The A matrix is:
[tex]\begin{bmatrix}{5} & {3} & {} \\ {-6} & {-3} & {} \\ {} & {} & {}\end{bmatrix}[/tex]We can write the system of equations in matrix form like this:
[tex]\begin{bmatrix}{5} & {3} & {} \\ {-6} & {-3} & {} \\ {} & {} & {}\end{bmatrix}\cdot\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{8} & {} & {} \\ {-6} & {} & {} \\ {} & {} & {}\end{bmatrix}[/tex]And we can express that as this:
[tex]A\cdot X=B[/tex]Then the solution is:
[tex]X=A^{-1}\cdot B[/tex]Then, we need to find the inverse of the function to find the solution, start by calculating the determinant:
[tex]\begin{gathered} \text{ Determinant:} \\ d(A)=(5)\cdot(-3)-(-6)\cdot(3)=-15+18 \\ d(A)=3 \end{gathered}[/tex]The inverse function is:
[tex]\begin{gathered} A^{-1}=\frac{1}{\det(A)}\begin{bmatrix}{-3} & {-3} & {} \\ {6} & {5} & {} \\ {} & {} & {}\end{bmatrix}=\frac{1}{3}\begin{bmatrix}{-3} & {-3} & {} \\ {6} & {5} & {} \\ {} & {} & {}\end{bmatrix} \\ A^{-1}=\begin{bmatrix}{-3/3} & {-3/3} & {} \\ {6/3} & {5/3} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{-1} & {-1} & {} \\ {2} & {5/3} & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}[/tex]Thus, the solution is:
[tex]\begin{gathered} \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{-1} & {-1} & {} \\ {2} & {5/3} & {} \\ {} & {} & {}\end{bmatrix}\cdot\begin{bmatrix}{8} & {} & {} \\ {-6} & {} & {} \\ {} & {} & {}\end{bmatrix} \\ \end{gathered}[/tex]Now, solve the multiplication:
[tex]\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{-1\cdot8+(-1)(-6)} & {} & {} \\ {2(8)+5/3\cdot(-6)} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{-2} & {} & {} \\ {6} & {} & {} \\ {} & {} & {}\end{bmatrix}[/tex]