Precalculus (6th Edition) Blitzer

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

Chapter 8 - Section 8.3 - Matrix Operations and Their Applications - Exercise Set - Page 919: 59

Answer

a) $AB=\left[ \begin{matrix} 0 & 0 & -1 & -1 & -5 & -5 \\ 0 & 3 & 3 & 1 & 1 & 0 \\ \end{matrix} \right]$ b) See the graph below:

Work Step by Step

(a) $\begin{align} & AB=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 3 & 3 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right] \\ & AB=\left[ \begin{matrix} 0\left( 0 \right)+\left( -1 \right)\left( 0 \right) & 0\left( 3 \right)+\left( -1 \right)\left( 0 \right) & 0\left( 3 \right)+\left( -1 \right)\left( 1 \right) & 0\left( 1 \right)+\left( -1 \right)\left( 1 \right) & 0\left( 1 \right)+\left( -1 \right)\left( 5 \right) & 0\left( 0 \right)+\left( -1 \right)\left( 5 \right) \\ 1\left( 0 \right)+0\left( 0 \right) & 1\left( 3 \right)+0\left( 0 \right) & 1\left( 3 \right)+0\left( 1 \right) & 1\left( 1 \right)+0\left( 1 \right) & 1\left( 1 \right)+0\left( 5 \right) & 1\left( 0 \right)+0\left( 5 \right) \\ \end{matrix} \right] \\ & AB=\left[ \begin{matrix} 0-0 & 0-0 & 0-1 & 0-1 & 0-5 & 0-5 \\ 0+0 & 3+0 & 3+0 & 1+0 & 1+0 & 0+0 \\ \end{matrix} \right] \\ & AB=\left[ \begin{matrix} 0 & 0 & -1 & -1 & -5 & -5 \\ 0 & 3 & 3 & 1 & 1 & 0 \\ \end{matrix} \right] \\ \end{align}$ (b) The graph for matrix A and AB is drawn as shown above. From the graph, it can be inferred that the shape of letter “L” has been changed due to matrix multiplication. It has shifted from the positive quadrant to the negative quadrant and it has become a horizontal “L” from the vertical.
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