Question

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.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{1}}}\\{\bf{5}}&{ - {\bf{2}}}\end{aligned}} \right)\). Compute \({\bf{3}}{I_{\bf{2}}} - A\) and \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\).

Short answer

\(\left( {\begin{aligned}{*{20}{c}}{ - 1}&1\\{ - 5}&5\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}{12}&{ - 3}\\{15}&{ - 6}\end{aligned}} \right)\)

Step-by-step solution

Find the matrix \(3{I_{\bf{2}}} - A\)

The value of \(3{I_2} - A\) can be calculated as follows:

\(\begin{aligned}{c}3{I_2} - A = 3\left( {\begin{aligned}{*{20}{c}}1&0\\0&1\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}3&0\\0&3\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 1}&1\\{ - 5}&5\end{aligned}} \right)\end{aligned}\)

Find the matrix \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\)

The value of \(\left( {3{I_2}} \right)A\) can be calculated as follows:

\(\begin{aligned}{c}\left( {3{I_2}} \right)A = 3\left( {{I_2}A} \right)\\ = 3\left( {\begin{aligned}{*{20}{c}}1&0\\0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = 3\left( {\begin{aligned}{*{20}{c}}{1 \times 4 + 0}&{1 \times \left( { - 1} \right) + 0}\\{0 + 1 \times 5}&{0 \times \left( { - 1} \right) + 1 \times \left( { - 2} \right)}\end{aligned}} \right)\\ = 3\left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{12}&{ - 3}\\{15}&{ - 6}\end{aligned}} \right)\end{aligned}\)

So, \(3{I_2} - A = \left( {\begin{aligned}{*{20}{c}}{ - 1}&1\\{ - 5}&5\end{aligned}} \right)\), and \(\left( {3{I_2}} \right)A = \left( {\begin{aligned}{*{20}{c}}{12}&{ - 3}\\{15}&{ - 6}\end{aligned}} \right)\).