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: 58

Answer

$x=8/7, \ y=5/7, \ z=17/7$ or $(8/7, \ 5/7, \ 17/7)$

Work Step by Step

Writing the system in matrix form, $AX=B,$ the solution is $X=A^{-1}B$. Here, $A=\left[\begin{array}{rrr}{3}&{3}&{1}\\{1}&{2}&{1}\\{2}&{-1}&{1}\end{array}\right]$, as in exercise $40$. There, we found $A^{-1}=\left[\begin{array}{rrr} {\displaystyle \frac{3}{7}}&{-\displaystyle \frac{4}{7}}&{\displaystyle \frac{1}{7}}\\ {\displaystyle \frac{1}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{2}{7}}\\ {-\displaystyle \frac{5}{7}}&{\displaystyle \frac{9}{7}}&{\displaystyle \frac{3}{7}}\end{array}\right]$ With $B=\left[\begin{array}{l} 8\\ 5\\ 4 \end{array}\right]$, the solution, $X=A^{-1}B$ equals $X=\left[\begin{array}{l} \frac{1}{7}(24-20+4)\\ \frac{1}{7}(8+5-8)\\ \frac{1}{7}(-40+45+12) \end{array}\right]=\left[\begin{array}{l} 8/7\\ 5/7\\ 17/7 \end{array}\right]$ $x=8/7, \ y=5/7, \ z=17/7$ or $(8/7, \ 5/7, \ 17/7)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.