College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 8 - Section 8.4 - Matrix Algebra - 8.4 Assess Your Understanding - Page 598: 48

Answer

$(-\displaystyle \frac{1}{2},-3)$

Work Step by Step

Solving $AX=B,\ \qquad X=A^{-1}B$. This problem is connected to exercise $34$, as the coefficient matrix is $A=\left[\begin{array}{rr}{-4}&{1}\\{6}&{-2}\end{array}\right]$, and $A^{-1}=\displaystyle \left[\begin{array}{ll}{-1}&{-\displaystyle \frac{1}{2}}\\{-3}&{-2}\end{array}\right]$ Writing the system $\left\{\begin{aligned}-4x+y&=5\\6x-2y&=-9\end{aligned}\right.$ in matrix form, $AX=B, $ the solution is $X=A^{-1}B=\left[\begin{array}{rr} {-1}&{-1/2}\\ {-3}&{-2}\end{array}\right]\left[\begin{array}{l} 5\\ -9 \end{array}\right]$ $\left[\begin{array}{l} x\\ y \end{array}\right]=\left[\begin{array}{l} -5+9/2\\ -15+18 \end{array}\right]=\left[\begin{array}{l} -1/2\\ 3 \end{array}\right]$ Solution: $(x,y)=(-\displaystyle \frac{1}{2},-3)$
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