Class 12 Maths Chapter 4 Determinants MCQs

Class 12 Maths Chapter 4 Determinants MCQs are available here to help the students of Class 12 score good marks in the CBSE board exam 2022-2023. Here, we provided multiple objective type questions that will help the students to practise for the exams. These multiple-choice questions form a useful source to prepare for exams as they are given as per the new CBSE guidelines for the academic year 2022-23. All the topics and subtopics of Chapter 4 Determinants of NCERT textbook are covered here in MCQs.

Get access to Class 12 Maths MCQs for all the chapters here at BYJU’S.

MCQs for Chapter 4 Determinants Class 12

Students can access several MCQs on matrices of Class 12 that cover various topics such as fundamentals of matrix and matrix algebra, i.e. mathematical operations on matrices.

Also, check:

Students must practise more MCQs to understand how to apply the concepts to solve problems in th exam. As we know, the chapter 4 determinants of Class 12 contain several topics and subtopics. They include finding the determinants up to order three only with real entries. Various properties of determinants, minors, cofactors and applications of determinants in finding the triangle area are included.

Download PDF – Chapter 4 Determinants MCQs

Besides, finding the adjoint and inverse of a square matrix, consistency and inconsistency of system of linear equations and solution of linear equations in two or three variables using the inverse of a matrix are included. All these topics are very important from the examination point of view. So, practising MCQs given below will help you get thorough with the formulas and applications of determinants.

MCQs for Class 12 Maths Chapter 4 Determinants with Solutions

\(\begin{array}{l}\text{1. If }\begin{vmatrix} 2x & 5\\ 8 &x \end{vmatrix}=\begin{vmatrix} 6 & -2\\ 7 &3 \end{vmatrix}, \text{ then the value of x is}\end{array} \)

(a) 3

(b) ±3

(c) ±6

(d) 6

Correct option: (c) ±6

Solution:

Given,

\(\begin{array}{l}\begin{vmatrix} 2x & 5\\ 8 &x \end{vmatrix}=\begin{vmatrix} 6 & -2\\ 7 &3 \end{vmatrix}\end{array} \)

So, (2x)(x) – (5)(8) = (6)(3) – (-2)(7)

2x2 – 40 = 18 + 14

2x2 = 32 + 40

2x2 = 72

x2 = 36

x = √36 = ±6

\(\begin{array}{l}\text{2. Value of k, for which  }A=\begin{bmatrix} k & 8\\ 4 &2k \end{bmatrix} \text{ is a singular matrix is}\end{array} \)

(a) 4

(b) -4

(c) ±4

(d) 0

Correct option: (c) ±4

Solution:

Given that matrix A is singular.

Thus, the determinant of A is 0.

So,

\(\begin{array}{l}|A|=\begin{vmatrix} k & 8\\ 4 &2k \end{vmatrix}=0\end{array} \)

k(2k) – 8(4) = 0

2k2 – 32 = 0

2k2 = 32

k2 = 16

k = ±4

3. If A is a square matrix of order 3 and |A| = 5, then the value of |2A′| is

(a) -10

(b) 10

(c) -40

(d) 40

Correct option: (d) 40

Solution:

According to the property of transpose of a matrix,

(kA′) = kA′

Also, from the property of determinant of a matrix,

|A′| = |A|

and |kA| = kn|A|, where n is the order of matrix A.

Thus, |2A′| = 23|A| {since A is a square matrix of order 3}

= 8 × 5

= 40

4. The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be

(a) 9

(b) 3

(c) -9

(d) 6

Correct option: (b) 3

Solution:

The formula of area of the triangle with vertices (x1, y1), (x2, y2), (x3, y3) is given by:

\(\begin{array}{l}\Delta =\frac{1}{2}\begin{vmatrix} x_1 &y_1 &1 \\ x_2 &y_2 &1 \\ x_3 & y_3 &1 \end{vmatrix}\end{array} \)

Thus, the area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is:

\(\begin{array}{l}\Delta =\frac{1}{2}\begin{vmatrix} -3 &0 &1 \\ 3 &0 &1 \\ 0 & k &1 \end{vmatrix}=9\end{array} \)
{given}

\(\begin{array}{l}\begin{vmatrix} -3 &0 &1 \\ 3 &0 &1 \\ 0 & k &1 \end{vmatrix}=18\end{array} \)

⇒ -3(0 – k) – 0+ 1(3k – 0) = 18

⇒ 3k + 3k = 18

⇒ 6k = 18

⇒ k = 3

5. Given that A is a square matrix of order 3 and |A| = -4, then |adj A| is equal to

(a) -4

(b) 4

(c) -16

(d) 16

Correct option: (d) 16

Solution:

Given that A is a square matrix of order 3 and |A| = -4.

We know that |adj A| = |A|n−1, where n is the order of matrix A.

So, |adj A| = (−4)3-1 = (-4)2 = 16

\(\begin{array}{l}\text{6. If  }A=\begin{bmatrix} 2 &\lambda &-3 \\ 0&2 &5 \\ 1 & 1 &3 \end{bmatrix}, \text{ then }A^{-1} \text{ exists if }\end{array} \)

(a) λ = 2

(b) λ ≠ 2

(c) λ ≠ – 2

(d) None of these

Correct option: (d) None of these

Solution:

Given,

\(\begin{array}{l}A=\begin{bmatrix} 2 &\lambda &-3 \\ 0&2 &5 \\ 1 & 1 &3 \end{bmatrix}\end{array} \)

The inverse of a matrix exists if its determinant is not equal to 0.

Consider,

\(\begin{array}{l}|A|=\begin{vmatrix} 2 &\lambda &-3 \\ 0&2 &5 \\ 1 & 1 &3 \end{vmatrix}\ne 0\end{array} \)

⇒ |A| = 2 (6 – 5) – λ (0 – 5) + (-3) (0 – 2) ≠ 0

⇒ 2 + 5λ + 6 ≠ 0

⇒ 5λ + 8 ≠ 0

⇒ 5λ ≠ -8

⇒ λ ≠ -8/5

Therefore, A-1 exists if and only if λ ≠ -8/5.

\(\begin{array}{l}\text{7. If  }A=\begin{bmatrix} 3 & 1\\ -1 &2 \end{bmatrix}, \text{ then }14A^{-1} \text{ is given by:}\end{array} \)
\(\begin{array}{l}(a) \ 14\begin{bmatrix} 2 & -1\\ 1 &3 \end{bmatrix}\end{array} \)
\(\begin{array}{l}(b) \ \begin{bmatrix} 4 & -2\\ 2 &6 \end{bmatrix}\end{array} \)
\(\begin{array}{l}(c) \ 2\begin{bmatrix} 2 & -1\\ 1 &-3 \end{bmatrix}\end{array} \)
\(\begin{array}{l}(d) \ 2\begin{bmatrix} -3 & -1\\ 1 &-2 \end{bmatrix}\end{array} \)
\(\begin{array}{l}\text{Correct option: (b)  }\begin{bmatrix} 4 & -2\\ 2 &6 \end{bmatrix}\end{array} \)

Solution:

Given,

\(\begin{array}{l}A=\begin{bmatrix} 3 & 1\\ -1 &2 \end{bmatrix}\end{array} \)
\(\begin{array}{l}\text{So, }|A|=\begin{vmatrix} 3 & 1\\ -1 &2 \end{vmatrix}\end{array} \)

= (3)(2) – (-1)(1)

= 6 + 1

= 7

Now,

\(\begin{array}{l}adjA=\begin{bmatrix} 2 & -1\\ 1 &3 \end{bmatrix}\end{array} \)

14A-1 = 14[adj A/ |A|] = (14/7) adj A

= 2 adj A

\(\begin{array}{l}=2\begin{bmatrix} 2 & -1\\ 1 &3 \end{bmatrix}\\=\begin{bmatrix} 4 & -2\\ 2 &6 \end{bmatrix}\end{array} \)

8. Which of the following is correct?

(a) Determinant is a square matrix.

(b) Determinant is a number associated with a matrix.

(c) Determinant is a number associated with a square matrix.

(d) None of these

Correct option: (c) Determinant is a number associated with a square matrix.

Solution:

We know that we can calculate determinant values only for square matrices.

Therefore, the determinant is a number associated with a square matrix.

9. Given that A = [aij] is a square matrix of order 3×3 and |A| = -7, then the value of ∑i=13 ai2 Ai2, where Aij denotes the cofactor of element aij is

(a) 7

(b) -7

(c) 0

(d) 49

Correct option: (b) -7

Solution:

Given,

|A| = -7

Order of matrix A is 3×3.

Now, ∑i=13 ai2 Ai2 = a12 A12 + a22 A22 + a32 A32

= |A|

= -7

10. If A is an invertible matrix of order 2, then det (A–1) is equal to

(a) det (A)

(b) 1/det (A)

(c) 1

(d) 0

Correct option: (b) 1/det (A)

Solution:

Given that the A is an invertible matrix of order 2.

If the matrix is invertible, then its determinant is not equal to 0.

We know that,

AA-1 = I, where I is the identity matrix

Taking determinant on both sides,

|AA-1| = |I|

|A| |A-1| = 1

|A-1| = 1/|A| {since A is non-singular, |A| ≠ 0}

Or

det(A-1) = 1/det(A)

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