CBSE Maths › Class 12 PYQs › Determinants

Determinants: 1 mark Questions (CBSE Class 12)

9 different 1 mark questions on Determinants from CBSE Class 12 Maths board exams 2026, newest first.

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

Q61 markMCQDeterminants CBSE 2026 · 65/3/1

If $B (adj B) = \frac{1}{3} 00 0 \frac{1}{3} 0 00 \frac{1}{3}$ , then the value of $det (B^{- 1})$ =

(A) $\frac{1}{3}$
(B) $\frac{1}{9}$
(C) $3$
(D) $9$

Show answer & solution

Answer: (C)

3

$B (adj B) = ∣ B ∣ I$ , so $∣ B ∣ = \frac{1}{3}$ .
$det (B^{- 1}) = \frac{1}{∣ B ∣} = 3$ .

Q171 markMCQDeterminants CBSE 2026 · 65/3/2

If $A (adj A) = 202600020260002026$ , then the value of $∣ adj A ∣$ is equal to :

(A) $2026$
(B) $(2026)^{- 1}$
(C) $(2026)^{- 2}$
(D) $(2026)^{2}$

Show answer & solution

Answer: (D)

(2026)^{2}

$A (adj A) = ∣ A ∣ I$ , so $∣ A ∣ = 2026$ and A is of order 3.
$∣ adj A ∣ = ∣ A ∣^{n - 1} = ∣ A ∣^{2} = (2026)^{2}$ .

Q101 markMCQDeterminants CBSE 2026 · 65/3/3

If $A (adj A) = 300030003$ , then the value of $∣2 A ∣$ is :

(A) $6$
(B) $54$
(C) $12$
(D) $24$

Show answer & solution

Answer: (D)

24

$A (adj A) = ∣ A ∣ I$ , so $∣ A ∣ = 3$ and A is of order 3.
$∣2 A ∣ = 2^{3} ∣ A ∣ = 8 \times 3 = 24$ .

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

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