CBSE Class 12 Mathematics Determinants MCQs Set 02

Question. The value of determinant \( \begin{vmatrix} a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c \end{vmatrix} \) 
(a) \( a^3 + b^3 + c^3 \)
(b) \( 3bc \)
(c) \( a^3 + b^3 + c^3 - 3abc \)
(d) None of the options
Answer: (c) \( a^3 + b^3 + c^3 - 3abc \)

Question. 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
Answer: (b) 3

Question. If A, B and C are angles of a triangle, then the determinant \( \begin{vmatrix} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{vmatrix} \) is equal to
(a) 0
(b) -1
(c) 1
(d) None of the options
Answer: (a) 0

Question. If \( f(x) = \begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{vmatrix} \), then
(a) \( f(a) = 0 \)
(b) \( f(b) = 0 \)
(c) \( f(0) = 0 \)
(d) \( f(1) = 0 \)
Answer: (c) \( f(0) = 0 \)
Solution: We have, \( f(x) = \begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{vmatrix} \)
\( \implies \) \( f(a) = \begin{vmatrix} 0 & 0 & a-b \\ 2a & 0 & a-c \\ a+b & a+c & 0 \end{vmatrix} = [(a-b)\{2a . (a+c)\}] \neq 0 \)
and \( f(b) = \begin{vmatrix} 0 & b-a & 0 \\ b+a & 0 & b-c \\ 2b & b+c & 0 \end{vmatrix} = -(b-a)[-2b(b-c)] = 2b(b-a)(b-c) \neq 0 \)
and \( f(0) = \begin{vmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{vmatrix} = a(bc) - b(ac) = abc - abc = 0 \)

Question. If \( A = \begin{bmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{bmatrix} \), then \( A^{-1} \) exists if
(a) \( \lambda = 2 \)
(b) \( \lambda \neq 2 \)
(c) \( \lambda \neq -2 \)
(d) None of the options
Answer: (d) None of the options

Question. If A and B are invertible matrices, then which of the following is not correct?
(a) \( \text{adj } A = |A| . A^{-1} \)
(b) \( \det(A)^{-1} = [\det(A)]^{-1} \)
(c) \( (AB)^{-1} = B^{-1}A^{-1} \)
(d) \( (A + B)^{-1} = B^{-1} + A^{-1} \)
Answer: (d) \( (A + B)^{-1} = B^{-1} + A^{-1} \)

Question. If x, y, z are all different from zero and \( \begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix} = 0 \), then value of \( x^{-1} + y^{-1} + z^{-1} \) is
(a) \( xyz \)
(b) \( x^{-1} y^{-1} z^{-1} \)
(c) \( -x -y -z \)
(d) -1
Answer: (d) -1

Question. The value of the determinant \( \begin{vmatrix} x & x+y & x+2y \\ x+2y & x & x+y \\ x+y & x+2y & x \end{vmatrix} \) is 
(a) \( 9x^2(x+y) \)
(b) \( 9y^2(x+y) \)
(c) \( 3y^2(x+y) \)
(d) \( 7x^2(x+y) \)
Answer: (b) \( 9y^2(x+y) \)

Question. There are two values of \( a \) which makes determinant \( \Delta = \begin{vmatrix} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = 86 \), then sum of these numbers is
(a) 4
(b) 5
(c) -4
(d) 9
Answer: (c) -4

Question. If a, b, c are in AP, then the determinant \( \Delta = \begin{vmatrix} x+2 & x+4 & x+2a \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{vmatrix} \) is
(a) 0
(b) 1
(c) \( x \)
(d) \( 2x \)
Answer: (a) 0

Question. If x, y, z are non-zero real numbers, then the inverse of matrix \( A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \) is
(a) \( \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)
(b) \( xyz \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)
(c) \( \frac{1}{xyz} \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \)
(d) \( \frac{1}{xyz} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Answer: (a) \( \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \)

Question. If area of triangle is 35 sq units with vertices (2, -6) (5, 4) and (k, 4), then k is
(a) 12
(b) -2
(c) -12, -2
(d) 12, -2
Answer: (d) 12, -2

Question. Let A be a square matrix of order \( 3 \times 3 \), then \( |KA| \) is equal to
(a) \( K|A| \)
(b) \( K^2|A| \)
(c) \( K^3|A| \)
(d) \( 3K|A| \)
Answer: (c) \( K^3|A| \)

Question. Let A be a nonsingular square matrix of order \( 3 \times 3 \). Then \( |\text{adj. } A| \) is equal to
(a) \( |A| \)
(b) \( |A|^2 \)
(c) \( |A|^3 \)
(d) \( 3|A| \)
Answer: (b) \( |A|^2 \)

Question. If A is an invertible matrix of order 2, then \(\det(A^{-1})\) is equal to
(a) \( \det(A) \)
(b) \( \frac{1}{\det(A)} \)
(c) 1
(d) 0
Answer: (b) \( \frac{1}{\det(A)} \)

Question. The value of \( \begin{vmatrix} 265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181 \end{vmatrix} \) is
(a) 0
(b) 1
(c) -1
(d) None of the options
Answer: (a) 0

Question. The value of \( \begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix} \) is
(a) 1
(b) 0
(c) \( a+b \)
(d) \( a-b \)
Answer: (b) 0

Question. The value of \( \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & \omega & 1 \end{vmatrix} \) is
(a) 1
(b) -1
(c) 0
(d) \( \omega \)
Answer: (c) 0

Question. The value of \( \begin{vmatrix} 0 & a-b & a-c \\ b-a & 0 & b-c \\ c-a & c-b & 0 \end{vmatrix} \) is
(a) a
(b) b
(c) 0
(d) None of the options
Answer: (c) 0

Question. The value of \( \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \) is
(a) \( (a-b)(b-c)(c-a) \)
(b) \( (b-a)(c-b)(c-a) \)
(c) \( a(b-c)(c-a) \)
(d) None of the options
Answer: (a) \( (a-b)(b-c)(c-a) \)

Assertion-Reason Questions [1 mark]

The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A) : Determinant is a number associated with a square matrix.
Reason (R) : Determinant is a square matrix.
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (c) A is true but R is false.
Solution: Clearly, Assertion (A) is true and Reason (R) is false.
Hence, (c) is the correct option.

Question. Assertion (A) : If \( A = \begin{bmatrix} 5-x & x+1 \\ 2 & 4 \end{bmatrix} \), then the matrix A is singular if x = 3.
Reason (R) : A square matrix is a singular matrix if its determinant is zero.
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation for A.
Solution: For singular matrix, we have \( \begin{vmatrix} 5-x & x+1 \\ 2 & 4 \end{vmatrix} = 0 \)
\( \implies \) \( 20 - 4x - 2x - 2 = 0 \)
\( \implies \) \( 18 - 6x = 0 \)
\( \implies \) \( 6x = 18 \)
\( \implies \) \( x = \frac{18}{6} = 3 \)
\( \implies \) \( x = 3 \)
So, Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation of Assertion(A).
Hence, (a) is the correct option.

Question. Assertion (A) : If A is a \( 3 \times 3 \) matrix, \( |A| \neq 0 \) and \( |5A| = K|A| \), then the value of K = 125.
Reason (R) : If A be any square matrix of order \( n \times n \) and k be any scalar then \( |KA| = K^n|A| \).
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation for A.
Solution: We have, A is a square matrix of order \( 3 \times 3 \)
\( \therefore |5A| = 5^3|A| = 125|A| \)
\( \implies \) \( 125|A| = K|A| \)
\( \implies \) \( K = 125 \)
So, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Hence, (a) is the correct option.

Question. Assertion (A) : If \( \begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} \) then \( x = \pm 6 \).
Reason (R) : If A is a skew-symmetric matrix of odd order, then \( |A| = 0 \).
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (b) Both A and R are true and R is not the correct explanation for A.
Solution: We have, \( \begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} \)
\( \implies \) \( x^2 - 36 = 36 - 36 \)
\( \implies \) \( x^2 - 36 = 0 \)
\( \implies \) \( x^2 = 36 \)
\( \implies \) \( x = \pm 6 \)
Clearly, both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Hence, (b) is the correct option.

Question. Assertion (A) : If \( A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix} \), then \( C_{22} = 1 \), where \( C_{ij} \) denotes the co-factor of \( i^{\text{th}} \) row and \( j^{\text{th}} \) column.
Reason (R) : The co-factor \( C_{ij} \) of \( a_{ij} \) in the matrix \( A = [a_{ij}]_{n \times n} \) equal to \( (-1)^{i+j}M_{ij} \).
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true and R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation for A.
Solution: We have, \( A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix} \)
\( \therefore M_{22} = -9 + 10 = 1 \)
\( \therefore C_{22} = (-1)^{2+2} . M_{22} = 1 \times 1 = 1 \)
Clearly, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Hence, (a) is the correct option.