Question

In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).

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

Short answer

\(\left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\)

Step-by-step solution

Find the value of \(A{b_{\bf{1}}}\)

Multiply matrix \(A\) with the first column of matrix \(B\).

\(\begin{aligned}{c}A{b_1} = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1\\2\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 \times 1 + \left( { - 2} \right) \times 2}\\{\left( { - 3} \right) \times 1 + 0}\\{3 \times 1 + 5 \times 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0\\{ - 3}\\{13}\end{aligned}} \right)\end{aligned}\)

Find the value of \(A{b_{\bf{2}}}\)

Multiply matrix \(A\) with the second column of matrix \(B\).

\(\begin{aligned}{c}A{b_2} = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 \times 3 + \left( { - 2} \right) \times \left( { - 1} \right)}\\{\left( { - 3} \right) \times 3 + 0}\\{3 \times 3 + 5 \times \left( { - 1} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{14}\\{ - 9}\\4\end{aligned}} \right)\end{aligned}\)

Write the product \(AB\)

The product \(AB\) can be written as follows:

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}{A{b_1}}&{A{b_2}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\end{aligned}\)

Find the product \(AB\) using row-column rule

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&3\\2&{ - 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 \times 1 + \left( { - 2} \right) \times 2}&{4 \times 3 + \left( { - 2} \right) \times \left( { - 1} \right)}\\{\left( { - 3} \right) \times 1 + 0}&{\left( { - 3} \right) \times 3 + 0}\\{3 \times 1 + 5 \times 2}&{3 \times 3 + 5 \times \left( { - 1} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\end{aligned}\)

So, \(AB = \left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\).