The coefficient matrix A is,
[tex]A=\begin{bmatrix}-4 & 10 \\ -1 & 2\end{bmatrix}[/tex]The constant matrix B is,
[tex]B=\begin{bmatrix}-2 \\ -3\end{bmatrix}[/tex]We have the equation,
[tex]AX=B[/tex]Solving for X, we have,
[tex]\begin{gathered} AX=B \\ X=A^{-1}B \end{gathered}[/tex]Now, if we have a 2 x 2 matrix of the form,
[tex]A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex]The inverse of this matrix is given by the formula,
[tex]A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}[/tex]So, let's find the inverse matrix using this formula. Shown below:
[tex]\begin{gathered} A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix} \\ A^{-1}=\frac{1}{-8--10}\begin{bmatrix}2 & -10 \\ 1 & -4\end{bmatrix} \\ A^{-1}=\frac{1}{2}\begin{bmatrix}2 & -10 \\ 1 & -4\end{bmatrix} \\ A^{-1}=\begin{bmatrix}1 & -5 \\ \frac{1}{2} & -2\end{bmatrix} \end{gathered}[/tex]Now, we calculate X,
[tex]\begin{gathered} X=A^{-1}B \\ X=\begin{bmatrix}1 & -5 \\ \frac{1}{2} & -2\end{bmatrix}\begin{bmatrix}-2 \\ -3\end{bmatrix} \\ X=\begin{bmatrix}(1)(-2)+(-5)(-3) \\ (\frac{1}{2})(-2)+(-2)(-3)\end{bmatrix} \\ X=\begin{bmatrix}-2+15 \\ -1+6\end{bmatrix} \\ X=\begin{bmatrix}13 \\ 5\end{bmatrix} \end{gathered}[/tex]Thus, the solution of the system is >>>
x = 13y = 5