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: 21


$$x_1=0,\quad x_2= -\frac{1}{7}, \quad x_3=\frac{3}{7}.$$

Work Step by Step

We have the system $$\left[ \begin {array}{ccc} 0&1&-2\\ -1&3&1 \\ 2&-2&4\end {array} \right] \left[\begin{array}{rrr}{x_1}\\ {x_2} \\{x_3} \end{array}\right]=\left[\begin{array}{rrr}{-1} \\ {0}\\{2} \end{array}\right].$$ The coefficients matrix $A= \left[ \begin {array}{ccc} 0&1&-2\\ -1&3&1 \\ 2&-2&4\end {array} \right] $ has the inverse $$A^{-1}=\left[ \begin {array}{ccc} 1&0&\frac{1}{2}\\ \frac{3}{7}&\frac{2}{7}&\frac{1}{7} \\ -\frac{2}{7}&\frac{1}{7}&\frac{1}{14}\end {array} \right] . $$ The system has the solution $$\left[\begin{array}{rrr}{x_1}\\ {x_2} \\{x_3} \end{array}\right]=\left[ \begin {array}{ccc} 1&0&\frac{1}{2}\\ \frac{3}{7}&\frac{2}{7}&\frac{1}{7} \\ -\frac{2}{7}&\frac{1}{7}&\frac{1}{14}\end {array} \right]\left[\begin{array}{rrr}{-1} \\ {0}\\{2} \end{array}\right]=\left[ \begin {array}{c} 0\\ -\frac{1}{7} \\ \frac{3}{7}\end {array} \right] .$$ That is $$x_1=0,\quad x_2= -\frac{1}{7}, \quad x_3=\frac{3}{7}.$$
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