CBSE Class 12 · Chapter 4 · Algebra · 2026-27

Determinants Class 12: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceEvaluating determinants

If $| \begin{matrix} x & - 2 \\ 4 & 5 \end{matrix} | = 23$ , then $x$ is

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Answer: (a)

3

Why: Expand: 5x − (−2)(4) = 5x + 8 = 23.

$5 x - (- 2) (4) = 5 x + 8 = 23 \Rightarrow x = 3$ .

Q2

·1 mark·Multiple choiceEvaluating determinants

If $| \begin{matrix} x & 2 \\ 8 & x \end{matrix} | = 0$ , then $x$ is

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Answer: (b)

\pm 4

Why: x² − 16 = 0.

$x^{2} - 16 = 0 \Rightarrow x = \pm 4$ .

Q3

·1 mark·Multiple choiceEvaluating determinants

The value of $| \begin{matrix} 1 & 2 & 0 \\ 0 & 3 & 1 \\ 2 & 0 & 1 \end{matrix} |$ is

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Answer: (c)

7

Why: Expand along R₁: 1(3 − 0) − 2(0 − 2) + 0.

$1 (3 \cdot 1 - 1 \cdot 0) - 2 (0 \cdot 1 - 1 \cdot 2) + 0 = 3 + 4 = 7$ .

Q4

·1 mark·Multiple choiceMinors, cofactors and adjoint

For $A = [\begin{matrix} 2 & 5 \\ - 1 & 3 \end{matrix}]$ , the cofactor $A_{12}$ of the element $a_{12} = 5$ is

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Answer: (b)

1

Why: Minor M₁₂ = −1; cofactor = (−1)¹⁺² M₁₂.

Deleting row 1 and column 2 leaves $- 1$ , so $M_{12} = - 1$ and $A_{12} = (- 1)^{1 + 2} (- 1) = 1$ .

Q5

·1 mark·Multiple choiceEvaluating determinants

The value of $| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{matrix} |$ is

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Answer: (d)

1

Why: Expand along R₁: 1(12 − 9) − 1(6 − 3) + 1(3 − 2).

$1 (2 \cdot 6 - 3 \cdot 3) - 1 (1 \cdot 6 - 3 \cdot 1) + 1 (1 \cdot 3 - 2 \cdot 1) = 3 - 3 + 1 = 1$ .

Q6

·1 mark·Multiple choiceMinors, cofactors and adjoint

$A$ is a square matrix of order $2$ with $| A | = 7$ . Then $| adj A |$ is

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Answer: (a)

7

Why: |adj A| = |A|ⁿ⁻¹ with n = 2.

$| adj A | = | A |^{2 - 1} = 7$ . (For $[\begin{matrix} a & b \\ c & d \end{matrix}]$ , $adj A = [\begin{matrix} d & - b \\ - c & a \end{matrix}]$ has the same determinant.)

Q7

·1 mark·Multiple choiceMinors, cofactors and adjoint

The adjoint of $[\begin{matrix} 4 & 1 \\ 3 & 2 \end{matrix}]$ is

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Answer: (c)

[\begin{matrix} 2 & - 1 \\ - 3 & 4 \end{matrix}]

Why: For a 2 × 2 matrix, swap the diagonal entries and change the signs of the others.

$adj [\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} d & - b \\ - c & a \end{matrix}]$ , so the adjoint is $[\begin{matrix} 2 & - 1 \\ - 3 & 4 \end{matrix}]$ .

Q8

·1 mark·Multiple choiceInverse of a matrix

The inverse of $[\begin{matrix} 3 & 1 \\ 5 & 2 \end{matrix}]$ is

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Answer: (d)

[\begin{matrix} 2 & - 1 \\ - 5 & 3 \end{matrix}]

Why: |A| = 6 − 5 = 1, so A⁻¹ = adj A.

$| A | = 1$ , so $A^{- 1} = \frac{1}{| A |} adj A = [\begin{matrix} 2 & - 1 \\ - 5 & 3 \end{matrix}]$ . Check: $[\begin{matrix} 3 & 1 \\ 5 & 2 \end{matrix}] [\begin{matrix} 2 & - 1 \\ - 5 & 3 \end{matrix}] = I$ .

Q9

·1 mark·Multiple choiceArea of a triangle

Using a determinant, find the area enclosed by the triangle whose corners are the origin, $(4, 0)$ and $(1, 5)$ .

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Answer: (a)

10

sq units

Why: Area = ½ |det[[0,0,1],[4,0,1],[1,5,1]]|.

$\frac{1}{2} | | \begin{matrix} 0 & 0 & 1 \\ 4 & 0 & 1 \\ 1 & 5 & 1 \end{matrix} | | = \frac{1}{2} | 0 - 0 + 1 (20 - 0) | = 10$ sq units.

Q10

·1 mark·Multiple choiceArea of a triangle

The area of the triangle with vertices $(k, 0)$ , $(6, 0)$ and $(0, 3)$ is $6$ sq units. Then $k$ is

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Answer: (c)

2

10

Why: ½ |3(6 − k)| = 6 gives |6 − k| = 4.

$\frac{1}{2} | | \begin{matrix} k & 0 & 1 \\ 6 & 0 & 1 \\ 0 & 3 & 1 \end{matrix} | | = \frac{1}{2} | k (0 - 3) + (18 - 0) | = \frac{3}{2} | 6 - k | = 6$ , so $| 6 - k | = 4$ : $k = 2$ or $k = 10$ .

Q11

·1 mark·Multiple choiceSystems of linear equations

The system $x + 2 y = 3$ , $2 x + 4 y = 7$ has

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Answer: (b) no solution

Why: |A| = 0 and (adj A)B ≠ O, so the system is inconsistent.

$A = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}]$ , $| A | = 0$ . $(adj A) B = [\begin{matrix} 4 & - 2 \\ - 2 & 1 \end{matrix}] [\begin{matrix} 3 \\ 7 \end{matrix}] = [\begin{matrix} - 2 \\ 1 \end{matrix}] \neq O$ : inconsistent, no solution.

Q12

·1 mark·Multiple choiceInverse of a matrix

The matrix $[\begin{matrix} k & 3 \\ 4 & 6 \end{matrix}]$ is singular when $k$ equals

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Answer: (d)

2

Why: Singular means determinant 0: 6k − 12 = 0.

$6 k - 12 = 0 \Rightarrow k = 2$ .

Q13

·1 mark·Multiple choiceMinors, cofactors and adjoint

$A$ is a square matrix of order $3$ with $| A | = - 3$ . Then $A (adj A)$ equals

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Answer: (b)

- 3 I

Why: A (adj A) = |A| I for every square matrix.

$A (adj A) = | A | I = - 3 I$ .

Q14

·1 mark·Multiple choiceArea of a triangle

Using determinants, the equation of the line through $(1, 2)$ and $(3, 6)$ is

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Answer: (a)

y = 2 x

Why: Set det[[x, y, 1], [1, 2, 1], [3, 6, 1]] = 0.

$| \begin{matrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{matrix} | = x (2 - 6) - y (1 - 3) + (6 - 6) = - 4 x + 2 y = 0$ , i.e. $y = 2 x$ .

Q15

·1 mark·Multiple choiceInverse of a matrix

If $A$ is invertible and $| A | = 4$ , then $| A^{- 1} |$ is

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Answer: (b)

\frac{1}{4}

Why: |A| |A⁻¹| = |AA⁻¹| = |I| = 1.

$| A | | A^{- 1} | = | I | = 1 \Rightarrow | A^{- 1} | = \frac{1}{4}$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 12 case study questions by chapter.

Case study 1: Triangular plot (4 marks)

On a village map, a triangular plot of land has corners $A (1, 2)$ , $B (7, 2)$ and $C (4, 8)$ , where $1$ unit is $10$ m. The panchayat wants to know its area and to place a well on the boundary $A C$ .

(i) Write the determinant that gives the area of the plot. [1 mark]

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Answer:

\frac{1}{2} | | \begin{matrix} 1 & 2 & 1 \\ 7 & 2 & 1 \\ 4 & 8 & 1 \end{matrix} | |

Area

= \frac{1}{2} | | \begin{matrix} 1 & 2 & 1 \\ 7 & 2 & 1 \\ 4 & 8 & 1 \end{matrix} | |

. A1

(ii) Find the area of the plot in square units. [1 mark]

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Answer:

18

sq units

\frac{1}{2} | 1 (2 - 8) - 2 (7 - 4) + 1 (56 - 8) | = \frac{1}{2} | - 6 - 6 + 48 | = 18

sq units. A1

(iii) The well

W (k, 5)

is to lie on

A C

. Use a determinant to find

k

. [2 marks]

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Answer:

k = \frac{5}{2}

A

W

C

collinear:

| \begin{matrix} 1 & 2 & 1 \\ k & 5 & 1 \\ 4 & 8 & 1 \end{matrix} | = 0

\Rightarrow 1 (5 - 8) - 2 (k - 4) + (8 k - 20) = 6 k - 15 = 0

, so

k = \frac{5}{2}

. A1

M

is the mid-point of

B C

. Find the area of

△ A B M

. [2 marks]

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Answer:

9

sq units

M = (\frac{11}{2}, 5)

M1; area

= \frac{1}{2} | 1 (2 - 5) - 2 (7 - \frac{11}{2}) + (35 - 11) | = \frac{1}{2} | - 3 - 3 + 24 | = 9

sq units. A1

Case study 2: Tea stall orders (4 marks)

At a tea stall, one group pays ₹ $60$ for $2$ cups of tea and $3$ samosas, and another pays ₹ $55$ for $3$ cups of tea and $2$ samosas. With tea at ₹ $x$ a cup and samosas at ₹ $y$ each, this is $A X = B$ with $A = [\begin{matrix} 2 & 3 \\ 3 & 2 \end{matrix}]$ , $X = [\begin{matrix} x \\ y \end{matrix}]$ , $B = [\begin{matrix} 60 \\ 55 \end{matrix}]$ .

(i) Find

| A |

. [1 mark]

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Answer:

- 5

| A | = 4 - 9 = - 5

. A1

(ii) Write

adj A

. [1 mark]

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Answer:

[\begin{matrix} 2 & - 3 \\ - 3 & 2 \end{matrix}]

adj A = [\begin{matrix} 2 & - 3 \\ - 3 & 2 \end{matrix}]

. A1

(iii) Using

A^{- 1}

, find the price of a cup of tea and of a samosa. [2 marks]

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Answer: Tea ₹

9

, samosa ₹

14

X = A^{- 1} B = - \frac{1}{5} [\begin{matrix} 2 & - 3 \\ - 3 & 2 \end{matrix}] [\begin{matrix} 60 \\ 55 \end{matrix}] = - \frac{1}{5} [\begin{matrix} - 45 \\ - 70 \end{matrix}]

= [\begin{matrix} 9 \\ 14 \end{matrix}]

: tea ₹

9

, samosa ₹

14

. A1

OR Verify that

A (adj A) = | A | I

. [2 marks]

Show answer

Answer: Both equal

- 5 I

[\begin{matrix} 2 & 3 \\ 3 & 2 \end{matrix}] [\begin{matrix} 2 & - 3 \\ - 3 & 2 \end{matrix}] = [\begin{matrix} 4 - 9 & - 6 + 6 \\ 6 - 6 & - 9 + 4 \end{matrix}]

= [\begin{matrix} - 5 & 0 \\ 0 & - 5 \end{matrix}] = - 5 I = | A | I

. A1

Next steps for Determinants

This free set is separate from the chapter's question bank. On the Determinants chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 58 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Determinants: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.