Respuesta :
Given:
The system of equations are
[tex]\begin{gathered} 2x-3y=-10 \\ 3x-6y=-8 \end{gathered}[/tex]Required:
To solve the given system of equations using an inverse matrix.
Explanation:
Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.
[tex]\begin{gathered} A=\begin{bmatrix}{2} & {-3} \\ {3} & {-6}\end{bmatrix} \\ X=\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix} \\ B=\begin{bmatrix}{-10} & {} \\ {-8} & {}\end{bmatrix} \end{gathered}[/tex]Then
[tex]\begin{bmatrix}{2} & {-3} \\ {3} & {-6}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{-10} & {} \\ {-8} & {}\end{bmatrix}----(1)[/tex]First, we need to calculate inverse of A. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:
[tex]\begin{gathered} A^{-1}=\frac{1}{|A|}\begin{bmatrix}{-6} & {3} \\ {-3} & {2}\end{bmatrix} \\ \\ =\frac{1}{-12-(-9)}\begin{bmatrix}{-6} & {3} \\ {-3} & {2}\end{bmatrix} \\ \\ =\frac{1}{-3}\begin{bmatrix}{-6} & {3} \\ {-3} & {2}\end{bmatrix} \\ \\ =\begin{bmatrix}{2} & {-1} \\ {1} & {-\frac{2}{3}}\end{bmatrix} \end{gathered}[/tex]Now we are ready to solve. Multiply both sides of the equation by inverse of A,
[tex]\begin{bmatrix}{2} & {-1} \\ {1} & {-\frac{2}{3}}\end{bmatrix}\begin{bmatrix}{2} & {-3} \\ {3} & {-6}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{2} & {-1} \\ {1} & {-\frac{2}{3}}\end{bmatrix}\begin{bmatrix}{-10} & {} \\ {-8} & {}\end{bmatrix}[/tex][tex]\begin{bmatrix}{4-3} & {-6+6} \\ {2-2} & {-3+4}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{-20+8} & {} \\ {-10+\frac{16}{3}} & {}\end{bmatrix}[/tex][tex]\begin{gathered} \begin{bmatrix}{1} & {0} \\ {0} & {1}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{-12} & {} \\ {-\frac{14}{3}} & {}\end{bmatrix} \\ \\ \begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{-12} & {} \\ {-\frac{14}{3}} & {}\end{bmatrix} \end{gathered}[/tex]Final Answer:
The solution is
[tex]\begin{gathered} x=-12 \\ \\ y=-\frac{14}{3} \end{gathered}[/tex]