Question

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\( - 2A\), \(B - 2A\), \(AC\), \(CD\).

Short answer

\( - 2A = \left( {\begin{aligned}{*{20}{c}}{ - 4}&0&2\\{ - 8}&{10}&{ - 4}\end{aligned}} \right)\),

\(B - 2A = \left( {\begin{aligned}{*{20}{c}}3&{ - 5}&3\\{ - 7}&6&{ - 7}\end{aligned}} \right)\),

\(AC\)is not defined and

\(CD = \left( {\begin{aligned}{*{20}{c}}1&{13}\\{ - 7}&{ - 6}\end{aligned}} \right)\).

Step-by-step solution

Find the matrix \( - 2A\)

The value of \( - 2A\) can be calculated as follows:

\(\begin{aligned}{c} - 2A = - 2\left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 4}&0&2\\{ - 8}&{10}&{ - 4}\end{aligned}} \right)\end{aligned}\)

Find the matrix \(B - 2A\)

The value of \(B - 2A\) can be calculated as follows:

\(\begin{aligned}{c}B - 2A = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right) - 2\left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}4&0&{ - 2}\\8&{ - 10}&4\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}3&{ - 5}&3\\{ - 7}&6&{ - 7}\end{aligned}} \right)\end{aligned}\)

Find the matrix \(AC\)

The order of matrix \(A\) is \(2 \times 3\), and the order of matrix \(C\) is \(2 \times 2\). The number of column of \(A\) does not match with the number of rows of \(C\). Therefore, the matrix \(AC\) is not defined.

Find the matrix \(CD\)

The product \(CD\) can be calculated as follows:

\(\begin{aligned}{c}CD = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right) \times \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{1 \times 3 + 2 \times \left( { - 1} \right)}&{1 \times 5 + 2 \times 4}\\{\left( { - 2} \right) \times 3 + 1 \times \left( { - 1} \right)}&{\left( { - 2} \right) \times 5 + 1 \times 4}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{3 - 2}&{5 + 8}\\{ - 6 - 1}&{ - 10 + 4}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&{13}\\{ - 7}&{ - 6}\end{aligned}} \right)\end{aligned}\)

So, \( - 2A = \left( {\begin{aligned}{*{20}{c}}{ - 4}&0&2\\{ - 8}&{10}&{ - 4}\end{aligned}} \right)\), \(B - 2A = \left( {\begin{aligned}{*{20}{c}}3&{ - 5}&3\\{ - 7}&6&{ - 7}\end{aligned}} \right)\), \(AC\) is not defined and \(CD = \left( {\begin{aligned}{*{20}{c}}1&{13}\\{ - 7}&{ - 6}\end{aligned}} \right)\).