CBSE Class 12 Mathematics Determinants Class Test Set 01

This class test on Chapter 4 Determinants for Class 12 Mathematics (CBSE) has 11 questions carrying 1 to 5 marks each, covering expansion, cofactors, adjoint, inverse and the matrix method. Complete worked solutions follow every question.

Class Test Questions with Solutions

Question 1. Write the value of the determinant \( \begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix} \).   [1]
Answer:
Let \( |A| = \begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix} \) and put p = a − b, q = b − c, r = c − a, so that p + q + r = 0.
Then \( |A| = \begin{vmatrix} p & q & r \\ q & r & p \\ r & p & q \end{vmatrix} \)
Expanding along R₁: |A| = p(rq − p²) − q(q² − pr) + r(pq − r²) = 3pqr − (p³ + q³ + r³)
Since p + q + r = 0, p³ + q³ + r³ = 3pqr.
∴ |A| = 3pqr − 3pqr = 0

Question 2. Find the value of x if the area of triangle is 35 square units with vertices (x, 4), (2, −6) and (5, 4).   [1]
Answer:
Given vertices are (x, 4), (2, −6) and (5, 4).
\( \text{Area of triangle} = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \)
\( 35 = \left|\frac{1}{2}\begin{vmatrix} x & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1 \end{vmatrix}\right| \)
⇒ ±70 = −10x + 12 + 38
Solving, we get x = −2, 12

Question 3. If A_ij is the cofactor of the element a_ij of the determinant \( \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \), then write the value of a₃₂·A₃₂.   [1]
Answer:
Let \( A = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \)
\( A_{32} = (-1)^{3+2}\begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} \)
= −(8 − 30) = 22
∴ a₃₂ · A₃₂ = 5 × 22 = 110

Question 4. If \( \begin{vmatrix} x-1 & x-2 \\ x & x-3 \end{vmatrix} = 0 \), then find the value of x.   [1]
Answer:
Given \( \begin{vmatrix} x-1 & x-2 \\ x & x-3 \end{vmatrix} = 0 \)
(x − 1)(x − 3) − x(x − 2) = 0
x² − 4x + 3 − (x² − 2x) = 0
⇒ −2x + 3 = 0 ⇒ \( x = \frac{3}{2} \)

Question 5. Let \( A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} \). Verify that (AB)⁻¹ = B⁻¹A⁻¹.   [2]
Answer:
Given \( A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix},\quad B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} \)
\( AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix} \)
\( (AB)^{-1} = \frac{1}{67\times 61 - 87\times 47}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \frac{-1}{2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} \)
Now,
\( B^{-1} = \frac{1}{6\times 9 - 8\times 7}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \frac{-1}{2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} \)
\( A^{-1} = \frac{1}{3\times 5 - 2\times 7}\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \)
\( B^{-1}A^{-1} = \frac{-1}{2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \frac{-1}{2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} \)
∴ (AB)⁻¹ = B⁻¹A⁻¹. Hence proved.

Question 6. Determine the values of x for which the matrix \( A = \begin{bmatrix} x+1 & -3 & 4 \\ -5 & x+2 & 2 \\ 4 & 1 & x-6 \end{bmatrix} \) is singular.   [2]
Answer:
Given \( A = \begin{bmatrix} x+1 & -3 & 4 \\ -5 & x+2 & 2 \\ 4 & 1 & x-6 \end{bmatrix} \)
A is singular if |A| = 0
∴ \( \begin{vmatrix} x+1 & -3 & 4 \\ -5 & x+2 & 2 \\ 4 & 1 & x-6 \end{vmatrix} = 0 \)
(x + 1)[(x + 2)(x − 6) − 2] + 3(−5x + 30 − 8) + 4(−5 − 4x − 8) = 0
⇒ x³ − 3x² − 49x = 0 ⇒ x(x² − 3x − 49) = 0
Solving, we get x = 0, \( \frac{1}{2}\left(3 \pm \sqrt{205}\right) \)

Question 7. If \( A = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \), find adj A and verify that A (adj A) = (adj A) A = |A| I₃.   [3]
Answer:
Given \( A = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
|A| = cos α (cos α − 0) + sin α (sin α − 0)
|A| = cos² α + sin² α = 1
Now, A₁₁ = cos α, A₁₂ = −sin α, A₁₃ = 0
A₂₁ = sin α, A₂₂ = cos α, A₂₃ = 0
A₃₁ = 0, A₃₂ = 0, A₃₃ = 1
∴ \( \text{adj } A = \begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Now,
\[ A\cdot \text{adj } A = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3 = |A|\cdot I_3 \quad [\text{since } |A| = 1] \]
\[ \text{adj } A\cdot A = \begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3 = |A|\cdot I_3 \quad (\text{since } |A| = 1) \]

Question 8. If \( A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix} \), show that \( A^T A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix} \).   [3]
Answer:
Given \( A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix} \)
|A| = 1 + tan² x = sec² x
Now \( A^{-1} = \frac{1}{|A|}\cdot \text{adj } A = \frac{1}{\sec^2 x}\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \)
Now,
\( A^T A^{-1} = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}\cdot \frac{1}{\sec^2 x}\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \)
\( = \frac{1}{1 + \tan^2 x}\begin{bmatrix} 1 - \tan^2 x & -2\tan x \\ 2\tan x & 1 - \tan^2 x \end{bmatrix} \)
\( = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix} \)

Question 9. Show that \( A = \begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} \) satisfies the equation x² − 6x + 17 = 0. Hence, find A⁻¹.   [3]
Answer:
Given \( A = \begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} \)
\( A^2 = \begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix} \)
∴ A² − 6A + 17I
\( = \begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix} - 6\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} + 17\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \)
Now, A² − 6A + 17I = O
⇒ A·A·A⁻¹ − 6A·A⁻¹ = −17I·A⁻¹   (post-multiplying by A⁻¹)
∴ \( A^{-1} = \frac{-1}{17}(A - 6I) \)   (∵ AA⁻¹ = I)
\( = \frac{-1}{17}\left(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}\right) = \frac{-1}{17}\begin{bmatrix} -4 & -3 \\ 3 & -2 \end{bmatrix} = \frac{1}{17}\begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix} \)

Question 10. (i) If \( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \), show that A⁻¹ = A².
(ii) If \( A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \) is such that A^T = A⁻¹, find α.   [3]
Answer:
(i) Given \( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
\( A^2 = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \)
|A| = 1(0 + 1) = 1   (expanding along R₃)
\( \text{adj } A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \)
∴ \( A^{-1} = \frac{1}{|A|}\,\text{adj } A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \)
Hence A⁻¹ = A².
(ii) Given \( A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \)
\( A^T = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \)
\( A^{-1} = \frac{1}{\cos^2\alpha + \sin^2\alpha}\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \)
Since A^T = A⁻¹,
∴ cos α = cos α and sin α = sin α for every α
⇒ α ∈ R

Question 11. If \( A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \), then find A⁻¹ and solve the following system of equations using it:   [5]
x + 2y − 3z = −4, 2x + 3y + 2z = 2, 3x − 3y − 4z = 11
Answer:
Given \( A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix} \)
|A| = 1(−12 + 6) − 2(−8 − 6) − 3(−6 − 9)
= −6 + 28 + 45 = 67
\( \text{adj } A = \begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \)
∴ \( A^{-1} = \frac{1}{|A|}\,\text{adj } A = \frac{1}{67}\begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix} \)
Also given,
x + 2y − 3z = −4
2x + 3y + 2z = 2
3x − 3y − 4z = 11
By using matrix method, AX = B,
where \( A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix},\; X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\; B = \begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix} \)
We know X = A⁻¹B
\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{67}\begin{bmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{bmatrix}\begin{bmatrix} -4 \\ 2 \\ 11 \end{bmatrix} = \frac{1}{67}\begin{bmatrix} 24+34+143 \\ -56+10-88 \\ 60+18-11 \end{bmatrix} = \frac{1}{67}\begin{bmatrix} 201 \\ -134 \\ 67 \end{bmatrix} \]
x = 3, y = −2, z = 1