Question

2. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{\bf{2}}\\{\bf{7}}&{\bf{4}}\end{aligned}} \right)\).

Short answer

The inverse of \(\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)\) is \(\left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\{3.5}&{ - 1.5}\end{aligned}} \right)\).

Step-by-step solution

Check if the matrix is invertible

\(\begin{aligned}{c}\det \left( {\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)} \right) = 3\left( 4 \right) - 2\left( 7 \right)\\ = 12 - 14\\\det \left( {\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)} \right) = - 2 \ne 0\end{aligned}\)

This implies that\(\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)\)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

\(\begin{aligned}{c}{\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)^{ - 1}} = \frac{1}{{ - 2}}\left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 7}&3\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\{\frac{7}{2}}&{ - \frac{3}{2}}\end{aligned}} \right)\\{\left( {\begin{aligned}{*{20}{c}}3&2\\7&4\end{aligned}} \right)^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}{ - 2}&1\\{3.5}&{ - 1.5}\end{aligned}} \right)\end{aligned}\)