Answer
\[\left[ {\begin{array}{*{20}{c}}
3&{ - 1} \\
{ - 5}&2
\end{array}} \right]\]
Work Step by Step
\[\begin{gathered}
\left[ {\begin{array}{*{20}{c}}
2&1 \\
5&3
\end{array}} \right] \hfill \\
{\text{Let a matrix A}} = \left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right],{\text{ its inverse is: }}{{\text{A}}^{ - 1}} = \frac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}
d&{ - b} \\
{ - c}&a
\end{array}} \right] \hfill \\
{\text{Then}} \hfill \\
{{\text{A}}^{ - 1}} = \frac{1}{{\left( 2 \right)\left( 3 \right) - \left( 1 \right)\left( 5 \right)}}\left[ {\begin{array}{*{20}{c}}
3&{ - 1} \\
{ - 5}&2
\end{array}} \right] \hfill \\
{\text{Simplifying}} \hfill \\
{{\text{A}}^{ - 1}} = \frac{1}{{6 - 5}}\left[ {\begin{array}{*{20}{c}}
3&{ - 1} \\
{ - 5}&2
\end{array}} \right] \hfill \\
{{\text{A}}^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
3&{ - 1} \\
{ - 5}&2
\end{array}} \right] \hfill \\
\end{gathered} \]
