Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 2 - Matrices - Review Exercises - Page 98: 22


$$x=-{\frac {23}{18}},\quad y= \frac {17}{9}, \quad z=-{\frac {17}{18}}.$$

Work Step by Step

We have the system $$\left[ \begin {array}{ccc} 0&1&2\\ 3&2&1 \\ 4&-3&-4\end {array} \right] \left[\begin{array}{rrr}{x}\\ {y} \\{z} \end{array}\right]=\left[\begin{array}{rrr}{0} \\ {-1}\\{-7} \end{array}\right].$$ The coefficients matrix $A= \left[ \begin {array}{ccc} 0&1&2\\ 3&2&1 \\ 4&-3&-4\end {array} \right] $ has the inverse $$A^{-1}=\left[ \begin {array}{ccc} {\frac {5}{18}}&\frac{1}{9}&\frac{1}{6} \\ -{\frac {8}{9}}&\frac{4}{9}&-\frac{1}{3}\\ { \frac {17}{18}}&-\frac{2}{9}&\frac{1}{6}\end {array} \right] . $$ The system has the solution $$\left[\begin{array}{rrr}{x}\\ {y} \\{z} \end{array}\right]=\left[ \begin {array}{ccc} {\frac {5}{18}}&\frac{1}{9}&\frac{1}{6} \\ -{\frac {8}{9}}&\frac{4}{9}&-\frac{1}{3}\\ { \frac {17}{18}}&-\frac{2}{9}&\frac{1}{6}\end {array} \right]\left[\begin{array}{rrr}{0} \\ {-1}\\{-7} \end{array}\right]=\left[ \begin {array}{c} -{\frac {23}{18}}\\ { \frac {17}{9}}\\ -{\frac {17}{18}}\end {array} \right] .$$ That is $$x=-{\frac {23}{18}},\quad y= \frac {17}{9}, \quad z=-{\frac {17}{18}}.$$
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