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 \(AB\).

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

Short answer

\(\left( {\begin{aligned}{*{20}{c}}{ - 7}&4\\7&{ - 6}\\{12}&{ - 7}\end{aligned}} \right)\)

Step-by-step solution

Find the value of \(A{b_{\bf{1}}}\)

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

\(\begin{aligned}{c}A{b_1} = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2\\5&4\\2&{ - 3}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}3\\{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{\left( { - 1} \right) \times 3 + 2 \times \left( { - 2} \right)}\\{5 \times 3 + 4 \times \left( { - 2} \right)}\\{2 \times 3 + \left( { - 3} \right) \times \left( { - 2} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 7}\\7\\{12}\end{aligned}} \right)\end{aligned}\)

Find the value of \(A{b_{\bf{2}}}\)

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

\(\begin{aligned}{c}A{b_2} = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2\\5&4\\2&{ - 3}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 2}\\1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{\left( { - 1} \right) \times \left( { - 2} \right) + 2 \times 1}\\{5 \times \left( { - 2} \right) + 4 \times 1}\\{2 \times \left( { - 2} \right) + \left( { - 3} \right) \times 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}4\\{ - 6}\\{ - 7}\end{aligned}} \right)\end{aligned}\)

Write the product \(AB\)

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

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}{A{b_1}}&{A{b_2}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 7}&4\\7&{ - 6}\\{12}&{ - 7}\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2\\5&4\\2&{ - 3}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}3&{ - 2}\\{ - 2}&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 1 \times \left( 3 \right) + 2 \times \left( { - 2} \right)}&{ - 1 \times \left( { - 2} \right) + 2 \times 1}\\{5 \times 3 + 4 \times \left( { - 2} \right)}&{5 \times \left( { - 2} \right) + 4 \times 1}\\{2 \times 3 + \left( { - 3} \right) \times \left( { - 2} \right)}&{2 \times \left( { - 2} \right) + \left( { - 3} \right) \times 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 7}&4\\7&{ - 6}\\{12}&{ - 7}\end{aligned}} \right)\end{aligned}\)

So, \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 7}&4\\7&{ - 6}\\{12}&{ - 7}\end{aligned}} \right)\).