CBSE Class 12 Mathematics Matrices Previous Year Questions Set 01

These previous year CBSE board questions on Matrices for Class 12 Mathematics come with complete step-by-step solutions and marking. They are grouped by marks, from 1-mark very short answers to 6-mark long answers.

Very Short Answer Questions (1 Mark)

Question 1. Find the value of x + y from the following equation: \( 2\begin{bmatrix} x & 5 \\ 7 & y - 3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \) [CBSE 2012, 1M]
Answer:
Given \( 2\begin{bmatrix} x & 5 \\ 7 & y - 3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix} \)
2x + 3 = 7 ⇒ x = 2
and 2(y − 3) + 2 = 14
⇒ y − 3 + 1 = 7 ⇒ y = 9
x + y = 2 + 9 = 11 [1]

Question 2. 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? [CBSE 2013, 1M]
Answer:
x = 2   [∵ A′ = −A] [1]

Question 3. If matrix \( A = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \) and \( A^2 = \lambda A \), then write the value of λ. [CBSE 2013, 1M]
Answer:
\( A^2 = \lambda A \)
\( A \cdot A = \lambda A \)
\( \begin{bmatrix} 18 & -18 \\ -18 & 18 \end{bmatrix} = \lambda \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} \)
18 = 3λ ⇒ λ = 6 [1]

Question 4. If A is a square matrix such that \( A^2 = A \), then write the value of \( 7A - (I + A)^3 \), where I is an identity matrix. [CBSE 2014, 1M]
Answer:
\( 7A - (I + A)^3 \)
\( = 7A - [(I + A)^2(I + A)] = 7A - [I^2 + A^2 + 2AI][I + A] \)
\( = 7A - [I + A + 2A][I + A] \)
\( = 7A - [I + 3A][I + A] \)
\( = 7A - [I \cdot I + IA + 3AI + 3A^2] \)
\( = 7A - [I + A + 3A + 3A] \)   (∵ \( A^2 = A \))
\( = 7A - [I + 7A] \)
= −I [1]

Question 5. If \( A = \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix} \), find α satisfying \( 0 \lt \alpha \lt \frac{\pi}{2} \) when \( A + A^T = \sqrt{2}I_2 \); where \( A^T \) is transpose of A. [CBSE 2016, 1M]
Answer:
\( A + A^T = \sqrt{2}I_2 \)
⇒ \( \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix} + \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix} = \sqrt{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) [½]
⇒ \( \begin{pmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{pmatrix} = \begin{pmatrix} \sqrt{2} & 0 \\ 0 & \sqrt{2} \end{pmatrix} \)
∵ \( 2\cos\alpha = \sqrt{2} \)
\( \cos\alpha = \frac{1}{\sqrt{2}} \)
\( \alpha = \frac{\pi}{4} \in \left(0, \frac{\pi}{2}\right) \) [½]

Question 6. 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'. [CBSE 2018, 1M]
Answer:
Matrix \( A = \begin{bmatrix} 0 & a & -3 \\ 2 & 0 & -1 \\ b & 1 & 0 \end{bmatrix} \)
\( A' = \begin{bmatrix} 0 & 2 & b \\ a & 0 & 1 \\ -3 & -1 & 0 \end{bmatrix} \)
If A is a skew-symmetric matrix, then
⇒ A = −A′ or a = −2, b = 3 [1]

Question 7. Given a skew-symmetric matrix \( A = \begin{bmatrix} 0 & a & 1 \\ -1 & b & 1 \\ -1 & c & 0 \end{bmatrix} \), the value of \( (a + b + c)^2 \) is ............... . [CBSE 2020, 1M]
Answer:
\( A = \begin{bmatrix} 0 & a & 1 \\ -1 & b & 1 \\ -1 & c & 0 \end{bmatrix} \)
∵ A is a skew-symmetric matrix, i.e. \( A^T = -A \)
so, a = 1, b = 0 and c = −1
Now \( (a + b + c)^2 = (1 + 0 - 1)^2 \)
\( (a + b + c)^2 = 0 \) [1]

Short Answer Questions (2 Marks)

Question 8. If \( A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \), show that (A − 2I)(A − 3I) = O. [CBSE 2019, 2M]
Answer:
\( A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \)
\( A - 2I = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -1 & -1 \end{bmatrix} \)
\( A - 3I = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} \) [1]
\( (A - 2I)(A - 3I) = \begin{bmatrix} 2 & 2 \\ -1 & -1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 2 - 2 & 4 - 4 \\ -1 + 1 & -2 + 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \) [1]

Question 9. 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 \( A^2 + I = kA \). [CBSE 2020, 2M]
Answer:
Given \( A = \begin{bmatrix} -3 & 2 \\ 1 & -1 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
∵ \( A^2 + I = kA \)
\( \begin{bmatrix} -3 & 2 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} -3 & 2 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = k\begin{bmatrix} -3 & 2 \\ 1 & -1 \end{bmatrix} \)
\( \begin{bmatrix} 11 & -8 \\ -4 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -3k & 2k \\ k & -k \end{bmatrix} \) [1]
\( \begin{bmatrix} 12 & -8 \\ -4 & 4 \end{bmatrix} = \begin{bmatrix} -3k & 2k \\ k & -k \end{bmatrix} \)
k = −4   (on comparing both sides) [1]

Long Answer Questions I (4 Marks)

Question 10. Express the following matrix as the sum of a symmetric and a skew-symmetric matrix, and verify your result: \( \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \) [CBSE 2010, 4M]
Answer:
Let \( A = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \)
\( A' = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \)
Now \( \frac{1}{2}(A + A') = \frac{1}{2}\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} = P \) (say) [1]
and \( \frac{1}{2}(A - A') = \frac{1}{2}\begin{bmatrix} 0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix} = Q \) (say) [1]
Here \( P' = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} = P \)
and \( Q' = \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix} = -Q \)
So P is symmetric and Q is skew-symmetric matrix. [1]
∵ \( P + Q = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0 \end{bmatrix} = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} = A \)
i.e. A = P + Q [1]

Question 11. For the following matrices A and B, verify that (AB)′ = B′A′.
\( A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \), B = [−1  2  1] [CBSE 2010, 4M]
Answer:
\( AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}\begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix} \)
\( (AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \) ...(i) [2]
\( B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \), A′ = [1  −4  3]
\( B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \) ...(ii)
By (i) and (ii),
(AB)′ = B′A′ [2]

Question 12. There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women and 4 children in family B. The recommended daily amount of calories is 2400 for men, 1900 for women, 1800 for children and 45 grams of proteins for men, 55 grams for women and 33 grams for children. Represent the above information using matrices. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the 2 families. [CBSE 2015, 4M]
Answer:
The numbers of members of the two families can be represented by the 2 × 3 matrix
\( F = \begin{array}{cc} & \begin{matrix} \;\,M & \;\,W & \;\,C \end{matrix} \\ \begin{matrix} A \\ B \end{matrix} & \begin{bmatrix} 4 & 6 & 2 \\ 2 & 2 & 4 \end{bmatrix} \end{array} \) [1]
and the recommended daily allowance of calories and proteins for each member can be represented by the 3 × 2 matrix
\( R = \begin{array}{cc} & \begin{matrix} \text{Calories} & \text{Proteins} \end{matrix} \\ \begin{matrix} M \\ W \\ C \end{matrix} & \begin{bmatrix} 2400 & 45 \\ 1900 & 55 \\ 1800 & 33 \end{bmatrix} \end{array} \) [1]
The total requirement of calories and proteins for each of the two families is given by the matrix multiplication:
\( FR = \begin{bmatrix} 4 & 6 & 2 \\ 2 & 2 & 4 \end{bmatrix}\begin{bmatrix} 2400 & 45 \\ 1900 & 55 \\ 1800 & 33 \end{bmatrix} = \begin{matrix} A \\ B \end{matrix} \begin{bmatrix} 24600 & 576 \\ 15800 & 332 \end{bmatrix} \) [2]

Question 13. If \( A = \begin{bmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0 \end{bmatrix} \), \( B = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} \), \( C = \begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} \), then calculate AC, BC and (A + B)C. Also verify that (A + B)C = AC + BC. [CBSE 2015, 4M]
Answer:
\( AC = \begin{bmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0 \end{bmatrix}\begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} \)
\( AC = \begin{bmatrix} 0 - 12 + 21 \\ -12 + 0 + 24 \\ 14 + 16 + 0 \end{bmatrix} = \begin{bmatrix} 9 \\ 12 \\ 30 \end{bmatrix} \) ...(i) [1]
\( BC = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix}\begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} \)
\( BC = \begin{bmatrix} 0 - 2 + 3 \\ 2 + 0 + 6 \\ 2 - 4 + 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 8 \\ -2 \end{bmatrix} \) ...(ii) [1]
\( (A + B)C = \left\{ \begin{bmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} \right\} \begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6 & 0 \end{bmatrix}\begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} \)
\( = \begin{bmatrix} 0 - 14 + 24 \\ -10 + 0 + 30 \\ 16 + 12 + 0 \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix} \) ...(iii) [1]
\( AC + BC = \begin{bmatrix} 9 \\ 12 \\ 30 \end{bmatrix} + \begin{bmatrix} 1 \\ 8 \\ -2 \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \\ 28 \end{bmatrix} \) ...(iv) [from equations (i) and (ii)]
From equations (iii) and (iv),
Hence (A + B)C = AC + BC [1]

Question 14. Find matrix A such that \( \begin{pmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{pmatrix}A = \begin{pmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \end{pmatrix} \). [CBSE 2017, 4M]
Answer:
Let matrix \( A = \begin{bmatrix} a & c \\ b & d \end{bmatrix} \)
\( \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}\begin{bmatrix} a & c \\ b & d \end{bmatrix} = \begin{bmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \end{bmatrix} \) [1]
\( \begin{bmatrix} 2a - b & 2c - d \\ a & c \\ -3a + 4b & -3c + 4d \end{bmatrix} = \begin{bmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \end{bmatrix} \) [1]
2a − b = −1 ...(i)    2c − d = −8 ...(iv)
a = 1 ...(ii)    c = −2 ...(v)
−3a + 4b = 9 ...(iii)    −3c + 4d = 22 ...(vi) [1]
From eq. (i) and (ii): a = 1, b = 3
From eq. (iv) and (v): c = −2, d = 4
Here the values of a, b and c, d satisfy eq. (iii) and (vi) respectively.
Hence, \( A = \begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix} \) [1]

Long Answer Questions II (6 Marks)

Question 15. If \( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \) and \( A^3 - 6A^2 + 7A + kI_3 = O \), find k. [CBSE 2016, 6M]
Answer:
\( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \)
\( A^2 = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \) [1½]
\( A^3 = A^2 \cdot A = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} \) [1½]
Now \( A^3 - 6A^2 + 7A + kI = O \)
\( \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - 6\begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} + 7\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} + k\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \) [2]
\( \begin{bmatrix} -2 + k & 0 & 0 \\ 0 & -2 + k & 0 \\ 0 & 0 & -2 + k \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
⇒ −2 + k = 0
∴ k = 2 [1]