In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Compute \(A - {\bf{5}}{I_{\bf{3}}}\) and \(\left( {{\bf{5}}{I_{\bf{3}}}} \right)A\)

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{9}}&{ - {\bf{1}}}&{\bf{3}}\\{ - {\bf{8}}}&{\bf{7}}&{ - {\bf{6}}}\\{ - {\bf{4}}}&{\bf{1}}&{\bf{8}}\end{aligned}} \right)\)

The value of \(A - 5{I_3}\) can be calculated as follows:

\(\begin{aligned}{c}A - 5{I_3} = \left( {\begin{aligned}{*{20}{c}}9&{ - 1}&3\\{ - 8}&7&{ - 6}\\{ - 4}&1&8\end{aligned}} \right) - 5\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{9 - 5}&{ - 1 - 0}&{3 - 0}\\{ - 8 - 0}&{7 - 5}&{ - 6 - 0}\\{ - 4 - 0}&{1 - 0}&{8 - 5}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}4&{ - 1}&3\\{ - 8}&2&{ - 6}\\{ - 4}&1&3\end{aligned}} \right)\end{aligned}\)

The value of \(\left( {5{I_3}} \right)A\) can be calculated as follows:

\(\begin{aligned}{c}\left( {5{I_3}} \right)A = 5\left( {{I_3}A} \right)\\ = 5\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}9&{ - 1}&3\\{ - 8}&7&{ - 6}\\{ - 4}&1&8\end{aligned}} \right)\\ = 5\left( {\begin{aligned}{*{20}{c}}9&{ - 1}&3\\{ - 8}&7&{ - 6}\\{ - 4}&1&8\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{45}&{ - 5}&{15}\\{ - 40}&{35}&{ - 30}\\{ - 20}&5&{40}\end{aligned}} \right)\end{aligned}\)

So, \(A - 5{I_3} = \left( {\begin{aligned}{*{20}{c}}4&{ - 1}&3\\{ - 8}&2&{ - 6}\\{ - 4}&1&3\end{aligned}} \right)\), and \(\left( {5{I_3}} \right)A = \left( {\begin{aligned}{*{20}{c}}{45}&{ - 5}&{15}\\{ - 40}&{35}&{ - 30}\\{ - 20}&5&{40}\end{aligned}} \right)\).