This can be written in the form AX=B.
$\begin{bmatrix}5 & 2\\3 & 2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\5\end{bmatrix}$
Where $A=\begin{bmatrix}5 & 2\\3 & 2\end{bmatrix}\;X=\begin{bmatrix}x\\y\end{bmatrix}\;B=\begin{bmatrix}3\\5\end{bmatrix}$
Let us now check whether A is a singular or non-singular matrix,by finding the value of the determinant
$|A|=(5\times 2-2\times 3)$
$\;\;\;=10-6=4\neq 0$
Hence it is non -singular and therefore inverse exists.
To find the inverse $A^{-1}$,let us first find the adjoint A.
This can be obtained by interchanging the element of $a_{11}$ and $a_{22}$ and interchanging the symbols of elements of $a_{21}$ and $a_{12}$
Therefore $adj A=\begin{bmatrix}2 & -2\\-3 & 5\end{bmatrix}$ and |A|=4.
We know $A^{-1}=\frac{1}{|A|}(adj \;A)$
$\qquad\qquad=\frac{1}{4}\begin{bmatrix}2 & -2\\-3 & 5\end{bmatrix}$
Since AX=B
X=$A^{-1}B$
Let us find $A^{-1}B=\frac{1}{4}\begin{bmatrix}2 &-2\\-3 & 5\end{bmatrix}\begin{bmatrix}3\\5\end{bmatrix}$
We can do matrix multiplication by multiplying the rows of matrix A by the column of matrix B
$A^{-1}B=\frac{1}{4}\begin{bmatrix}6-10\\-9+25\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}$
$\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}= \frac{1}{4}\begin{bmatrix}-4\\16\end{bmatrix}$
$\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}-1\\4\end{bmatrix}$
Hence x=-1 and y=4.
