CBSE Maths › Class 12 PYQs › Determinants

Determinants: 5 marks Questions (CBSE Class 12)

8 different 5 marks questions on Determinants from CBSE Class 12 Maths board exams 2026, newest first.

Q335 marksLong AnswerDeterminants CBSE 2026 · 65/1/1

If $A = 0 - 2 1 2 - 1 - 1 1 - 2 0$ , find $A^{- 1}$ and use it to solve the following system of equations :
$- 2 y + z = 7, 2 x - y - z = 8, x - 2 y = 10$

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Answer:

A^{- 1} = 22 - 3 11 - 2 32 - 4

;

x = 0, y = - 5, z = - 3

$∣ A ∣ = 0 (0 - 2) - 2 (0 + 2) + 1 (2 + 1) = - 1$ .
Cofactors: $C_{11} = - 2, C_{12} = - 2, C_{13} = 3, C_{21} = - 1, C_{22} = - 1, C_{23} = 2, C_{31} = - 3, C_{32} = - 2, C_{33} = 4$ .
$adj A = - 2 - 2 3 - 1 - 1 2 - 3 - 2 4$ , so $A^{- 1} = \frac{1}{∣ A ∣} adj A = 22 - 3 11 - 2 32 - 4$ .
The system is $A^{'} X = B$ with $X = x y z$ , $B = 7810$ , since the coefficient matrix $021 - 2 - 1 - 2 1 - 1 0 = A^{'}$ .
$X = (A^{'})^{- 1} B = (A^{- 1})^{'} B = 213212 - 3 - 2 - 4 7810 = 0 - 5 - 3$ .
So $x = 0, y = - 5, z = - 3$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q33 (OR) (OR)5 marksLong AnswerDeterminants CBSE 2026 · 65/1/1

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

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Answer:

x = 0, \frac{π}{6}

Singular means the determinant is 0. Expand along $C_{3}$ .
Cofactors: $C_{13} = (- 7) (7) - (4) (- 11) = - 5$ , $C_{23} = - [(3) (7) - (- 1) (- 11)] = - 10$ , $C_{33} = 12 - 7 = 5$ .
$- 5 sin 3 x - 10 cos 2 x + 10 = 0 \Rightarrow sin 3 x + 2 cos 2 x - 2 = 0$ .
With $s = sin x$ : $3 s - 4 s^{3} + 2 (1 - 2 s^{2}) - 2 = 0 \Rightarrow s (4 s^{2} + 4 s - 3) = 0 \Rightarrow s (2 s - 1) (2 s + 3) = 0$ .
$sin x = 0$ or $sin x = \frac{1}{2}$ ( $sin x = - \frac{3}{2}$ is impossible).
In $[0, \frac{π}{2}]$ : $x = 0$ or $x = \frac{π}{6}$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q325 marksLong AnswerDeterminants CBSE 2026 · 65/2/1

If $P = 120 - 1 31 042$ and $Q = 2 - 4 2 22 - 1 - 4 - 4 5$ , find (QP) and hence solve the following system of equations using matrices :
$x - y = 3$ , $2 x + 3 y + 4 z = 17$ , $y + 2 z = 7$

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Answer:

QP = 6 I

;

x = 2

y = - 1

z = 4

$QP = 600060006 = 6 I$ , so $P^{- 1} = \frac{1}{6} Q$ .
The system is $P X = B$ with $X = x y z$ , $B = 3177$ .
$X = P^{- 1} B = \frac{1}{6} 6 + 34 - 28 - 12 + 34 - 28 6 - 17 + 35 = \frac{1}{6} 12 - 6 24 = 2 - 1 4$ .
$x = 2$ , $y = - 1$ , $z = 4$ .

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

Q32 (OR) (OR)5 marksLong AnswerDeterminants CBSE 2026 · 65/2/1

Obtain the value of $Δ = 1 + x 11 1 1 + y 1 11 1 + z$ in terms of $x$ , y and z.
Further, if $Δ = 0$ and $x$ , y, z are non-zero real numbers, prove that $x^{- 1} + y^{- 1} + z^{- 1} = - 1$ .

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Answer:

Δ = x y z + x y + y z + z x

; proved.

Expanding along $R_{1}$ : $Δ = (1 + x) [(1 + y) (1 + z) - 1] - 1 [(1 + z) - 1] + 1 [1 - (1 + y)]$ .
$= (1 + x) (y + z + y z) - z - y = x y z + x y + y z + z x$ .
If $Δ = 0$ : $x y + y z + z x = - x y z$ .
Dividing by $x y z \neq = 0$ : $\frac{1}{z} + \frac{1}{x} + \frac{1}{y} = - 1$ , i.e. $x^{- 1} + y^{- 1} + z^{- 1} = - 1$ .

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

Q345 marksLong AnswerDeterminants CBSE 2026 · 65/3/1

On the inauguration day of a new showroom, a lucky draw was organized and some vouchers of ₹ 1,000 and ₹ 500 were given to the lucky draw winners.
A total of 60 vouchers were given on the day. The number of ₹ 1,000 vouchers added to 3 times the number of ₹ 500 vouchers, gives 100. Express the given information as a system of linear equations in two variables. Hence, find the number of vouchers of each type by matrix method.

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Answer: 40 vouchers of ₹ 1,000 and 20 vouchers of ₹ 500

Let x = number of ₹ 1,000 vouchers, y = number of ₹ 500 vouchers.
$x + y = 60$ , $x + 3 y = 100$ .
$A X = B$ with $A = [1113]$ , $X = [x y]$ , $B = [60100]$ .
$∣ A ∣ = 2 \neq = 0$ , $adj A = [3 - 1 - 1 1]$ , $A^{- 1} = \frac{1}{2} [3 - 1 - 1 1]$ .
$X = A^{- 1} B = \frac{1}{2} [180 - 100 - 60 + 100] = [4020]$ .
So 40 vouchers of ₹ 1,000 and 20 vouchers of ₹ 500.

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

Q325 marksLong AnswerDeterminants CBSE 2026 · 65/5/1

A man goes to buy fruits from the market. The shopkeeper informs him that 4 apples, 3 oranges and 2 bananas cost ₹ 60; 2 apples, 4 oranges and 6 bananas cost ₹ 90; whereas 6 apples, 2 oranges and 3 bananas cost ₹ 70. Using matrix method, find the cost of one fruit of each kind.

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Answer: Apple ₹ 5, orange ₹ 8, banana ₹ 8.

Let the costs be x (apple), y (orange), z (banana): $4 x + 3 y + 2 z = 60$ , $2 x + 4 y + 6 z = 90$ , $6 x + 2 y + 3 z = 70$ .
$A = 426342263$ , $∣ A ∣ = 4 (0) - 3 (6 - 36) + 2 (4 - 24) = 50 \neq = 0$ .
$adj A = 030 - 20 - 5 010 10 - 20 10$ .
$X = A^{- 1} B = \frac{1}{50} 030 - 20 - 5 010 10 - 20 10 609070 = \frac{1}{50} 250400400 = 588$ .
Apple ₹ 5, orange ₹ 8, banana ₹ 8.

Q325 marksLong AnswerDeterminants CBSE 2026 · 65/5/2

Find cost (per kg) of each fertilizer A, B and C that the farmer needs to buy, such that 1 kg each of fertilizer A and C added to 2 kg of B costs him ₹ 400. Also, cost of each kg of fertilizer B and C added together is equal to cost of 1 kg of fertilizer A. However, cost of 3 kg of fertilizer B added to ₹ 200 is the same as cost of 1 kg of fertilizer A and C together. Use matrix method to find the solution.

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Answer: A: ₹ 180 per kg, B: ₹ 40 per kg, C: ₹ 140 per kg.

Let the costs per kg of A, B, C be ₹ x, ₹ y, ₹ z: $x + 2 y + z = 400$ , $x - y - z = 0$ , $x - 3 y + z = 200$ .
$P = 111 2 - 1 - 3 1 - 1 1$ , $∣ P ∣ = 1 (- 1 - 3) - 2 (1 + 1) + 1 (- 3 + 1) = - 10 \neq = 0$ .
$adj P = - 4 - 2 - 2 - 5 05 - 1 2 - 3$ .
$X = P^{- 1} B = - \frac{1}{10} - 4 - 2 - 2 - 5 05 - 1 2 - 3 4000200 = - \frac{1}{10} - 1800 - 400 - 1400 = 18040140$ .
A costs ₹ 180, B ₹ 40 and C ₹ 140 per kg.

Q325 marksLong AnswerDeterminants CBSE 2026 · 65/5/3

Three students A, B and C go to a book-store to buy art books, story books and puzzle solving books. A buys one of each type of book for a total of ₹ 21. B buys 4 art books, 3 story books and 2 puzzle solving books for ₹ 60. C buys 6 art books, 2 story books and 3 puzzle solving books and pays ₹ 10 more than B. Use matrix method to find the cost of each type of book.

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Answer: Art book ₹ 5, story book ₹ 8, puzzle solving book ₹ 8.

Let the costs be ₹ x (art), ₹ y (story), ₹ z (puzzle): $x + y + z = 21$ , $4 x + 3 y + 2 z = 60$ , $6 x + 2 y + 3 z = 70$ .
$A = 146132123$ , $∣ A ∣ = 1 (9 - 4) - 1 (12 - 12) + 1 (8 - 18) = - 5 \neq = 0$ .
$adj A = 50 - 10 - 1 - 3 4 - 1 2 - 1$ .
$X = A^{- 1} B = - \frac{1}{5} 50 - 10 - 1 - 3 4 - 1 2 - 1 216070 = - \frac{1}{5} - 25 - 40 - 40 = 588$ .
Art book ₹ 5, story book ₹ 8, puzzle solving book ₹ 8.

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