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 918: 58

Answer

a) The product of A and B is $\left[ \begin{matrix} 0 & -3 & -3 & -1 & -1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right]$. b) The graph is shown below.

Work Step by Step

(a) To multiply the two provided matrices, we will use row-by-column operation, $A=\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ $B=\left[ \begin{matrix} 0 & 3 & 3 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right]$ Calculate the product as follows: $\begin{align} & AB=\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 3 & 3 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1\left( 0 \right)+0\left( 0 \right) & -1\left( 3 \right)+0\left( 0 \right) & -1\left( 3 \right)+0\left( 1 \right) & -1\left( 1 \right)+0\left( 1 \right) & -1\left( 1 \right)+0\left( 5 \right) & -1\left( 0 \right)+0\left( 5 \right) \\ 0\left( 0 \right)+1\left( 0 \right) & 0\left( 3 \right)+1\left( 0 \right) & 0\left( 3 \right)+1\left( 1 \right) & 0\left( 1 \right)+1\left( 1 \right) & 0\left( 1 \right)+5\left( 1 \right) & 0\left( 0 \right)+5\left( 1 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0 & -3 & -3 & -1 & -1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right] \end{align}$ Thus, the product is $AB=\left[ \begin{matrix} 0 & -3 & -3 & -1 & -1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right]$. (b) The matrix obtained by multiplying the matrices A and B is as follows: $AB=\left[ \begin{matrix} 0 & -3 & -3 & -1 & -1 & 0 \\ 0 & 0 & 1 & 1 & 5 & 5 \\ \end{matrix} \right]$ Each column of the matrix will represent the coordinates of the plot. Plot the above obtained coordinates from the matrix AB and trace them to obtain the figure. In the graph, it can be seen that the letter L gets reflected across the y-axis. Due to the multiplication of a new matrix, the coordinates are changed and as a result the graph is also changed.
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