Solve system of linear equations, using matrix method.

The given system of equations can be written in the form of AX = B, where

$A=\left[\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right], X=\left[\begin{array}{c}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}-2 \\ 3\end{array}\right]$.

Now,

$|A|=8+3=11 \neq 0$

Thus, A is non-singular. Therefore, its inverse exists.

Now,

$A^{-1}=\frac{1}{|A|}$ adj $A=\frac{1}{11}\left[\begin{array}{cc}4 & 1 \\ -3 & 2\end{array}\right]$

$\therefore X=A^{-1} B=\frac{1}{11}\left[\begin{array}{cc}4 & 1 \\ -3 & 2\end{array}\right]\left[\begin{array}{c}-2 \\ 3\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}-8+3 \\ 6+6\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}-5 \\ 12\end{array}\right]=\left[\begin{array}{c}-\frac{5}{11} \\ \frac{12}{11}\end{array}\right]$

Hence, $x=\frac{-5}{11}$ and $y=\frac{12}{11}$.