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: 20


$$x=\frac {18}{11},\quad y= -{\frac {19}{11}}.$$

Work Step by Step

We have the system $$\left[\begin{array}{rrr}{2} & {-1} \\{3}& {4} \end{array}\right]\left[\begin{array}{rrr}{x}\\ {y} \end{array}\right]=\left[\begin{array}{rrr}{5} \\ {-2} \end{array}\right].$$ The coefficients matrix $A=\left[\begin{array}{rrr}{2} & {-1} \\{3}& {4} \end{array}\right]$ has the inverse $$A^{-1}=\left[ \begin {array}{cc} \frac{4}{11}&\frac{1}{11}\\ -\frac{3}{11}&\frac{2}{11} \end {array} \right] . $$ The system has the solution $$\left[\begin{array}{rrr}{x}\\ {y} \end{array}\right]=\left[ \begin {array}{cc} \frac{4}{11}&\frac{1}{11}\\ -\frac{3}{11}&\frac{2}{11} \end {array} \right]\left[\begin{array}{rrr}{5} \\ {-2} \end{array}\right]=\left[ \begin {array}{c} {\frac {18}{11}}\\ -{ \frac {19}{11}}\end {array} \right] .$$ That is $$x=\frac {18}{11},\quad y= -{ \frac {19}{11}}.$$
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