Question

Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{4}}}\\{\bf{7}}&{ - {\bf{8}}}\end{aligned}} \right)\).

Short answer

The inverse of \(\left( {\begin{aligned}{*{20}{c}}3&{ - 4}\\7&{ - 8}\end{aligned}} \right)\) is \(\left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\{ - 1.75}&{0.75}\end{aligned}} \right)\).

Step-by-step solution

Check if the matrix is invertible

\(\begin{aligned}{c}\det \left( {\left( {\begin{aligned}{*{20}{c}}3&{ - 4}\\7&{ - 8}\end{aligned}} \right)} \right) = 3\left( { - 8} \right) - \left( { - 4} \right)\left( 7 \right)\\ = - 24 + 28\\\det \left( {\left( {\begin{aligned}{*{20}{c}}3&{ - 4}\\7&{ - 8}\end{aligned}} \right)} \right) = 4 \ne 0\end{aligned}\)

This implies that\(\left( {\begin{aligned}{*{20}{c}}3&{ - 4}\\7&{ - 8}\end{aligned}} \right)\)is invertible.

Use the formula

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

Write the inverse matrix

\(\begin{aligned}{c}{\left( {\begin{aligned}{*{20}{c}}3&{ - 4}\\7&{ - 8}\end{aligned}} \right)^{ - 1}} = \frac{1}{4}\left( {\begin{aligned}{*{20}{c}}{ - 8}&4\\{ - 7}&3\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\{ - \frac{7}{4}}&{\frac{3}{4}}\end{aligned}} \right)\\{\left( {\begin{aligned}{*{20}{c}}3&{ - 4}\\7&{ - 8}\end{aligned}} \right)^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\{ - 1.75}&{0.75}\end{aligned}} \right)\end{aligned}\)