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: 26

Answer

$X=\left[\begin{array}{ll} 29/2 & 25/2\\ -3 & 27/2\\ -27/2 & 8 \end{array}\right]$

Work Step by Step

$4B+3A=-2X$ Multiply both sides with $-\displaystyle \frac{1}{2} .$ $-\displaystyle \frac{1}{2}(4B+3A)=-\frac{1}{2}(-2X)$ $X=-\displaystyle \frac{1}{2}(4B+3A)$ $X=-\displaystyle \frac{1}{2}\cdot(4\left[\begin{array}{ll} -5 & -1\\ 0 & 0\\ 3 & -4 \end{array}\right]+3\displaystyle \left[\begin{array}{ll} -3 & -7\\ 2 & -9\\ 5 & 0 \end{array}\right])$ $X=-\displaystyle \frac{1}{2}\cdot(\left[\begin{array}{ll} -20 & -4\\ 0 & 0\\ 12 & -16 \end{array}\right]+\displaystyle \left[\begin{array}{ll} -9 & -21\\ 6 & -27\\ 15 & 0 \end{array}\right])$ $X=-\displaystyle \frac{1}{2}\cdot\left[\begin{array}{ll} -29 & -25\\ 6 & -27\\ 27 & -16 \end{array}\right]$ $X=\left[\begin{array}{ll} 29/2 & 25/2\\ -3 & 27/2\\ -27/2 & 8 \end{array}\right]$
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