Determinants MCQ with Solutions

Determinants MCQ with answers and solutions from NCERT Class 12 Maths. Revise all 7 important questions step by step with clear explanations. You can also watch the video solutions for better understanding. This is in continuation of the previous topic on Matrices MCQs.

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Ready? Okay, let’s begin with the first question!

Determinants MCQ#1 (NCERT Exercise 4.1 – Question 8)

📍If $\left\lvert \begin{array}{cc} x & 2 \\ 18 & x \end{array} \right\rvert = \left\lvert \begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array} \right\rvert$

then $x$ is equal to

Options:

(A) 6
(B) $\pm 6$
(C) –6
(D) 0

Answer:
✅ Correct option: (B) $\pm 6$

Explanation:

$\left\lvert \begin{array}{cc} x & 2 \\ 18 & x \end{array} \right\rvert = x^2 - 36$

$\left\lvert \begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array} \right\rvert = 0$

So,

$x^2 - 36 = 0$

$x^2 = 36 \Rightarrow x = \pm 6$

Watch Video Solution on YouTube

Let me tell you all of these 7 Determinants MCQ are so simple and easy to understand and therefore scoring too. But then as they say, Math requires practice.

So, my suggestion to you is to practice these NCERT Determinants MCQ a good number of times and feel confident. And if you ever need any clarification, just drop a comment — I’ll be more than happy to help.

Determinants MCQ#2 (NCERT Exercise 4.2 – Question 5)

📍If area of triangle is 35 sq units with vertices $(2,-6), (5,4), (k,4)$ . Then $k$ is

Options:

(A) 12
(B) –2
(C) –12, –2
(D) 12, –2

Answer:
✅ Correct Option: (D) $12, -2$

Explanation:

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is

$\text{Area} = \tfrac{1}{2} \left\lvert \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right\rvert$

So,

$35 = \tfrac{1}{2} \left\lvert \begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array} \right\rvert$

Multiply both sides by 2:

$70 = \left\lvert \begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array} \right\rvert$

Now expand the determinant (along row 1):

$\left\lvert \begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array} \right\rvert = 2(4-4) - (-6)(5-k) + 1(20-4k)$

$= 0 + (30 - 6k) + (20 - 4k) = 50 - 10k$

So,

$\lvert 50 - 10k \rvert = 70$

Case 1: $50 - 10k = 70 \Rightarrow -10k = 20 \Rightarrow k = -2$

Case 2: $50 - 10k = -70 \Rightarrow -10k = -120 \Rightarrow k = 12$

Watch Video Solution on YouTube

Determinants MCQ#3 (NCERT Exercise 4.3 – Question 5)

📍If $\Delta = \left\lvert \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right\rvert$
and $A_{ij}$ is the cofactor of $a_{ij}$ , then value of $\Delta$ is given by

Options:

(A) $a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33}$
(B) $a_{11}A_{11} + a_{12}A_{21} + a_{13}A_{31}$
(C) $a_{21}A_{12} + a_{22}A_{22} + a_{23}A_{32}$
(D) $a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$

Answer:
✅ Correct Option: (D) $a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$

Explanation:

We know that the determinant of a matrix can be expanded along the first row as

$\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$

It can also be expanded along the first column as

$\Delta = a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$

It is clear that the value of a determinant can be obtained by multiplying the elements of any one row (or column) with the cofactors of the corresponding elements of the same row (or column).
But, if we multiply the elements of one row (or column) with the cofactors of elements of a different row (or column), the result will always be zero.

Watch Video Solution on YouTube

You know there are about 100 MCQs in NCERT textbooks for Class 12 (both Part 1 and Part 2). I shall be taking one Chapter at a time and show you the explanations for each of the questions.

Moreover, I have also covered these solutions in the video form and they are freely available on my YouTube channel, @Mathsbetter. And here also, in each of the 7 Determinants MCQ on this page, I have provided the direct links to each and every MCQ for you to understand the solutions in a very simple and easy manner.

Determinants MCQ#4 (NCERT Exercise 4.4 – Question 17)

📍Let $A$ be a nonsingular square matrix of order $3 \times 3$ . Then $\lvert \text{adj } A \rvert$ is equal to

Options:

(A) $\lvert A \rvert$
(B) $\lvert A \rvert^2$
(C) $\lvert A \rvert^3$
(D) $3 \lvert A \rvert$

Answer:
✅ Correct Option: (B) $\lvert A \rvert^2$

Explanation:

The property used here is that the determinant of the adjoint of a square matrix of order $n$ is equal to the determinant of the matrix raised to the power $(n-1)$ .

In other words, for a square matrix of order $n$ ,

$\lvert \text{adj } A \rvert = \lvert A \rvert^{n-1}$

Here, $n = 3$ . So,

$\lvert \text{adj } A \rvert = \lvert A \rvert^{2}$

Watch Video Solution on YouTube

Determinants MCQ#5 (NCERT Exercise 4.4 – Question 18)

📍If $A$ is an invertible matrix of order $2$ , then

$\det(A^{-1})$ is equal to

Options:

(A) $\det(A)$
(B) $\tfrac{1}{\det(A)}$
(C) 1
(D) 0

Answer:
✅ Correct Option: (B) $\tfrac{1}{\det(A)}$

Explanation:

The property used here is that the determinant of the inverse of a matrix is equal to the reciprocal of the determinant of the matrix.

In other words, for any invertible matrix $A$ ,

$\det(A^{-1}) = \tfrac{1}{\det(A)}$

Watch Video Solution on YouTube

If you have been following this website as a resource for some of the important questions. For example, how to Integrate Square Root of tan x, or may be a complete guide to look at using Matrix Method in solving a system of linear equations.

Trust me, similarly, this series on NCERT Class 12 MCQs is going to help you in many ways.

Moving onto the next Determinants MCQ in this part of the series.

Determinants MCQ#6 (NCERT Miscellaneous Exercise – Question 8)

📍If $A = \begin{bmatrix} x & 0 & 0 \\[2pt] 0 & y & 0 \\[2pt] 0 & 0 & z \end{bmatrix}$ where $x,y,z$ are nonzero real numbers, then the inverse of $A$ is

Options:

(A) $\begin{bmatrix} x^{-1} & 0 & 0 \\[2pt] 0 & y^{-1} & 0 \\[2pt] 0 & 0 & z^{-1} \end{bmatrix}$
(B) $xyz\begin{bmatrix} x^{-1} & 0 & 0 \\[2pt] 0 & y^{-1} & 0 \\[2pt] 0 & 0 & z^{-1} \end{bmatrix}$
(C) $\tfrac{1}{xyz}\begin{bmatrix} x & 0 & 0 \\[2pt] 0 & y & 0 \\[2pt] 0 & 0 & z \end{bmatrix}$
(D) $\tfrac{1}{xyz}\begin{bmatrix} 1 & 0 & 0 \\[2pt] 0 & 1 & 0 \\[2pt] 0 & 0 & 1 \end{bmatrix}$

Answer:
✅ Correct option: (A)

Explanation:

For a diagonal matrix the inverse is obtained by taking reciprocals of diagonal entries.

$A^{-1} = \begin{bmatrix} x^{-1} & 0 & 0 \\[2pt] 0 & y^{-1} & 0 \\[2pt] 0 & 0 & z^{-1} \end{bmatrix}$

Watch Video Solution on YouTube

Determinant MCQ#7 (NCERT Miscellaneous Exercise – Question 9)

📍Let $A = \left\lvert \begin{array}{ccc} 1 & \sin\theta & 1 \\[2pt] -\sin\theta & 1 & \sin\theta \\[2pt] -1 & -\sin\theta & 1 \end{array} \right\rvert$ , where $0 \le \theta \le 2\pi$ . Then

Options:

(A) $\det(A) = 0$
(B) $\det(A) \in (2,\infty)$
(C) $\det(A) \in (2,4)$
(D) $\det(A) \in [2,4]$

Answer:
✅ Correct option: (D)

Explanation:

Compute the determinant (expand or use a symbolic simplification). The determinant simplifies to

$\det(A) = 2\big(1+\sin^2\theta\big)$ .

Since $0 \le \sin^2\theta \le 1$ , we get

$2 \le \det(A) \le 4$ .

So the determinant lies in the closed interval $[2,4]$ .

Watch Video Solution on YouTube

Quick Answer Key

Q1 → (B)
Q2 → (D)
Q3 → (D)
Q4 → (B)
Q5 → (B)
Q6 → (A)
Q7 → (D)

Closing Note

That completes all the important NCERT Class 12 Maths Determinants MCQ from chapter 4.
I hope the explanations and video solutions made things clearer and gave you more confidence. Stay tuned for the next chapter, where we’ll cover all the important Relations and Functions MCQ (Chapter 1 from NCERT) with the same clarity and video support.

👉 Make sure to practice these questions again to strengthen your concepts.
👉 Watch the video solutions for a clearer and faster revision.

Keep practicing, and you’ll master determinants easily! 🚀

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Determinants MCQ#1 (NCERT Exercise 4.1 – Question 8)

Determinants MCQ#2 (NCERT Exercise 4.2 – Question 5)

Determinants MCQ#3 (NCERT Exercise 4.3 – Question 5)

Determinants MCQ#4 (NCERT Exercise 4.4 – Question 17)

Determinants MCQ#5 (NCERT Exercise 4.4 – Question 18)

Determinants MCQ#6 (NCERT Miscellaneous Exercise – Question 8)

Determinant MCQ#7 (NCERT Miscellaneous Exercise – Question 9)

Quick Answer Key

Closing Note

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