Question

1. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{6}}\\{\bf{5}}&{\bf{4}}\end{aligned}} \right)\).

Short answer

The inverse of \(\left( {\begin{aligned}{*{20}{c}}8&6\\5&4\end{aligned}} \right)\) is \(\left( {\begin{aligned}{*{20}{c}}2&{ - 3}\\{ - 2.5}&4\end{aligned}} \right)\).

Step-by-step solution

Check if the matrix is invertible

\(\begin{aligned}{c}\det \left( {\left( {\begin{aligned}{*{20}{c}}8&6\\5&4\end{aligned}} \right)} \right) = 8\left( 4 \right) - 6\left( 5 \right)\\ = 32 - 30\\\det \left( {\left( {\begin{aligned}{*{20}{c}}8&6\\5&4\end{aligned}} \right)} \right) = 2 \ne 0\end{aligned}\)

This implies that\(\left( {\begin{aligned}{*{20}{c}}8&6\\5&4\end{aligned}} \right)\)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

\(\begin{aligned}{c}{\left( {\begin{aligned}{*{20}{c}}8&6\\5&4\end{aligned}} \right)^{ - 1}} = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}4&{ - 6}\\{ - 5}&8\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}2&{ - 3}\\{ - \frac{5}{2}}&4\end{aligned}} \right)\\{\left( {\begin{aligned}{*{20}{c}}8&6\\5&4\end{aligned}} \right)^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}2&{ - 3}\\{ - 2.5}&4\end{aligned}} \right)\end{aligned}\)