For two matrices A=bmatrix3&-6&-1\\2&-5&-1\\-2&4&1bmatrix and B=bmatrix1&-2&-1\\0&-1&-1\\2&0&3bmatrix, find the product AB and hence solve the system of equations: 3x-6y-z=3 2x-5y-z+2=0 -2x+4y+z=5

MathsCBSE Class 12 · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2025-26: question 32

From the official CBSE sample question paper for 2025-26, section D, question 32, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32 · 5 marks · Long answerMatrices

For two matrices $A = 32 - 2 - 6 - 5 4 - 1 - 1 1$ and $B = 102 - 2 - 1 0 - 1 - 1 3$ , find the product $A B$ and hence solve the system of equations:

$3 x - 6 y - z = 3$
$2 x - 5 y - z + 2 = 0$
$- 2 x + 4 y + z = 5$

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

If $A B = I$ , what is $A^{- 1}$ ?

Hint 2

Rewrite the second equation as $2 x - 5 y - z = - 2$ and put the system as $A X = C$ ; then $X = A^{- 1} C$ .

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.

$A B = 32 - 2 - 6 - 5 4 - 1 - 1 1 102 - 2 - 1 0 - 1 - 1 3 = 3 - 0 - 2 2 - 0 - 2 - 2 + 0 + 2 - 6 + 6 + 0 - 4 + 5 + 0 4 - 4 + 0 - 3 + 6 - 3 - 2 + 5 - 3 2 - 4 + 3 = 100010001 = I$ .
1 mark
Hence $A^{- 1} = B$ (and $B^{- 1} = A$ ).
½ mark
The system is $3 x - 6 y - z = 3$ , $2 x - 5 y - z = - 2$ , $- 2 x + 4 y + z = 5$ , i.e. $A X = C$ .
½ mark
where $X = x y z$ , $C = 3 - 2 5$ .
½ mark
$∣ A ∣ = 3 (- 5 + 4) + 6 (2 - 2) - 1 (8 - 10) = - 3 + 0 + 2 = - 1 \neq = 0$ , so the system has a unique solution.
½ mark
$X = A^{- 1} C = B C$ .
½ mark
$B C = 102 - 2 - 1 0 - 1 - 1 3 3 - 2 5 = 3 + 4 - 5 0 + 2 - 5 6 + 0 + 15 = 2 - 3 21$ .
1 mark
Therefore $x = 2, y = - 3, z = 21$ .
½ mark

Answer: $A B = I$ ; $x = 2, y = - 3, z = 21$

Where the marks are

Computes $A B = I$ (1 mark)
Concludes $A^{- 1} = B$ (½ mark)
Writes system as $A X = C$ (½ mark)
Correct $X$ and $C$ (½ mark)
$∣ A ∣ \neq = 0$ , unique solution (½ mark)
Uses $X = B C$ (½ mark)
Computes $B C$ (1 mark)
States $x = 2, y = - 3, z = 21$ (½ mark)

Revise the chapter

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

Practise this chapterBoard-style questions from Matrices, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2025-26: every question, by section.

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CBSE Class 12 Mathematics sample paper 2025-26: question 32

Hints

Step-by-step solution

Revise the chapter

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