) unit matrix
(d) square matrix
Answer: (d)

Question. If \( A = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} \), \( n \in N \), then \( A^{4n} \) equals 
(a) \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
(b) \( \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} \)
(c) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
(d) \( \begin{bmatrix} 0 & 0 \\ i & i \end{bmatrix} \)
Answer: (c)

Question. For any two matrices \( A \) and \( B \), we have
(a) \( AB = BA \)
(b) \( AB \neq BA \)
(c) \( AB = O \)
(d) None of these
Answer: (b)

Question. On using elementary row operation \( R_1 \rightarrow R_1 - 3R_2 \) in the following matrix equation: \( \begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \), we have:
(a) \( \begin{bmatrix} -5 & -7 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -7 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \)
(b) \( \begin{bmatrix} -5 & -7 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} -1 & -3 \\ 1 & 1 \end{bmatrix} \)
(c) \( \begin{bmatrix} -5 & -7 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & -7 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \)
(d) \( \begin{bmatrix} 4 & 2 \\ -5 & -7 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -3 & -3 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \)
Answer: (a)