College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 6 - Matrices and Determinants - Exercise Set 6.3 - Page 624: 25

Answer

$X= \left[\begin{array}{ll} 27/2 & 31/2\\ -4 & 18\\ -29/2 & 6 \end{array}\right]$

Work Step by Step

$4A+3B=-2X$ Multiply both sides with $-\displaystyle \frac{1}{2} .$ $-\displaystyle \frac{1}{2} (4A+3B)=-\frac{1}{2}(-2X)$ $X=-\displaystyle \frac{1}{2} .(4A+3B)$ $X=-\displaystyle \frac{1}{2}\cdot(4\left[\begin{array}{ll} -3 & -7\\ 2 & -9\\ 5 & 0 \end{array}\right]+3\displaystyle \left[\begin{array}{ll} -5 & -1\\ 0 & 0\\ 3 & -4 \end{array}\right])$ $X=-\displaystyle \frac{1}{2}\cdot(\left[\begin{array}{ll} -12 & -28\\ 8 & -36\\ 20 & 0 \end{array}\right]+\displaystyle \left[\begin{array}{ll} -15 & -3\\ 0 & 0\\ 9 & -12 \end{array}\right])$ $X=-\displaystyle \frac{1}{2}\cdot\left[\begin{array}{ll} -27 & -31\\ 8 & -36\\ 29 & -12 \end{array}\right]$ $X= \left[\begin{array}{ll} 27/2 & 31/2\\ -4 & 18\\ -29/2 & 6 \end{array}\right]$
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