Answer
\[\left[ {\begin{array}{*{20}{c}}
{ - 1/5}&{ - 2/5} \\
{2/5}&{ - 1/5}
\end{array}} \right]\]
Work Step by Step
\[\begin{gathered}
{\text{A}} = \left[ {\begin{array}{*{20}{c}}
{ - 1}&2 \\
{ - 2}&{ - 1}
\end{array}} \right] \hfill \\
{\text{Let a matrix A}} = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right],{\text{ its inverse is: }}{{\text{A}}^{ - 1}} = \frac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}
d&{ - b} \\
{ - c}&a
\end{array}} \right] \hfill \\
{\text{Then}} \hfill \\
{{\text{A}}^{ - 1}} = \frac{1}{{\left( { - 1} \right)\left( { - 1} \right) - \left( 2 \right)\left( { - 2} \right)}}\left[ {\begin{array}{*{20}{c}}
{ - 1}&{ - 2} \\
2&{ - 1}
\end{array}} \right] \hfill \\
{\text{Simplifying}} \hfill \\
{{\text{A}}^{ - 1}} = \frac{1}{5}\left[ {\begin{array}{*{20}{c}}
{ - 1}&{ - 2} \\
2&{ - 1}
\end{array}} \right] \hfill \\
{{\text{A}}^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
{ - 1/5}&{ - 2/5} \\
{2/5}&{ - 1/5}
\end{array}} \right] \hfill \\
\end{gathered} \]
