Determinants 4 marks Questions, CBSE Class 12 PYQs 2026

1 different 4 marks questions on Determinants from CBSE Class 12 Maths board exams 2026, newest first.

Q374 marksCase StudyDeterminants CBSE 2026 · 65/4/1

A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.
On the basis of the above information, answer the following questions :
(i) Write the equations representing the various dimensions and express them as the matrix equation AX = B. (1)
(ii) Find if $A^{- 1}$ exists. Justify your answer. (1)
(iii) Find $A^{- 1}$ . (2)
OR
(iii) Find $A^{2} + 7 I$ . (2)

Show answer & solution

Answer: (i)

l + b - h = 3

2 l + 3 b + h = 10

3 l - b - 7 h = 1

;

123 13 - 1 - 1 1 - 7 l b h = 3101

(ii) Yes,

∣ A ∣ = 8 \neq = 0

(iii)

A^{- 1} = \frac{1}{8} - 20 17 - 11 8 - 4 4 4 - 3 1

; OR

A^{2} + 7 I = 711 - 20 5177 7 - 6 52

(i) Let length l, breadth b, height h (cm). $l + b = h + 3$ , $2 l + 3 b + h = 10$ , $b + 7 h = 3 l - 1$ .
So $l + b - h = 3$ , $2 l + 3 b + h = 10$ , $3 l - b - 7 h = 1$ : $A = 123 13 - 1 - 1 1 - 7$ , $X = l b h$ , $B = 3101$ .
(ii) $∣ A ∣ = 1 (- 21 + 1) - 1 (- 14 - 3) - 1 (- 2 - 9) = - 20 + 17 + 11 = 8 \neq = 0$ , so A is non-singular and $A^{- 1}$ exists.
(iii) Cofactors: $C_{11} = - 20, C_{12} = 17, C_{13} = - 11, C_{21} = 8, C_{22} = - 4, C_{23} = 4, C_{31} = 4, C_{32} = - 3, C_{33} = 1$ .
$A^{- 1} = \frac{1}{∣ A ∣} adj A = \frac{1}{8} - 20 17 - 11 8 - 4 4 4 - 3 1$ . (This gives $l = 3$ , $b = 1$ , $h = 1$ .)
OR (iii) $A^{2} = 011 - 20 5107 7 - 6 45$ , so $A^{2} + 7 I = 711 - 20 5177 7 - 6 52$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Determinants: 4 marks Questions (CBSE Class 12)