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: 21

Answer

$X=\left[\begin{array}{ll} 1/3 & 13/3\\ -4/3 & 6\\ -7/3 & -4/3 \end{array}\right]$

Work Step by Step

$ 3X+2A=B\qquad$ ... subtract 2A from both sides $ 3X=B-2A\qquad$ ... multiply both sides with $\displaystyle \frac{1}{3}$ $ X=\displaystyle \frac{1}{3}(B-2A)$ $X=\displaystyle \frac{1}{3}\cdot(\left[\begin{array}{ll} -5 & -1\\ 0 & 0\\ 3 & -4 \end{array}\right]-2\displaystyle \cdot\left[\begin{array}{ll} -3 & -7\\ 2 & -9\\ 5 & 0 \end{array}\right]) $ $X=\displaystyle \frac{1}{3}\cdot\left[\begin{array}{ll} -5-2(-3) & -1-2(-7)\\ 0-2(2) & 0-2(-9)\\ 3-2(5) & -4-2(0) \end{array}\right]$ $X= \displaystyle \frac{1}{3}\cdot\left[\begin{array}{ll} 1 & 13\\ -4 & 18\\ -7 & -4 \end{array}\right]$ $X=\left[\begin{array}{ll} 1/3 & 13/3\\ -4/3 & 6\\ -7/3 & -4/3 \end{array}\right]$
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