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: 24

Answer

$$x_1=1 ,\quad x_2= -1 .$$

Work Step by Step

We have the system $$\left[ \begin {array}{ccc} 3&2\\ 1& 4 \end {array} \right] \left[\begin{array}{rrr}{x_1}\\ {x_2} \end{array}\right]=\left[\begin{array}{rrr}{1} \\ {-3} \end{array}\right].$$ The coefficients matrix $A= \left[ \begin {array}{ccc} 3&2\\ 1& 4 \end {array} \right] $ has the inverse $$A^{-1}=\left[ \begin {array}{cc} \frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{10}&\frac{3}{10} \end {array} \right] . $$ The system has the solution $$\left[\begin{array}{rrr}{x_1}\\ {x_2} \end{array}\right]=\left[ \begin {array}{cc} \frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{10}&\frac{3}{10} \end {array} \right]\left[\begin{array}{rrr}{1} \\ {-3} \end{array}\right]=\left[ \begin {array}{c} {1 }\\ -1 \end {array} \right] .$$ That is $$x_1=1 ,\quad x_2= -1 .$$
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