Algebra

Matrices & Determinants

Medium

core

If $A = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix}$ , then

Match List-I with List-II

List-I List-II
(A) det (A) (I) $-\frac{1}{3}$
(B) det $(A^{-1})$ (II) $-12$
(C) det (2A) (III) $-3$
(D) det $(3A^T)$ (IV) $-27$

Choose the correct answer from the options given below:

List-I	List-II
(A) det (A)	(I) $-\frac{1}{3}$
(B) det $(A^{-1})$	(II) $-12$
(C) det (2A)	(III) $-3$
(D) det $(3A^T)$	(IV) $-27$

(A) - (I), (B) - (III), (C) - (II), (D) - (IV)

(A) - (III), (B) - (I), (C) - (II), (D) - (IV)

(A) - (III), (B) - (I), (C) - (IV), (D) - (II)

(A) - (I), (B) - (IV), (C) - (II), (D) - (III)

✅ Correct Option: 2

Given $A = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix}$

For a 2×2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the determinant is $(a \times d) - (b \times c)$ .

det(A) $= (1 \times 5) - (2 \times 4)$

det(A) $= 5 - 8$

det(A) $= -3$

(A) matches with (III)

For the inverse matrix: det(A⁻¹) $= \frac{1}{\text{det}(A)}$

det(A⁻¹) $= \frac{1}{-3}$

det(A⁻¹) $= -\frac{1}{3}$

(B) matches with (I)

When multiplying a matrix by scalar $k$ , the property det( $kA$ ) $= k^n \times$ det( $A$ ) applies, where $n$ is the order of the matrix.

For a 2×2 matrix with scalar 2:

det(2A) $= 2^2 \times$ det(A)

det(2A) $= 4 \times (-3)$

det(2A) $= -12$

The transpose property: det( $A^T$ ) $=$ det( $A$ ) $= -3$

Combining with the scalar property:

det(3 $A^T$ ) $= 3^2 \times$ det( $A^T$ )

det(3 $A^T$ ) $= 9 \times (-3)$

det(3 $A^T$ ) $= -27$

(D) matches with (IV)

List-I	Matches	List-II
(A) det(A) $= -3$	→	(III)
(B) det(A⁻¹) $= -\frac{1}{3}$	→	(I)
(C) det(2A) $= -12$	→	(II)
(D) det(3 $A^T$ ) $= -27$	→	(IV)

2026: 25 May Shift 1

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2025: 30 May Shift 2

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2026: 29 May Shift 1

Which of the following are correct?

(A) If $A=\begin{bmatrix}7 & 5 \\ 4 & 3\end{bmatrix}$ , then $A^{-1}=\begin{bmatrix}3 & -5 \\ -4 & 7\end{bmatrix}$

(B) If $A=\begin{bmatrix}3 & 5 \\ 6 & 10\end{bmatrix}$ , then $A$ is a non-singular matrix

(D) If $A=\begin{bmatrix}6 & x \\ y & 0\end{bmatrix}$ and $A=A^T$ , then $x=y$

Choose the correct answer from the options given below:

2025: 3 June Shift 2

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2025: 16 May Shift 1

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2025: 15 May Shift 2

If $A = \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix}$ , then the matrix $A^2 - 6A + 14$ I is (where I is an identity matrix of order 2)

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