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 917: 20

Answer

$X=\left[ \begin{matrix} -\frac{2}{3} & -2 \\ -\frac{2}{3} & 3 \\ -\frac{2}{3} & -\frac{4}{3} \\ \end{matrix} \right]$

Work Step by Step

Consider the matrix equation, $3X+A=B$ This implies that, $3X=B-A$ or $X=\frac{B-A}{3}$. Now we will consider, $\begin{align} & X=\frac{B-A}{3} \\ & =\frac{1}{3}\left( \left[ \begin{matrix} -5 & -1 \\ 0 & 0 \\ 3 & -4 \\ \end{matrix} \right]-\left[ \begin{matrix} -3 & -7 \\ 2 & -9 \\ 5 & 0 \\ \end{matrix} \right] \right) \\ & =\frac{1}{3}\left( \left[ \begin{matrix} -2 & -6 \\ -2 & 9 \\ -2 & -4 \\ \end{matrix} \right] \right) \\ & =\left[ \begin{matrix} -\frac{2}{3} & -2 \\ -\frac{2}{3} & 3 \\ -\frac{2}{3} & -\frac{4}{3} \\ \end{matrix} \right] \end{align}$
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