Answer:
[tex]x\text{ = }\frac{11}{3},\text{ y = 7}[/tex]Explanation:
Here, we want to solve the given simultaneous equation by matrix notation
Mathematically, we have it that:
[tex]\begin{bmatrix}{-6} & {2} \\ {3} & {-2}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}\text{ = }\begin{bmatrix}{-8} & {} \\ {-3} & {}\end{bmatrix}[/tex][tex]\begin{gathered} if\text{ matrix A = }\begin{bmatrix}{a} & {b} & {} & {} \\ {c} & {d} & {} & {} \\ {} & {} & {} & {} \\ {} & {} & {} & {}\end{bmatrix} \\ \\ Then\text{ A}^{-1}\text{ = }\frac{1}{ad-bc}\begin{bmatrix}{d} & {-b} \\ {-c} & {a}\end{bmatrix} \end{gathered}[/tex]where in this case: a = -6 , b = 2 , c = 3 and d = -2
Substituting the values, we have it that:
[tex]\begin{gathered} A^{-1}\text{ = }\frac{1}{-6(-2)-2(3)}\text{ }\begin{bmatrix}{-2} & {-2} \\ {-3} & {-6}\end{bmatrix} \\ \\ A^{-1}\text{ = }\frac{1}{6}\begin{bmatrix}{-2} & {-2} \\ {-3} & {-6}\end{bmatrix} \end{gathered}[/tex]We proceed as follows:
Now, we multiply the inverse with the solution vector as follows:
[tex]\begin{gathered} \begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\text{ }\frac{1}{6}\begin{bmatrix}{-2} & {-2} \\ {-3} & {-6}\text{ }\end{bmatrix}\begin{bmatrix}{-8} & {} \\ {-3} & {}\end{bmatrix} \\ \\ \end{gathered}[/tex]Finally, we have the product as:
[tex]\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}\text{ = }\begin{bmatrix}{\frac{11}{3}} & {} \\ {7} & {}\end{bmatrix}[/tex]
