Determinants: CBSE Class 12 Previous Year Questions

18 different questions from Determinants (NCERT Chapter 4) asked in CBSE Class 12 Maths board exams 2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions9 questions 4 marks questions1 questions 5 marks questions8 questions

Most asked Determinants questions

Q41 markMCQDeterminants CBSE 2026 · 65/1/1

If $Δ_{1} = 100020003$ and $Δ_{2} = 010200006$ , then

(A) $Δ_{1} = 2 Δ_{2}$
(B) $Δ_{2} = - 2 Δ_{1}$
(C) $Δ_{1} = Δ_{2}$
(D) $Δ_{2} = - Δ_{1}$

Show answer & solution

Answer: (B)

Δ_{2} = - 2 Δ_{1}

$Δ_{1} = 1 \times 2 \times 3 = 6$ (diagonal determinant).
Expanding $Δ_{2}$ along $R_{1}$ : $Δ_{2} = 0 - 2 (1 \times 6 - 0 \times 0) + 0 = - 12$ .
So $Δ_{2} = - 12 = - 2 \times 6 = - 2 Δ_{1}$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q51 markMCQDeterminants CBSE 2026 · 65/1/1

One of the values of $x$ for which $cos x - cos x sin x sin x = 1$ is

(A)0
(B) $\frac{π}{4}$
(C) $\frac{π}{3}$
(D) $\frac{π}{2}$

Show answer & solution

Answer: (B)

\frac{π}{4}

$cos x - cos x sin x sin x = cos x sin x + sin x cos x = sin 2 x$ .
$sin 2 x = 1 \Rightarrow 2 x = \frac{π}{2} \Rightarrow x = \frac{π}{4}$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q61 markMCQDeterminants CBSE 2026 · 65/2/1

If $- 1 - 2 0 - 2 a 4 5 - 1 2 a = - 86$ , then the sum of all possible values of a is

(A)4
(B)5
(C) $- 4$
(D)9

Show answer & solution

Answer: (C)

- 4

Expanding along $R_{1}$ : $- 1 (2 a^{2} + 4) + 2 (- 4 a - 0) + 5 (- 8 - 0) = - 2 a^{2} - 8 a - 44$ .
$- 2 a^{2} - 8 a - 44 = - 86 \Rightarrow a^{2} + 4 a - 21 = 0 \Rightarrow (a + 7) (a - 3) = 0$ .
$a = - 7$ or $a = 3$ ; sum $= - 4$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q41 markMCQDeterminants CBSE 2026 · 65/4/1

If points (2, 3), (0, 4) and (p, 2) are collinear, then the value of p is :

(A) $\frac{4}{7}$
(B) $- \frac{3}{7}$
(C)4
(D) $- 4$

Show answer & solution

Answer: (C) 4

For collinear points the area of the triangle is zero: $\frac{1}{2} 20 p 342111 = 0$ .
Expanding: $2 (4 - 2) - 3 (0 - p) + 1 (0 - 4 p) = 0$ .
$4 + 3 p - 4 p = 0 \Rightarrow p = 4$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q51 markMCQDeterminants CBSE 2026 · 65/5/1

If A is a non-singular matrix, then which of the following is not true ?

(A)adj A is singular
(B) $(adj A)^{- 1} = (adj A^{- 1})$
(C) $∣ A ∣ \neq = 0$
(D) $A^{- 1}$ exists

Show answer & solution

Answer: (A) adj A is singular

For an $n \times n$ matrix, $∣ adj A ∣ = ∣ A ∣^{n - 1} \neq = 0$ when $∣ A ∣ \neq = 0$ , so adj A is non-singular.
(B), (C), (D) are all true for a non-singular matrix, so (A) is not true.

Also asked in: 2026 65/5/2, 2026 65/5/3

Q71 markMCQDeterminants CBSE 2026 · 65/5/1

If the area of $△ A B C$ with vertices $A (3, 1)$ , $B (- 2, 1)$ and $C (0, k)$ is 5 sq. units, then values of k are :

(A)3, 1
(B) $- 1$ , 3
(C) $- 1$ , 2
(D)0, 2

Show answer & solution

Answer: (B)

- 1

, 3

Area $= \frac{1}{2} 3 - 2 0 11 k 111 = \frac{1}{2} ∣3 (1 - k) - 1 (- 2 - 0) + 1 (- 2 k) ∣ = \frac{1}{2} ∣5 - 5 k ∣$ .
$\frac{1}{2} ∣5 - 5 k ∣ = 5 \Rightarrow ∣1 - k ∣ = 2 \Rightarrow k = - 1$ or $k = 3$ .

Also asked in: 2026 65/5/2, 2026 65/5/3

Q374 marksCase StudyDeterminants CBSE 2026 · 65/4/1

A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.
On the basis of the above information, answer the following questions :
(i) Write the equations representing the various dimensions and express them as the matrix equation AX = B. (1)
(ii) Find if $A^{- 1}$ exists. Justify your answer. (1)
(iii) Find $A^{- 1}$ . (2)
OR
(iii) Find $A^{2} + 7 I$ . (2)

Show answer & solution

Answer: (i)

l + b - h = 3

2 l + 3 b + h = 10

3 l - b - 7 h = 1

;

123 13 - 1 - 1 1 - 7 l b h = 3101

(ii) Yes,

∣ A ∣ = 8 \neq = 0

(iii)

A^{- 1} = \frac{1}{8} - 20 17 - 11 8 - 4 4 4 - 3 1

; OR

A^{2} + 7 I = 711 - 20 5177 7 - 6 52

(i) Let length l, breadth b, height h (cm). $l + b = h + 3$ , $2 l + 3 b + h = 10$ , $b + 7 h = 3 l - 1$ .
So $l + b - h = 3$ , $2 l + 3 b + h = 10$ , $3 l - b - 7 h = 1$ : $A = 123 13 - 1 - 1 1 - 7$ , $X = l b h$ , $B = 3101$ .
(ii) $∣ A ∣ = 1 (- 21 + 1) - 1 (- 14 - 3) - 1 (- 2 - 9) = - 20 + 17 + 11 = 8 \neq = 0$ , so A is non-singular and $A^{- 1}$ exists.
(iii) Cofactors: $C_{11} = - 20, C_{12} = 17, C_{13} = - 11, C_{21} = 8, C_{22} = - 4, C_{23} = 4, C_{31} = 4, C_{32} = - 3, C_{33} = 1$ .
$A^{- 1} = \frac{1}{∣ A ∣} adj A = \frac{1}{8} - 20 17 - 11 8 - 4 4 4 - 3 1$ . (This gives $l = 3$ , $b = 1$ , $h = 1$ .)
OR (iii) $A^{2} = 011 - 20 5107 7 - 6 45$ , so $A^{2} + 7 I = 711 - 20 5177 7 - 6 52$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q335 marksLong AnswerDeterminants CBSE 2026 · 65/1/1

If $A = 0 - 2 1 2 - 1 - 1 1 - 2 0$ , find $A^{- 1}$ and use it to solve the following system of equations :
$- 2 y + z = 7, 2 x - y - z = 8, x - 2 y = 10$

Show answer & solution

Answer:

A^{- 1} = 22 - 3 11 - 2 32 - 4

;

x = 0, y = - 5, z = - 3

$∣ A ∣ = 0 (0 - 2) - 2 (0 + 2) + 1 (2 + 1) = - 1$ .
Cofactors: $C_{11} = - 2, C_{12} = - 2, C_{13} = 3, C_{21} = - 1, C_{22} = - 1, C_{23} = 2, C_{31} = - 3, C_{32} = - 2, C_{33} = 4$ .
$adj A = - 2 - 2 3 - 1 - 1 2 - 3 - 2 4$ , so $A^{- 1} = \frac{1}{∣ A ∣} adj A = 22 - 3 11 - 2 32 - 4$ .
The system is $A^{'} X = B$ with $X = x y z$ , $B = 7810$ , since the coefficient matrix $021 - 2 - 1 - 2 1 - 1 0 = A^{'}$ .
$X = (A^{'})^{- 1} B = (A^{- 1})^{'} B = 213212 - 3 - 2 - 4 7810 = 0 - 5 - 3$ .
So $x = 0, y = - 5, z = - 3$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q33 (OR) (OR)5 marksLong AnswerDeterminants CBSE 2026 · 65/1/1

If $3 - 7 - 11 - 1 47 sin 3 x cos 2 x 2$ is a singular matrix, then find all values of $x$ where $x \in [0, \frac{π}{2}]$ .

Show answer & solution

Answer:

x = 0, \frac{π}{6}

Singular means the determinant is 0. Expand along $C_{3}$ .
Cofactors: $C_{13} = (- 7) (7) - (4) (- 11) = - 5$ , $C_{23} = - [(3) (7) - (- 1) (- 11)] = - 10$ , $C_{33} = 12 - 7 = 5$ .
$- 5 sin 3 x - 10 cos 2 x + 10 = 0 \Rightarrow sin 3 x + 2 cos 2 x - 2 = 0$ .
With $s = sin x$ : $3 s - 4 s^{3} + 2 (1 - 2 s^{2}) - 2 = 0 \Rightarrow s (4 s^{2} + 4 s - 3) = 0 \Rightarrow s (2 s - 1) (2 s + 3) = 0$ .
$sin x = 0$ or $sin x = \frac{1}{2}$ ( $sin x = - \frac{3}{2}$ is impossible).
In $[0, \frac{π}{2}]$ : $x = 0$ or $x = \frac{π}{6}$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q325 marksLong AnswerDeterminants CBSE 2026 · 65/2/1

If $P = 120 - 1 31 042$ and $Q = 2 - 4 2 22 - 1 - 4 - 4 5$ , find (QP) and hence solve the following system of equations using matrices :
$x - y = 3$ , $2 x + 3 y + 4 z = 17$ , $y + 2 z = 7$

Show answer & solution

Answer:

QP = 6 I

;

x = 2

y = - 1

z = 4

$QP = 600060006 = 6 I$ , so $P^{- 1} = \frac{1}{6} Q$ .
The system is $P X = B$ with $X = x y z$ , $B = 3177$ .
$X = P^{- 1} B = \frac{1}{6} 6 + 34 - 28 - 12 + 34 - 28 6 - 17 + 35 = \frac{1}{6} 12 - 6 24 = 2 - 1 4$ .
$x = 2$ , $y = - 1$ , $z = 4$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

← Matrices Continuity and Differentiability →

Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on MonoMath CBSE →