CBSE Class 12 Mathematics Matrices Worksheet Set 01

Very Short Answer Questions [1 mark]

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

Question. If matrix \( A = [1 \quad 2 \quad 3] \), then write \( AA' \), where \( A' \) is the transpose of matrix A. [CBSE Delhi 2009]
Answer: Given \( A = [1 \quad 2 \quad 3] \)
\( \implies \) \( A' = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \)
Hence, \( AA' = (1 \times 1 + 2 \times 2 + 3 \times 3) = (14) \)

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
\( \therefore A' = -A \)
\( \implies \) \( \begin{bmatrix} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -a & -3 \\ -2 & -b & +1 \\ -c & -1 & 0 \end{bmatrix} \)
By equating corresponding elements, we get
\( a = -2, c = -3 \) and \( b = -b \)
\( \implies \) \( b = 0 \)
\( \therefore a = -2, b = 0 \) 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
\( \implies \) \( \cos \alpha = 1, \sin \alpha = 0 \)
\( \implies \) \( \alpha = 0 \)

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: \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} (1)(3) + (2)(2) & (1)(1) + (2)(5) \\ (3)(3) + (4)(2) & (3)(1) + (4)(5) \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} 7 & 11 \\ 17 & 23 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \)
Equating the corresponding elements, we get
\( k = 17 \)

Question. Write a square matrix of order 2, which is both symmetric and skew symmetric.
Answer: Square matrix of order 2, which is both symmetric and skew symmetric is
\( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

Question. From the following matrix equation, find the value of x :
\( \begin{bmatrix} x + y & 4 \\ -5 & 3y \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ -5 & 6 \end{bmatrix} \)
Answer: Given matrix equation
\( \begin{bmatrix} x + y & 4 \\ -5 & 3y \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ -5 & 6 \end{bmatrix} \)
Equating the corresponding elements, we get,
\( x + y = 3 \) and \( 3y = 6 \)
i.e., \( y = 2 \) and \( x = 1 \)
\( \therefore x = 1, y = 2 \).

Question. Write the order of the product matrix.
\( \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix} \)
Answer: Order is \( 3 \times 3 \) because it is product of two matrices having order \( 3 \times 1 \) and \( 1 \times 3 \).

Question. For a \( 2 \times 2 \) matrix, \( A = [a_{ij}] \), whose elements are given by \( a_{ij} = \frac{i}{j} \), write the value of \( a_{12} \).
Answer: \( \because a_{ij} = \frac{i}{j} \)
\( \implies \) \( a_{12} = \frac{1}{2} \) [Here \( i = 1 \) and \( j = 2 \)]

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 \( A^T = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} \), then find \( A^T - B^T \).
Answer: Given: \( B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} \)
\( \implies \) \( B^T = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} \)
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 \( A = -A' \).
\( \implies \) \( \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & -1 & x \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -x \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix} \)
Equating the corresponding elements, we get \( x = 2 \).

Question. If A 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 is a square matrix and \( |A| = 2 \), then write the value of \( |AA'| \), where \( A' \) is the transpose of matrix A.
Answer: \( |AA'| = |A| \cdot |A'| = |A| \cdot |A| = |A|^2 = 2^2 = 4 \).
[Note: \( |AB| = |A| \cdot |B| \) and \( |A| = |A^T| \), where A and B are square matrices.]

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 \( 3 \times 4 \).

Short Answer Questions [2 marks]

Question. If \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} \), then write the value of \( x \). 
Answer: Given \( \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} 2 \times 1 + 3 \times (-2) & 2 \times (-3) + 3 \times 4 \\ 5 \times 1 + 7 \times (-2) & 5 \times (-3) + 7 \times 4 \end{bmatrix} = \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} -4 & 6 \\ -9 & 13 \end{bmatrix} = \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} \)
Equating the corresponding elements, we get \( x = 13 \).

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, \( 2 \begin{bmatrix} x & 5 \\ 7 & y - 3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} 2x & 10 \\ 14 & 2y - 6 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} 2x + 3 & 6 \\ 15 & 2y - 4 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
Equating the corresponding elements, we get
\( 2x + 3 = 7 \) and \( 2y - 4 = 14 \)
\( \implies \) \( x = \frac{7 - 3}{2} \) and \( y = \frac{14 + 4}{2} \)
\( \implies \) \( x = 2 \) and \( y = 9 \quad \therefore x + y = 2 + 9 = 11 \)

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

Question. Show that \( A'A \) and \( AA' \) are both symmetric matrices for any matrix A. 
Answer: Let \( P = A'A \)
\( P' = (A'A)' \)
\( = A'(A')' = A'A = P \quad [\because (AB)' = B'A'] \)
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. If matrix \( A = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \) and \( A^2 = \lambda A \), then write the value of \( \lambda \). 
Answer: Here, \( A = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \)
Given, \( A^2 = \lambda A \)
\( \implies \) \( \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} = \lambda \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} 18 & -18 \\ -18 & 18 \end{bmatrix} = \lambda \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \)
\( \implies \) \( 6 \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} = \lambda \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \)
\( \implies \) \( \lambda = 6 \)

Question. If A is a square matrix such that \( A^2 = I \), then find the simplified value of \( (A - I)^3 + (A + I)^3 - 7A \).
Answer: We have \( A^2 = I \)
Now, \( (A - I)^3 + (A + I)^3 - 7A = A^3 - 3A^2I + 3AI^2 - I^3 + A^3 + 3A^2I + 3AI^2 + I^3 - 7A \)
\( = 2A^3 + 6AI^2 - 7A \)
\( = 2A^3 + 6AI - 7A \quad [\because I^2 = I] \)
\( = 2A^2 \cdot A + 6A - 7A \quad [\because AI = A] \)
\( = 2I \cdot A + 6A - 7A \quad [\because A^2 = I] \)
\( = 2A + 6A - 7A = A \quad [\because IA = A] \)

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 \( b \). 
Answer: We have \( A = \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix} \)
\( \because \) A is symmetric matrix.
\( \implies \) \( A^T = A \)
\( \implies \) \( \begin{bmatrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix} \)
Equating the corresponding elements, we get
\( 2b = 3 \) and \( 3a = -2 \)
\( \implies \) \( b = \frac{3}{2} \) and \( a = -\frac{2}{3} \)

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 \( B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix} \)
\( \therefore BA = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}_{3 \times 2} \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix}_{2 \times 3} \)
\( [b_{ij}] = \begin{bmatrix} 2 - 12 & -4 + 6 & 6 + 15 \\ 4 - 20 & -8 + 10 & 12 + 25 \\ 2 - 4 & -4 + 2 & 6 + 5 \end{bmatrix}_{3 \times 3} \)
\( \implies \) \( [b_{ij}] = \begin{bmatrix} -10 & 2 & 21 \\ -16 & 2 & 37 \\ -2 & -2 & 11 \end{bmatrix}_{3 \times 3} \)
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: \( \implies \) \( \begin{bmatrix} 2x & 10 \\ 14 & 2y - 6 \end{bmatrix} + \begin{bmatrix} -3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} 2x - 3 & 10 - 4 \\ 14 + 1 & 2y - 6 + 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} 2x - 3 & 6 \\ 15 & 2y - 4 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
We know that two matrices of same order are equal if the corresponding entries are equal.
i.e., \( 2x - 3 = 7 \)
\( \implies \) \( 2x = 10 \)
\( \implies \) \( x = 5 \)
and \( 2y - 4 = 14 \)
\( \implies \) \( 2y = 18 \)
\( \implies \) \( y = 9 \)
\( \therefore x - y = 5 - 9 = -4 \)

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 \( (AB - BA)^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T \quad (\because (AB)^T = B^T A^T) \)
\( = BA - AB = -(AB - BA) \)
\( \implies \) \( AB - BA \) 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 \( A = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}, A^T = \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix} \)
\( \implies \) \( A + A^T = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} + \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 8 & 14 \end{bmatrix} \)
\( \implies \) \( (A + A^T)^T = \begin{bmatrix} 4 & 8 \\ 8 & 14 \end{bmatrix}^T = \begin{bmatrix} 4 & 8 \\ 8 & 14 \end{bmatrix} = A + A^T \)
Hence, \( A + A^T \) is a symmetric matrix.

Long Answer Questions-I [4 marks]

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

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 \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \)
\( \therefore A + B = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} = \begin{bmatrix} 1 + a & 0 \\ 2 + b & -2 \end{bmatrix} \)
\( \implies \) \( (A + B)^2 = \begin{bmatrix} 1 + a & 0 \\ 2 + b & -2 \end{bmatrix} \begin{bmatrix} 1 + a & 0 \\ 2 + b & -2 \end{bmatrix} = \begin{bmatrix} 1 + a^2 + 2a & 0 \\ 2 + 2a + b + ab - 4 - 2b & -2 \end{bmatrix} = \begin{bmatrix} a^2 + 2a + 1 & 0 \\ 2a - b + ab - 2 & 4 \end{bmatrix} \)
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
\( a^2 + 2a + 1 = a^2 + b - 1 \)
\( \implies \) \( 2a - b = -2 \) ...(i)
\( a - 1 = 0 \)
\( \implies \) \( a = 1 \) ...(ii)
\( 2a - b + ab - 2 = ab - b \)
\( \implies \) \( 2a - 2 = 0 \) ...(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 \( CD - AB = 0 \) gives
\( \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} = O \)
or \( \begin{bmatrix} 2a + 5c & 2b + 5d \\ 3a + 8c & 3b + 8d \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \) 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 \( D = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix} \)

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
\( A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A'), \quad ...(i) \quad \left[ \because \frac{1}{2}(A + A') + \frac{1}{2}(A - A') = \frac{2A}{2} = A \right] \)
where, \( A + A' \) and \( A - A' \) are symmetric and skew symmetric matrices respectively.
Now, \( A + A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} \)
\( A - A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} = \begin{bmatrix} 0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0 \end{bmatrix} \)
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, \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}_{2 \times 2} A \cdot \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}_{2 \times 2} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}_{2 \times 2} \)
Let \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}_{2 \times 2} \)
\( \therefore \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} 2a + c & 2b + d \\ 3a + 2c & 3b + 2d \end{bmatrix} \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \implies \) \( \begin{bmatrix} -6a - 3c + 10b + 5d & 4a + 2c - 6b - 3d \\ -9a - 6c + 15b + 10d & 6a + 4c - 9b - 6d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \implies \) \( -6a - 3c + 10b + 5d = 1 \) ...(i)
\( \implies \) \( 4a + 2c - 6b - 3d = 0 \) ...(ii)
\( \implies \) \( -9a - 6c + 15b + 10d = 0 \) ...(iii)
\( \implies \) \( 6a + 4c - 9b - 6d = 1 \) ...(iv)
On adding equations (i) and (iv), we get
\( c + b - d = 2 \)
\( \implies \) \( d = c + b - 2 \) ...(v)
On adding equations (ii) and (iii), we get
\( -5a - 4c + 9b + 7d = 0 \) ...(vi)
On adding equations (vi) and (iv), we get
\( a + 0 + 0 + d = 1 \)
\( \implies \) \( d = 1 - a \) ...(vii)
From equations (v) and (vii),
\( c + b - 2 = 1 - a \)
\( \implies \) \( a + b + c = 3 \) ...(viii)
\( \implies \) \( a = 3 - b - c \)
Now, using the values of a and d in equation (iii), we get
\( -9(3 - b - c) - 6c + 15b + 10(-2 + b + c) = 0 \)
\( \implies \) \( -27 + 9b + 9c - 6c + 15b - 20 + 10b + 10c = 0 \)
\( \implies \) \( 34b + 13c = 47 \) ...(ix)
Now, using the values of a and d in equation (ii), we get
\( 4(3 - b - c) + 2c - 6b - 3(b + c - 2) = 0 \)
\( \implies \) \( 12 - 4b - 4c + 2c - 6b - 3b - 3c + 6 = 0 \)
\( \implies \) \( -13b - 5c = -18 \) ...(x)
On multiplying equation (ix) by 5 and equation (x) by 13, then adding, we get
\[ \begin{array}{r@{\quad}l} & -169b - 65c = -234 \\ & 170b + 65c = 235 \\ \hline & b = 1 \end{array} \]
\( \implies \) \( -13 \times 1 - 5c = -18 \) [From equation (x)]
\( \implies \) \( -5c = -18 + 13 = -5 \)
\( \implies \) \( c = 1 \)
\( \therefore a = 3 - 1 - 1 = 1 \) 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, \( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \)
\( \therefore A^2 = A \cdot A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4 + 1 & -2 - 2 \\ -2 - 2 & 1 + 4 \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix} \) ...(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 \( A^{-1} \)
\( A^{-1} \cdot A^2 = A^{-1} \cdot (4A - 3I) \)
\( \implies \) \( (A^{-1} \cdot A) \cdot A = 4A^{-1} \cdot A - 3A^{-1} \cdot I \)
\( \implies \) \( IA = 4I - 3A^{-1} \)
\( \implies \) \( A = 4I - 3A^{-1} \quad [\because AA^{-1} = I, A^{-1}I = A^{-1}] \)
\( \implies \) \( 3A^{-1} = 4I - A \)
\( \implies \) \( A^{-1} = \frac{1}{3} \left( 4 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \right) = \frac{1}{3} \left( \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \right) \quad \)
\( \implies \) \( \quad \frac{1}{3} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2/3 & 1/3 \\ 1/3 & 2/3 \end{bmatrix} \)

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, \( A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \)
\( \implies \) \( A^2 = A \times A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} \)
Now, \( A^2 - 4A + 7I = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - 4 \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \) (zero matrix)
\( \implies \) \( A^2 - 4A + 7I = 0 \)
\( \implies \) \( A^2 = 4A - 7I \)
\( \implies \) \( A \cdot A^2 = 4A \cdot A - 7A \cdot I \) [Pre multiplying by A]
\( \implies \) \( A^3 = 4A^2 - 7A \) [\( AI = A \)]
\( \implies \) \( A^3 = 4(4A - 7I) - 7A \) [Putting the value of \( A^2 \)]
\( \implies \) \( A^3 = 16A - 28I - 7A \)
\( \implies \) \( A^3 = 9A - 28I \)
\( \implies \) \( A \cdot A^3 = 9A \cdot A - 28A \cdot I \) [Pre multiplying by A]
\( \implies \) \( A^4 = 9A^2 - 28A \)
\( \implies \) \( A^4 = 9(4A - 7I) - 28A \) [Putting the value of \( A^2 \)]
\( \implies \) \( A^4 = 8A - 63I \)
\( \implies \) \( A \cdot A^4 = 8A^2 - 63A \) [Pre multiplying by A]
\( \implies \) \( A^5 = 8(4A - 7I) - 63A = -31A - 56I \)
\( = -31 \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} - 56 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -118 & -93 \\ 31 & -118 \end{bmatrix} \)

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, \( P^T = \left( \frac{1}{2}(A + A^T) \right)^T \quad [\because (KT)^T = K \cdot A^T] \)
\( = \frac{1}{2}[A^T + (A^T)^T] \quad [\because (A + B)^T = A^T + B^T] \)
\( = \frac{1}{2}(A^T + A) \quad [\because (A^T)^T = A] \)
\( = \frac{1}{2}(A + A^T) = P \)
\( \therefore P \) is symmetric matrix.
Also, \( Q^T = \frac{1}{2}(A - A^T)^T = \frac{1}{2}[A^T - (A^T)^T] = \frac{1}{2}[A^T - A] = -\frac{1}{2}[A - A^T] = -Q \)
\( \therefore Q \) 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,
\( A^T = (R + S)^T = R^T + S^T \)
\( \implies \) \( A^T = R - S \quad [\because R^T = R \text{ and } S^T = -S] \)
Now, \( A = R + S \) and \( A^T = R - S \)
\( \implies \) \( R = \frac{1}{2}[A + A^T] = P, \ S = \frac{1}{2}(A - A^T) = Q \)
Hence, A is uniquely expressible as the sum of a symmetric and a skew-symmetric matrix.