CBSE Class 12 Mathematics Matrices Class Test Set 01

This Class 12 CBSE Mathematics class test on Matrices has 11 questions of 1, 2 and 3 marks with complete solutions. It covers equality of matrices, matrix operations, transpose, symmetric and skew-symmetric matrices and the inverse of a 2 × 2 matrix.

Class Test Questions with Solutions

Question 1. If matrix \( A = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \) and \( A^2 = \lambda A \), then write the value of λ. [1]
Answer:
Given \( A = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \)
and \( A^2 = \lambda A \) ...(1)
∵ \( A^2 = A \cdot A = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \cdot \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 9 + 9 & -9 - 9 \\ -9 - 9 & 9 + 9 \end{bmatrix} \)
\( = \begin{bmatrix} 18 & -18 \\ -18 & 18 \end{bmatrix} \)
Put the value of \( A^2 \) in equation (1):
\( \begin{bmatrix} 18 & -18 \\ -18 & 18 \end{bmatrix} = \lambda \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \)
\( 6\begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} = \lambda \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \)
Solving, we get λ = 6

Question 2. Find the value of x + y from the following equation: \( 2\begin{bmatrix} x & 5 \\ 7 & y - 3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \) [1]
Answer:
\( 2\begin{bmatrix} x & 5 \\ 7 & y - 3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
\( \begin{bmatrix} 2x & 10 \\ 14 & 2y - 6 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
Solving, we get
2x + 3 = 7 ⇒ x = 2
and 2y − 6 + 2 = 14 ⇒ y = 9
then x + y = 2 + 9 = 11

Question 3. Find x, y, z and w if \( \begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \). [2]
Answer:
Given \( \begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \).
By definition of equality, we have
x − y = −1 ...(i)    2x − y = 0 ...(ii)
2x + z = 5 ...(iii)    3z + w = 13 ...(iv)
To find x and y, subtract (ii) from (i) to get
x − 2x = −1 ⇒ x = 1 and
then from (i), y = x + 1 = 1 + 1 = 2.
To find z and w, substitute x = 1 in (iii) to obtain z = 5 − 2x = 5 − 2 = 3 and
then from (iv), w = 13 − 3z = 13 − 3 × 3 = 4.
Hence, x = 1, y = 2, z = 3 and w = 4.

Question 4. Given \( A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} \), compute \( A^{-1} \) and show that \( 2A^{-1} = 9I - A \). [2]
Answer:
Given, \( A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} \)
|A| = 14 − (12) = 2 ≠ 0
Hence, A is invertible.
\( A^{-1} = \frac{\text{adj}(A)}{|A|}, \ |A| \neq 0 \)

\( \text{adj}\,A = \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} \) [½]
Hence, \( A^{-1} = \frac{\text{adj}(A)}{|A|} \)
\( = \frac{1}{2}\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} \frac{7}{2} & \frac{3}{2} \\ 2 & 1 \end{bmatrix} \) [½]
Now RHS = 9I − A
\( = 9\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} \)
\( = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} \)
\( = \begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} \)
\( = 2\begin{bmatrix} \frac{7}{2} & \frac{3}{2} \\ 2 & 1 \end{bmatrix} = 2A^{-1} \) = LHS. Hence proved. [1]

Question 5. Construct a matrix \( A = [a_{ij}] \) of order 2 × 3, in which elements are \( a_{ij} = \frac{1}{2}|2i - 3j| \). [2]
Answer:
In general a 2 × 3 matrix is given by
\( A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix} \)
Now \( a_{ij} = \frac{1}{2}|2i - 3j| \), i = 1, 2; j = 1, 2, 3
\( a_{11} = \frac{1}{2}|2 - 3| = \frac{1}{2} \)
\( a_{12} = \frac{1}{2}|2 - 6| = 2 \)
\( a_{13} = \frac{1}{2}|2 - 9| = \frac{7}{2} \)
\( a_{21} = \frac{1}{2}|4 - 3| = \frac{1}{2} \)
\( a_{22} = \frac{1}{2}|4 - 6| = 1 \)
\( a_{23} = \frac{1}{2}|4 - 9| = \frac{5}{2} \)
Hence the required matrix is given by
\( A = \begin{bmatrix} 1/2 & 2 & 7/2 \\ 1/2 & 1 & 5/2 \end{bmatrix} \)

Question 6. If \( \begin{bmatrix} a + b & 2 \\ 7 & ab \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 7 & 8 \\ -3 & 4 \end{bmatrix} \), then find the value of a and b. [2]
Answer:
Given \( \begin{bmatrix} a + b & 2 \\ 7 & ab \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 7 & 8 \\ -3 & 4 \end{bmatrix} \)
a + b = 6 ...(i)
ab = 8 ⇒ \( b = \frac{8}{a} \) ...(ii)
Put the value of b in equation (i):
\( a + \frac{8}{a} = 6 \)
\( a^2 - 6a + 8 = 0 \)
Solving, we get a = 2, 4
Then put the value of a in equation (ii):
a = 4, b = 2 or a = 2, b = 4

Question 7. If \( A = \begin{bmatrix} 4 & 2 & -5 \\ 1 & 0 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 6 & -7 & 0 \\ -1 & 2 & 5 \\ 1 & 0 & 2 \end{bmatrix} \), then prove that \( (AB)^T = B^TA^T \). [3]
Answer:
L.H.S.
Here \( AB = \begin{bmatrix} 4 & 2 & -5 \\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 6 & -7 & 0 \\ -1 & 2 & 5 \\ 1 & 0 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 24 - 2 - 5 & -28 + 4 - 0 & 0 + 10 - 10 \\ 6 + 0 + 3 & -7 + 0 + 0 & 0 + 0 + 6 \end{bmatrix} \)
\( = \begin{bmatrix} 17 & -24 & 0 \\ 9 & -7 & 6 \end{bmatrix} \)
⇒ \( (AB)^T = \begin{bmatrix} 17 & 9 \\ -24 & -7 \\ 0 & 6 \end{bmatrix} \) ...(1)
R.H.S.
Now \( B^TA^T = \begin{bmatrix} 6 & -1 & 1 \\ -7 & 2 & 0 \\ 0 & 5 & 2 \end{bmatrix}\begin{bmatrix} 4 & 1 \\ 2 & 0 \\ -5 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 24 - 2 - 5 & 6 + 0 + 3 \\ -28 + 4 + 0 & -7 + 0 + 0 \\ 0 + 10 - 10 & 0 + 0 + 6 \end{bmatrix} \)
\( = \begin{bmatrix} 17 & 9 \\ -24 & -7 \\ 0 & 6 \end{bmatrix} \) ...(2)
From (1) and (2), we have \( (AB)^T = B^TA^T \)
L.H.S. = R.H.S. Hence proved.

Question 8. If \( A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \) and \( f(A) = A^2 - 5A + 7I \), then find f(A). [3]
Answer:
Given \( A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \)
\( A^2 = A \cdot A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} \)
\( = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \)
\( f(A) = A^2 - 5A + 7I \)
\( = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + 7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

Question 9. Express the matrix \( B = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \) as the sum of a symmetric and a skew-symmetric matrix. [3]
Answer:
Given \( B = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \)
Here \( B^T = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix} \)
Let \( P = \frac{1}{2}(B + B^T) = \frac{1}{2}\begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix} = \begin{bmatrix} 2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix} \)
Now \( P' = \begin{bmatrix} 2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix} = P \)
Also let \( Q = \frac{1}{2}(B - B^T) = \begin{bmatrix} 0 & -\frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix} \)
\( Q^T = \begin{bmatrix} 0 & \frac{1}{2} & \frac{5}{2} \\ -\frac{1}{2} & 0 & -3 \\ -\frac{5}{2} & 3 & 0 \end{bmatrix} = -Q \)
∴ \( Q = \frac{1}{2}(B - B^T) \) is a skew-symmetric matrix.
Now \( P + Q = \begin{bmatrix} 2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix} + \begin{bmatrix} 0 & -\frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = B \)

Question 10. Find the value of x, if \( \begin{bmatrix} 1 & x & 1 \end{bmatrix}\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = O \). [3]
Answer:
We have \( \begin{bmatrix} 1 & x & 1 \end{bmatrix}\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = O \)
⇒ \( \begin{bmatrix} 1 + 2x + 15 & 3 + 5x + 3 & 2 + x + 2 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = O \)
⇒ \( \begin{bmatrix} 16 + 2x & 5x + 6 & x + 4 \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} = O \)
⇒ \( [16 + 2x + 10x + 12 + x^2 + 4x] = [0] \)
⇒ \( x^2 + 16x + 28 = 0 \)
Solving, we get
x = −2, −14

Question 11. Let \( A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} \), \( C = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \). Find a matrix D such that CD − AB = O. [3]
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
\( \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix} = O \)
\( \begin{bmatrix} 2a + 5c & 2b + 5d \\ 3a + 8c & 3b + 8d \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 43 & 22 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \begin{bmatrix} 2a + 5c - 3 & 2b + 5d \\ 3a + 8c - 43 & 3b + 8d - 22 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
By equality of matrices we get
2a + 5c − 3 = 0 ...(1)
3a + 8c − 43 = 0 ...(2)
2b + 5d = 0 ...(3)
3b + 8d − 22 = 0 ...(4)
Solving (1) and (2), we get a = −191, c = 77
and solving (3) and (4), we get b = −110, d = 44
Therefore \( D = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix} \)