3. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{5}}\\{ - {\bf{7}}}&{ - {\bf{5}}}\end{aligned}} \right)\).

\(\begin{aligned}{c}\det \left( {\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)} \right) = 8\left( { - 5} \right) - 5\left( { - 7} \right)\\ = - 40 + 35\\\det \left( {\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)} \right) = - 5 \ne 0\end{aligned}\)

This implies that\(\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)\)is invertible.

\({\left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{aligned}{*{20}{c}}d&{ - b}\\{ - c}&a\end{aligned}} \right)\) when \(ad - bc \ne 0\).

\(\begin{aligned}{c}{\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)^{ - 1}} = \frac{1}{{ - 5}}\left( {\begin{aligned}{*{20}{c}}{ - 5}&{ - 5}\\7&8\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&1\\{ - \frac{7}{5}}&{ - \frac{8}{5}}\end{aligned}} \right)\\{\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}1&1\\{ - 1.4}&{ - 1.6}\end{aligned}} \right)\end{aligned}\)