If bmatrix 3 & -1 & 3x \\ -7 & 4 & 2x \\ -11 & 7 & 2 bmatrix is a singular matrix, then find all values of x where x [0, (π)/(2) ].

MathsCBSE Class 12 · Board exam 2026, set 65/1/1 · 5 marks · Long answer

CBSE Class 12 Mathematics 2026: question 33(b)

Set in the CBSE Class 12 board exam of 2026, paper 65/1/1, section D, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerDeterminants

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

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 65/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

What is the determinant of a singular matrix?

Hint 2

Expand along the third column. Then write $sin 3 x$ and $cos 2 x$ in terms of $sin x$ only.

Hint 3

You will get a cubic in $sin x$ . Factor out $sin x$ , then factorise the quadratic left over and keep only roots with $∣ sin x ∣ \leq 1$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

A singular matrix has zero determinant, so
$3 - 7 - 11 - 1 47 sin 3 x cos 2 x 2 = 0$ .
1 mark
Expand along $C_{3}$ :
$sin 3 x (- 49 + 44) - cos 2 x (21 - 11) + 2 (12 - 7) = 0$
$\Rightarrow - 5 sin 3 x - 10 cos 2 x + 10 = 0 \Rightarrow sin 3 x + 2 cos 2 x = 2$ .
1 mark
Use $sin 3 x = 3 sin x - 4 sin^{3} x$ and $cos 2 x = 1 - 2 sin^{2} x$ :
$3 sin x - 4 sin^{3} x + 2 - 4 sin^{2} x = 2$
$\Rightarrow 4 sin^{3} x + 4 sin^{2} x - 3 sin x = 0$ .
1 mark
Factorise:
$sin x (4 sin^{2} x + 4 sin x - 3) = 0 \Rightarrow sin x (2 sin x - 1) (2 sin x + 3) = 0$ .
So $sin x = 0$ , $sin x = \frac{1}{2}$ or $sin x = - \frac{3}{2}$ .
1 mark
$sin x = - \frac{3}{2}$ is impossible because $∣ sin x ∣ \leq 1$ .
For $x \in [0, \frac{π}{2}]$ : $sin x = 0 \Rightarrow x = 0$ , and $sin x = \frac{1}{2} \Rightarrow x = \frac{π}{6}$ .
1 mark

Answer: $x = 0$ or $x = \frac{π}{6}$

Where the marks are

Sets the determinant equal to 0 because the matrix is singular (1 mark)
Expands the determinant to get $sin 3 x + 2 cos 2 x = 2$ (1 mark)
Converts to the cubic $4 sin^{3} x + 4 sin^{2} x - 3 sin x = 0$ (1 mark)
Factorises as $sin x (2 sin x - 1) (2 sin x + 3) = 0$ (1 mark)
Rejects $sin x = - 3/2$ and gets $x = 0, π /6$ (1 mark)

Revise the chapter

Chapter: Determinants (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Determinants, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 65/1/1: every question, by section.

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CBSE Class 12 Mathematics 2026: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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