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: 52



Work Step by Step

This problem is connected to exercise $36$, as the coefficient matrix of the system $\left\{\begin{array}{l}{b x+3 y=14 b+3}\\{bx+2y=2b+2}\end{array}\right., b\neq 0$ is $A=\left[\begin{array}{ll}{b}&{3}\\{b}&{2}\end{array}\right]$, and we found $A^{-1}=\displaystyle \left[\begin{array}{rr}{-\dfrac{2}{b}}&{\displaystyle \frac{3}{b}}\\{1}&{-1}\end{array}\right]$ Writing the system in matrix form, $AX=B,$ the solution is $X=A^{-1}B$ $\left[\begin{array}{l} x\\ y \end{array}\right]=\left[\begin{array}{rr} {-\displaystyle \frac{2}{b}}&{\displaystyle \frac{3}{b}}\\ {1}&{-1}\end{array}\right] \left[\begin{array}{l} 2b+3\\ 2b+2 \end{array}\right]$ $\left[\begin{array}{l} x\\ y \end{array}\right]=\left[\begin{array}{l} -\frac{2}{b}(2b+3)+\frac{3}{b}(2b+2)\\ 2b+3-(2b+2) \end{array}\right]=\left[\begin{array}{l} -4-\frac{6}{b}+6+\frac{6}{b}\\ 1 \end{array}\right]=\left[\begin{array}{l} 2\\ 1 \end{array}\right]$ Solution: $(x,y)=(2,1)$
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