Answer
$AB \ne I_2$, so the given matrices are not inverses of each other.
Work Step by Step
Recall the property of a matrix that when two matrices are the inverses of each other, then their products are called as the identity matrix. Let us consider two matrices A and B are having $n \times n$ matrix form, then $A$ and $B$ are inverse of each other if $AB=I_n$ and $BA=I_n$
We are given the matrices $A=\begin{bmatrix} 2 & 1 \\ 3& 2 \end{bmatrix} $ and $B=\begin{bmatrix} 2 & 1 \\ -3& 2 \end{bmatrix} $ . Since, the matrices $A$ and $B$ are of the same dimensions $2 \times 2$. So, it is possible to multiply them and their product will also be a $2 \times 2$ matrix.
Therefore, we have:
$AB=\begin{bmatrix} 2 & 1 \\ 3& 2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ -3& 2 \end{bmatrix}\\=\begin{bmatrix} 4-3 & 2+2 \\ 6-6&3+4 \end{bmatrix} \\=\begin{bmatrix} 1 & 4 \\ 0&7 \end{bmatrix}\\ \ne I_2$
We can conclude that $AB \ne I_2$, so the given matrices are not inverses of each other.
