If \(\begin{vmatrix}x & -2\\ 4 & 5\end{vmatrix} = 23\), then \(x\) is
- (a)\(3\)
- (b)\(5\)
- (c)\(-3\)
- (d)\(\dfrac{31}{5}\)
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Answer: (a) \(3\)
Why: Expand: 5x − (−2)(4) = 5x + 8 = 23.
\(5x - (-2)(4) = 5x + 8 = 23 \Rightarrow x = 3\).
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)
- No calculator needed
- Free, no sign-in
Multiple-choice questions
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
If \(\begin{vmatrix}x & 2\\ 8 & x\end{vmatrix} = 0\), then \(x\) is
- (a)\(4\) only
- (b)\(\pm 4\)
- (c)\(\pm 2\)
- (d)\(\pm 16\)
Show answer
Answer: (b) \(\pm 4\)
Why: x² − 16 = 0.
\(x^2 - 16 = 0 \Rightarrow x = \pm 4\).
The value of \(\begin{vmatrix}1 & 2 & 0\\ 0 & 3 & 1\\ 2 & 0 & 1\end{vmatrix}\) is
- (a)\(-1\)
- (b)\(5\)
- (c)\(7\)
- (d)\(11\)
Show answer
Answer: (c) \(7\)
Why: Expand along R₁: 1(3 − 0) − 2(0 − 2) + 0.
\(1(3 \cdot 1 - 1 \cdot 0) - 2(0 \cdot 1 - 1 \cdot 2) + 0 = 3 + 4 = 7\).
For \(A = \begin{bmatrix}2 & 5\\ -1 & 3\end{bmatrix}\), the cofactor \(A_{12}\) of the element \(a_{12} = 5\) is
- (a)\(-1\)
- (b)\(1\)
- (c)\(5\)
- (d)\(-5\)
Show answer
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 \(A_{12} = (-1)^{1+2}(-1) = 1\).
The value of \(\begin{vmatrix}1 & 1 & 1\\ 1 & 2 & 3\\ 1 & 3 & 6\end{vmatrix}\) is
- (a)\(0\)
- (b)\(2\)
- (c)\(-1\)
- (d)\(1\)
Show answer
Answer: (d) \(1\)
Why: Expand along R₁: 1(12 − 9) − 1(6 − 3) + 1(3 − 2).
\(1(2 \cdot 6 - 3 \cdot 3) - 1(1 \cdot 6 - 3 \cdot 1) + 1(1 \cdot 3 - 2 \cdot 1) = 3 - 3 + 1 = 1\).
\(A\) is a square matrix of order \(2\) with \(|A| = 7\). Then \(|\mathrm{adj}\,A|\) is
- (a)\(7\)
- (b)\(49\)
- (c)\(\dfrac17\)
- (d)\(14\)
Show answer
Answer: (a) \(7\)
Why: |adj A| = |A|ⁿ⁻¹ with n = 2.
\(|\mathrm{adj}\,A| = |A|^{2-1} = 7\). (For \(\begin{bmatrix}a & b\\ c & d\end{bmatrix}\), \(\mathrm{adj}\,A = \begin{bmatrix}d & -b\\ -c & a\end{bmatrix}\) has the same determinant.)
The adjoint of \(\begin{bmatrix}4 & 1\\ 3 & 2\end{bmatrix}\) is
- (a)\(\begin{bmatrix}2 & 3\\ 1 & 4\end{bmatrix}\)
- (b)\(\begin{bmatrix}-2 & 1\\ 3 & -4\end{bmatrix}\)
- (c)\(\begin{bmatrix}2 & -1\\ -3 & 4\end{bmatrix}\)
- (d)\(\begin{bmatrix}4 & -3\\ -1 & 2\end{bmatrix}\)
Show answer
Answer: (c) \(\begin{bmatrix}2 & -1\\ -3 & 4\end{bmatrix}\)
Why: For a 2 × 2 matrix, swap the diagonal entries and change the signs of the others.
\(\mathrm{adj}\begin{bmatrix}a & b\\ c & d\end{bmatrix} = \begin{bmatrix}d & -b\\ -c & a\end{bmatrix}\), so the adjoint is \(\begin{bmatrix}2 & -1\\ -3 & 4\end{bmatrix}\).
The inverse of \(\begin{bmatrix}3 & 1\\ 5 & 2\end{bmatrix}\) is
- (a)\(\begin{bmatrix}2 & 1\\ 5 & 3\end{bmatrix}\)
- (b)\(\begin{bmatrix}-2 & 1\\ 5 & -3\end{bmatrix}\)
- (c)\(\begin{bmatrix}3 & -1\\ -5 & 2\end{bmatrix}\)
- (d)\(\begin{bmatrix}2 & -1\\ -5 & 3\end{bmatrix}\)
Show answer
Answer: (d) \(\begin{bmatrix}2 & -1\\ -5 & 3\end{bmatrix}\)
Why: |A| = 6 − 5 = 1, so A⁻¹ = adj A.
\(|A| = 1\), so \(A^{-1} = \dfrac{1}{|A|}\mathrm{adj}\,A = \begin{bmatrix}2 & -1\\ -5 & 3\end{bmatrix}\). Check: \(\begin{bmatrix}3 & 1\\ 5 & 2\end{bmatrix}\begin{bmatrix}2 & -1\\ -5 & 3\end{bmatrix} = I\).
Using a determinant, find the area enclosed by the triangle whose corners are the origin, \((4, 0)\) and \((1, 5)\).
- (a)\(10\) sq units
- (b)\(20\) sq units
- (c)\(5\) sq units
- (d)\(9\) sq units
Show answer
Answer: (a) \(10\) sq units
Why: Area = ½ |det[[0,0,1],[4,0,1],[1,5,1]]|.
\(\dfrac12\left|\begin{vmatrix}0 & 0 & 1\\ 4 & 0 & 1\\ 1 & 5 & 1\end{vmatrix}\right| = \dfrac12|0 - 0 + 1(20 - 0)| = 10\) sq units.
The area of the triangle with vertices \((k, 0)\), \((6, 0)\) and \((0, 3)\) is \(6\) sq units. Then \(k\) is
- (a)\(10\) only
- (b)\(2\) only
- (c)\(2\) or \(10\)
- (d)\(\pm 4\)
Show answer
Answer: (c) \(2\) or \(10\)
Why: ½ |3(6 − k)| = 6 gives |6 − k| = 4.
\(\dfrac12\left|\begin{vmatrix}k & 0 & 1\\ 6 & 0 & 1\\ 0 & 3 & 1\end{vmatrix}\right| = \dfrac12|k(0 - 3) + (18 - 0)| = \dfrac32|6 - k| = 6\), so \(|6 - k| = 4\): \(k = 2\) or \(k = 10\).
The system \(x + 2y = 3\), \(2x + 4y = 7\) has
- (a)a unique solution
- (b)no solution
- (c)infinitely many solutions
- (d)exactly two solutions
Show answer
Answer: (b) no solution
Why: |A| = 0 and (adj A)B ≠ O, so the system is inconsistent.
\(A = \begin{bmatrix}1 & 2\\ 2 & 4\end{bmatrix}\), \(|A| = 0\). \((\mathrm{adj}\,A)B = \begin{bmatrix}4 & -2\\ -2 & 1\end{bmatrix}\begin{bmatrix}3\\ 7\end{bmatrix} = \begin{bmatrix}-2\\ 1\end{bmatrix} \ne O\): inconsistent, no solution.
The matrix \(\begin{bmatrix}k & 3\\ 4 & 6\end{bmatrix}\) is singular when \(k\) equals
- (a)\(-2\)
- (b)\(8\)
- (c)\(\dfrac12\)
- (d)\(2\)
Show answer
Answer: (d) \(2\)
Why: Singular means determinant 0: 6k − 12 = 0.
\(6k - 12 = 0 \Rightarrow k = 2\).
\(A\) is a square matrix of order \(3\) with \(|A| = -3\). Then \(A(\mathrm{adj}\,A)\) equals
- (a)\(9I\)
- (b)\(-3I\)
- (c)\(-27I\)
- (d)\(I\)
Show answer
Answer: (b) \(-3I\)
Why: A (adj A) = |A| I for every square matrix.
\(A(\mathrm{adj}\,A) = |A|\,I = -3I\).
Using determinants, the equation of the line through \((1, 2)\) and \((3, 6)\) is
- (a)\(y = 2x\)
- (b)\(y = 3x - 1\)
- (c)\(2y = x\)
- (d)\(y = x + 1\)
Show answer
Answer: (a) \(y = 2x\)
Why: Set det[[x, y, 1], [1, 2, 1], [3, 6, 1]] = 0.
\(\begin{vmatrix}x & y & 1\\ 1 & 2 & 1\\ 3 & 6 & 1\end{vmatrix} = x(2 - 6) - y(1 - 3) + (6 - 6) = -4x + 2y = 0\), i.e. \(y = 2x\).
If \(A\) is invertible and \(|A| = 4\), then \(|A^{-1}|\) is
- (a)\(4\)
- (b)\(\dfrac14\)
- (c)\(16\)
- (d)\(-4\)
Show answer
Answer: (b) \(\dfrac14\)
Why: |A| |A⁻¹| = |AA⁻¹| = |I| = 1.
\(|A|\,|A^{-1}| = |I| = 1 \Rightarrow |A^{-1}| = \dfrac14\).
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.
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 \(AC\).
(i) Write the determinant that gives the area of the plot. [1 mark]
Show answer
Answer: \(\dfrac12\left|\begin{vmatrix}1 & 2 & 1\\ 7 & 2 & 1\\ 4 & 8 & 1\end{vmatrix}\right|\)
Area \(= \dfrac12\left|\begin{vmatrix}1 & 2 & 1\\ 7 & 2 & 1\\ 4 & 8 & 1\end{vmatrix}\right|\). A1
(ii) Find the area of the plot in square units. [1 mark]
Show answer
Answer: \(18\) sq units
\(\dfrac12|1(2 - 8) - 2(7 - 4) + 1(56 - 8)| = \dfrac12|-6 - 6 + 48| = 18\) sq units. A1
(iii) The well \(W(k, 5)\) is to lie on \(AC\). Use a determinant to find \(k\). [2 marks]
Show answer
Answer: \(k = \dfrac52\)
\(A\), \(W\), \(C\) collinear: \(\begin{vmatrix}1 & 2 & 1\\ k & 5 & 1\\ 4 & 8 & 1\end{vmatrix} = 0\) M1 \(\Rightarrow 1(5 - 8) - 2(k - 4) + (8k - 20) = 6k - 15 = 0\), so \(k = \dfrac52\). A1
OR \(M\) is the mid-point of \(BC\). Find the area of \(\triangle ABM\). [2 marks]
Show answer
Answer: \(9\) sq units
\(M = \left(\dfrac{11}{2}, 5\right)\) M1; area \(= \dfrac12|1(2 - 5) - 2(7 - \tfrac{11}{2}) + (35 - 11)| = \dfrac12|-3 - 3 + 24| = 9\) 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 \(AX = B\) with \(A = \begin{bmatrix}2 & 3\\ 3 & 2\end{bmatrix}\), \(X = \begin{bmatrix}x\\ y\end{bmatrix}\), \(B = \begin{bmatrix}60\\ 55\end{bmatrix}\).
(i) Find \(|A|\). [1 mark]
Show answer
Answer: \(-5\)
\(|A| = 4 - 9 = -5\). A1
(ii) Write \(\mathrm{adj}\,A\). [1 mark]
Show answer
Answer: \(\begin{bmatrix}2 & -3\\ -3 & 2\end{bmatrix}\)
\(\mathrm{adj}\,A = \begin{bmatrix}2 & -3\\ -3 & 2\end{bmatrix}\). A1
(iii) Using \(A^{-1}\), find the price of a cup of tea and of a samosa. [2 marks]
Show answer
Answer: Tea ₹\(9\), samosa ₹\(14\)
\(X = A^{-1}B = -\dfrac15\begin{bmatrix}2 & -3\\ -3 & 2\end{bmatrix}\begin{bmatrix}60\\ 55\end{bmatrix} = -\dfrac15\begin{bmatrix}-45\\ -70\end{bmatrix}\) M1 \(= \begin{bmatrix}9\\ 14\end{bmatrix}\): tea ₹\(9\), samosa ₹\(14\). A1
OR Verify that \(A(\mathrm{adj}\,A) = |A|\,I\). [2 marks]
Show answer
Answer: Both equal \(-5I\)
\(\begin{bmatrix}2 & 3\\ 3 & 2\end{bmatrix}\begin{bmatrix}2 & -3\\ -3 & 2\end{bmatrix} = \begin{bmatrix}4 - 9 & -6 + 6\\ 6 - 6 & -9 + 4\end{bmatrix}\) M1 \(= \begin{bmatrix}-5 & 0\\ 0 & -5\end{bmatrix} = -5I = |A|\,I\). 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 Essentials adds all 40 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 and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
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 re-checked by computer algebra and 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.