We are given the following 2x2 matrix
[tex]A=\begin{bmatrix}{-3} & {-2} \\ {4} & {8}\end{bmatrix}[/tex]
We are asked to find the inverse of matrix A.
Recall that the inverse of a 2x2 matrix is given by
[tex]A^{-1}=\frac{1}{ad-bc}\times\begin{bmatrix}{d} & {-b} \\ {-c} & {a}\end{bmatrix}[/tex]
Where
a = -3
b = -2
c = 4
d = 8
Let us substitute these values into the above equation
[tex]A^{-1}=\frac{1}{(-3)(8)-(-2)(4)}\times\begin{bmatrix}{8} & {-(-2)} \\ {-(4)} & {-3}\end{bmatrix}[/tex]
Now simplify
[tex]\begin{gathered} A^{-1}=\frac{1}{-24+8}\times\begin{bmatrix}{8} & {2} \\ {-4} & {-3}\end{bmatrix} \\ A^{-1}=\frac{1}{-16}\times\begin{bmatrix}{8} & {2} \\ {-4} & {-3}\end{bmatrix} \\ A^{-1}=\begin{bmatrix}{\frac{8}{-16}} & {\frac{2}{-16}} \\ {\frac{-4}{-16}} & {\frac{-3}{-16}}\end{bmatrix} \\ A^{-1}=\begin{bmatrix}{-\frac{1}{2}} & {-\frac{1}{8}} \\ {\frac{1}{4}} & {\frac{3}{16}}\end{bmatrix} \end{gathered}[/tex]
Therefore, the inverse of the matrix A is
[tex]A^{-1}=\begin{bmatrix}{-\frac{1}{2}} & {-\frac{1}{8}} \\ {\frac{1}{4}} & {\frac{3}{16}}\end{bmatrix}[/tex]
