Question

Find the matrix C whose inverse is \({C^{ - {\bf{1}}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{\bf{5}}\\{\bf{6}}&{\bf{7}}\end{aligned}} \right)\).

Short answer

Matrix \(C = \left( {\begin{aligned}{*{20}{c}}{{{ - 7} \mathord{\left/

{\vphantom {{ - 7} 2}} \right.

\kern-\nulldelimiterspace} 2}}&{{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}}\\3&{ - 2}\end{aligned}} \right)\).

Step-by-step solution

Use the inverse property

Note that C is invertible. Since \({C^{ - 1}}\) exists, therefore,

\({\left( {{C^{ - 1}}} \right)^{ - 1}} = C\)

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 \(ad - bc \ne 0\).

Find matrix C

\(\begin{aligned}{c}C = {\left( {\begin{aligned}{*{20}{c}}4&5\\6&7\end{aligned}} \right)^{ - 1}}\\ = \frac{1}{{4(7) - 5\left( 6 \right)}}\left( {\begin{aligned}{*{20}{c}}7&{ - 5}\\{ - 6}&4\end{aligned}} \right)\\ = \frac{1}{{28 - 30}}\left( {\begin{aligned}{*{20}{c}}7&{ - 5}\\{ - 6}&4\end{aligned}} \right)\\ = - \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}7&{ - 5}\\{ - 6}&4\end{aligned}} \right)\\C = \left( {\begin{aligned}{*{20}{c}}{{{ - 7} \mathord{\left/

{\vphantom {{ - 7} 2}} \right.

\kern-\nulldelimiterspace} 2}}&{{5 \mathord{\left/

{\vphantom {5 2}} \right.

\kern-\nulldelimiterspace} 2}}\\3&{ - 2}\end{aligned}} \right)\end{aligned}\)