Given that $A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}$, matrix $A$ is

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

The correct answer is Option (2) → $\begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}$ ##

We know that,

$(A^{-1})^{-1} = A$

Given, $A^{-1} = \begin{bmatrix} \frac{2}{7} & \frac{1}{7} \\ \frac{-3}{7} & \frac{2}{7} \end{bmatrix}$

$\Rightarrow A = \frac{\text{adj}(A^{-1})}{|A^{-1}|}$

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