Question

solve the equation
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Answer 1

For a matrix $A=\begin{vmatrix}a & b \\c & d\end{vmatrix}$,  $A^{-1}=\dfrac{1}{\det A}\begin{vmatrix}d & -b \\-c & a\end{vmatrix}$.

$A=\begin{vmatrix}-1 & 5 \\-3 & 2\end{vmatrix}\\\\\det A=-2-(-15)=13\implies A^{-1}\text{ exists}\\\\A^{-1}=\dfrac{1}{13}\begin{vmatrix}2 & -5 \\3 & -1\end{vmatrix}=\begin{vmatrix}\dfrac{2}{13} & -\dfrac{5}{13} \\[1em]\dfrac{3}{13} & -\dfrac{1}{13}\end{vmatrix}$