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Found in: Page 93

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# 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)

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

See the step by step solution

## Step 1: 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}

## Step 2: 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}

## Step 3: 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}

## Step 3: 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).