Let $A =\begin{bmatrix}2&3\\1&2\end{bmatrix}$ and $B = \begin{bmatrix}4&-6\\-2&4\end{bmatrix}$ then

(A) $det (A^T) = 1$
(B) $AB = I$, where I is the identity matrix of order 2.
(C) $A^{-1} =\begin{bmatrix}2&-3\\-1&2\end{bmatrix}$
(D) $adj (B) = \begin{bmatrix}4&2\\6&4\end{bmatrix}$

Choose the correct answer from the options given below:

(A) and (C) only

The correct answer is Option (2) → (B), (C) and (D) only

$A=\begin{bmatrix}2&3\\1&2\end{bmatrix},\ B=\begin{bmatrix}4&-6\\-2&4\end{bmatrix}$

(A) $\det(A^{T})=\det(A)=2\cdot2-3\cdot1=1$ → True ✅

(B) $AB=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}4&-6\\-2&4\end{bmatrix} =\begin{bmatrix}(8-6)&(-12+12)\\(4-4)&(-6+8)\end{bmatrix} =\begin{bmatrix}2&0\\0&2\end{bmatrix}\ne I$ → False ❌

(C) $A^{-1}=\frac{1}{\det A}\begin{bmatrix}2&-3\\-1&2\end{bmatrix}=\begin{bmatrix}2&-3\\-1&2\end{bmatrix}$ → True ✅

(D) $\text{adj}(B)=\begin{bmatrix}4&6\\2&4\end{bmatrix}$ (since $\text{adj}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$)

→ Given matrix $\begin{bmatrix}4&2\\6&4\end{bmatrix}$ ≠ correct → False ❌

Correct statements: (A) and (C)