Match List-I with List-II

List-I (Matrix A)	List-II (Determinant of Adjoint of A)
(A) $\begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$	(I) 9
(B) $\begin{bmatrix} 5 & -1 \\ 4 & 2 \end{bmatrix}$	(II) 8
(C) $\begin{bmatrix} 6 & -1 \\ 2 & 1 \end{bmatrix}$	(III) 14
(D) $\begin{bmatrix} 4 & 1 \\ 3 & 3 \end{bmatrix}$	(IV) 2

Choose the correct answer from the options given below:

For a 2×2 matrix A, the relationship between the determinant of the adjoint and the determinant of the matrix is:

$\text{det(adj(A))} = [\text{det}(A)]^{n-1}$

where n is the order of the matrix.

Since the matrices are 2×2, n = 2:

$\text{det(adj(A))} = [\text{det}(A)]^{2-1}$

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

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

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For (A) $\begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$:

$\text{det}(A) = (3)(2) - (1)(4)$

$\text{det}(A) = 6 - 4$

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

Matches with (IV)

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For (B) $\begin{bmatrix} 5 & -1 \\ 4 & 2 \end{bmatrix}$:

$\text{det}(B) = (5)(2) - (-1)(4)$

$\text{det}(B) = 10 + 4$

$\text{det}(B) = 14$

Matches with (III)

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For (C) $\begin{bmatrix} 6 & -1 \\ 2 & 1 \end{bmatrix}$:

$\text{det}(C) = (6)(1) - (-1)(2)$

$\text{det}(C) = 6 + 2$

$\text{det}(C) = 8$

Matches with (II)

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For (D) $\begin{bmatrix} 4 & 1 \\ 3 & 3 \end{bmatrix}$:

$\text{det}(D) = (4)(3) - (1)(3)$

$\text{det}(D) = 12 - 3$

$\text{det}(D) = 9$

Matches with (I)

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The correct matching is:

(A) → (IV)

(B) → (III)

(C) → (II)

(D) → (I)