Match List-I with List-II List-I (Matrix A) List-II (Determinant of a…

The correct answer is Option (4) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

List-I (Matrix A)	List-II (Determinant of adj A)
(A) $\begin{bmatrix}2&1\\-1&2 \end{bmatrix}$	(III) 5
(B) $\begin{bmatrix}3&4\\3&6 \end{bmatrix}$	(II) 6
(C) $\begin{bmatrix}3&7\\-2&-4 \end{bmatrix}$	(IV) 2
(D) $\begin{bmatrix}4&3\\3&3\end{bmatrix}$	(I) 3

Formula: For a square matrix $A$ of order $n$, $\det(\text{adj } A) = (\det A)^{n-1}$

(A) $A = \begin{bmatrix}2 & 1 \\ -1 & 2\end{bmatrix}$, $\det A = 2*2 - (-1*1) = 4 + 1 = 5$, $n=2$

So, $\det(\text{adj } A) = (\det A)^{2-1} = 5^1 = 5 \Rightarrow$ matches (III)

(B) $A = \begin{bmatrix}3 & 4 \\ 3 & 6\end{bmatrix}$, $\det A = 3*6 - 3*4 = 18 - 12 = 6$, $n=2$

So, $\det(\text{adj } A) = 6^1 = 6 \Rightarrow$ matches (II)

(C) $A = \begin{bmatrix}3 & 7 \\ -2 & -4\end{bmatrix}$, $\det A = 3*(-4) - (-2*7) = -12 + 14 = 2$, $n=2$

So, $\det(\text{adj } A) = 2^1 = 2 \Rightarrow$ matches (IV)

(D) $A = \begin{bmatrix}4 & 3 \\ 3 & 3\end{bmatrix}$, $\det A = 4*3 - 3*3 = 12 - 9 = 3$, $n=2$

So, $\det(\text{adj } A) = 3^1 = 3 \Rightarrow$ matches (I)