Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Review Exercises - Page 950: 21

Answer

The matrix $ AB $ is, $ AB=\left[ \begin{matrix} -1 & -16 \\ 8 & 1 \\ \end{matrix} \right]$

Work Step by Step

Here we need to find $ AB $. Therefore consider, $\begin{align} & AB=\left[ \begin{array}{*{35}{l}} 2 & -1 & 2 \\ 5 & 3 & -1 \\ \end{array} \right]\left[ \begin{array}{*{35}{l}} 0 & -2 \\ 3 & 2 \\ 1 & -5 \\ \end{array} \right] \\ & =\left[ \begin{matrix} 2\left( 0 \right)-1\left( 3 \right)+2\left( 1 \right) & 2\left( -2 \right)-1\left( 2 \right)+2\left( -5 \right) \\ 5\left( 0 \right)+3\left( 3 \right)-1\left( 1 \right) & 5\left( -2 \right)+3\left( 2 \right)-1\left( -5 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0-3+2 & -4-2-10 \\ 0+9-1 & -10+6+5 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & -16 \\ 8 & 1 \\ \end{matrix} \right] \end{align}$ Thus, $ AB=\left[ \begin{matrix} -1 & -16 \\ 8 & 1 \\ \end{matrix} \right]$
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