Answer :
Given:
[tex]\begin{gathered} -2x-4y=1 \\ -6x-9y=-8 \end{gathered}[/tex]To find:
The solutions
Explanation:
It can be written as,
[tex]\begin{bmatrix}{-2} & {-4} \\ {-6} & {-9}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{1} & {} \\ {-8} & {}\end{bmatrix}[/tex]It is of the form,
[tex]\begin{gathered} AX=B \\ X=A^{-1}B \end{gathered}[/tex]Let us find the inverse matrix of A,
[tex]\begin{gathered} A^{-1}=\frac{1}{|A|}adjA \\ =\frac{1}{[-2(-9)-(-4)(-6)]}\begin{bmatrix}{-9} & 4 \\ {6} & {-2}\end{bmatrix}^ \\ A^{-1}=\frac{1}{-6}\begin{bmatrix}{-9} & 4{} \\ {6} & {-2}\end{bmatrix} \\ A^{-1}=\frac{-1}{6}\begin{bmatrix}{-9} & {4} \\ {6} & {-2}\end{bmatrix} \end{gathered}[/tex]Then, find the matrix X,
[tex]\begin{gathered} X=A^{-1}B \\ \begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=-\frac{1}{6}\begin{bmatrix}{-9} & {4} \\ {6} & {-2}\end{bmatrix}\begin{bmatrix}{1} & {} \\ {-8} & {}\end{bmatrix} \\ =-\frac{1}{6}\begin{bmatrix}{-9-32} & {} \\ {6+16} & {}\end{bmatrix} \\ =-\frac{1}{6}\begin{bmatrix}{-41} & {} \\ {22} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{\frac{41}{6}} & {} \\ -{\frac{11}{3}} & {}\end{bmatrix} \end{gathered}[/tex]Therefore, the solutions are,
[tex]\begin{gathered} x=\frac{41}{6} \\ y=-\frac{11}{3} \end{gathered}[/tex]Final answer:
The inverse matrix of A is,
[tex]\frac{1}{A}=\frac{-1}{6}\begin{bmatrix}{-9} & {4} \\ {6} & {-2}\end{bmatrix}[/tex]The solutions are,
[tex]\begin{gathered} x=\frac{41}{6} \\ y=-\frac{11}{3} \end{gathered}[/tex]