$\left|A\right|=2\times \left(-2\right)-1\times 3=-7$

since $\left|A\right|\ne 0$ then $A^{-1}$ exists.

$A^{-1}=\frac{1}{\left(-7\right)}\left[\begin{array}{cc}\hfill -2& \hfill -3\\ \hfill -1& \hfill 2\end{array}\right]=\frac{1}{7}\left[\begin{array}{cc}\hfill 2& \hfill 3\\ \hfill 1& \hfill -2\end{array}\right]$

$X=A^{-1}B=\frac{1}{7}\left[\begin{array}{cc}\hfill 2& \hfill 3\\ \hfill 1& \hfill -2\end{array}\right]\left[\begin{array}{c}\hfill 5\\ \hfill -1\end{array}\right]=\frac{1}{7}\left[\begin{array}{c}\hfill 7\hfill \\ \hfill 7\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right]$ . Hence $x_1=1,x_2=1.$

Task!

$AX=B$ is $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 4\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 11\hfill \end{array}\right]$

$\left|A\right|=2\times 4-3\times 5=-7$   and   $A^{-1}=-\frac{1}{7}\left[\begin{array}{cc}\hfill 4\hfill & \hfill -3\hfill \\ \hfill -5\hfill & \hfill 2\hfill \end{array}\right]$

Using $X=A^{-1}B$ :

$X=\left[\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right]=-\frac{1}{7}\left[\begin{array}{cc}\hfill 4\hfill & \hfill -3\hfill \\ \hfill -5\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 11\hfill \end{array}\right]=-\frac{1}{7}\left[\begin{array}{c}\hfill 4\times 3-3\times 11\hfill \\ \hfill -5\times 3+2\times 11\hfill \end{array}\right]=-\frac{1}{7}\left[\begin{array}{c}\hfill -21\hfill \\ \hfill 7\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 3\hfill \\ \hfill -1\hfill \end{array}\right]$

So $x_1=3,x_2=-1$