Match List-I with List-II

List-I	List-II
(Matrix A)	(Determinant of adj A)
(A) $\begin{bmatrix} 2 & 1 \\ 0 & -1 \end{bmatrix}$	(I) 6
(B) $\begin{bmatrix} 0 & 1 \\ 4 & -1 \end{bmatrix}$	(II) 5
(C) $\begin{bmatrix} 1 & 2 \\ -3 & -1 \end{bmatrix}$	(III) -4
(D) $\begin{bmatrix} 4 & -2 \\ 3 & 0 \end{bmatrix}$	(IV) -2

Choose the correct answer from the options given below:

For any 2×2 matrix A:

$\text{det(adj A)} = \text{det(A)}$

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

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Matrix A = $\begin{bmatrix} 2 & 1 \\ 0 & -1 \end{bmatrix}$

$\text{det(A)} = (2)(-1) - (1)(0)$

$\text{det(A)} = -2 - 0$

$\text{det(A)} = -2$

Therefore, $\text{det(adj A)} = -2$, which matches with (IV).

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Matrix B = $\begin{bmatrix} 0 & 1 \\ 4 & -1 \end{bmatrix}$

$\text{det(B)} = (0)(-1) - (1)(4)$

$\text{det(B)} = 0 - 4$

$\text{det(B)} = -4$

Therefore, $\text{det(adj B)} = -4$, which matches with (III).

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Matrix C = $\begin{bmatrix} 1 & 2 \\ -3 & -1 \end{bmatrix}$

$\text{det(C)} = (1)(-1) - (2)(-3)$

$\text{det(C)} = -1 + 6$

$\text{det(C)} = 5$

Therefore, $\text{det(adj C)} = 5$, which matches with (II).

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Matrix D = $\begin{bmatrix} 4 & -2 \\ 3 & 0 \end{bmatrix}$

$\text{det(D)} = (4)(0) - (-2)(3)$

$\text{det(D)} = 0 + 6$

$\text{det(D)} = 6$

Therefore, $\text{det(adj D)} = 6$, which matches with (I).

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The correct matching is:

(A) → (IV)

(B) → (III)

(C) → (II)

(D) → (I)