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

(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:

Checking Option (A): $\det(A^T) = 1$

We know that $\det(A^T) = \det(A)$ always. So let's find $\det(A)$:

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

Therefore, $\det(A^T) = \det(A) = 1$ ✅

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Checking Option (B): $AB = I$

$AB = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 4 & -6 \\ -2 & 4 \end{bmatrix}$

$= \begin{bmatrix} (2)(4)+(3)(-2) & (2)(-6)+(3)(4) \\ (1)(4)+(2)(-2) & (1)(-6)+(2)(4) \end{bmatrix}$

$= \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

$= 2I \neq I$ ❌

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Checking Option (C): $A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is:

$A^{-1} = \dfrac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

This means: swap the diagonal elements, flip the signs of the off-diagonal elements, and divide by the determinant.

Since $\det(A) = 1$:

$A^{-1} = \dfrac{1}{1} \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$

This matches! ✅

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Checking Option (D): $\text{adj}(B) = \begin{bmatrix} 4 & 2 \\ 6 & 4 \end{bmatrix}$

For a $2 \times 2$ matrix, $\text{adj}(B) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ (same swap-and-negate step as the inverse, just without dividing by the determinant).

For $B = \begin{bmatrix} 4 & -6 \\ -2 & 4 \end{bmatrix}$:

$\text{adj}(B) = \begin{bmatrix} 4 & 6 \\ 2 & 4 \end{bmatrix}$

The given option has $6$ and $2$ in the wrong positions. ❌

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Options (A) and (C) are correct.