Elementary Linear Algebra 7th Edition

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

Chapter 3 - Determinants - 3.4 Applications of Determinants - 3.4 Exercises - Page 136: 19

Answer

The system can be solved by using Cramer's Rule: $x_1=2$ and $x_2=-2$

Work Step by Step

The coefficient matrix is $A=\left[\begin {array}{cc} 3&4\\ 5&3 \end{array}\right]$ and $|A|=-11 \neq0 $. Then $ A$ is a nonsingular matrix such that $AX=B$, where $X=\left[\begin {array}{c} x_1\\ x_2 \end{array}\right]$ and $B=\left[\begin {array}{c} -2\\ 4 \end{array}\right]$ since $A$ is a nonsingular matrix, then we can use Cramer's Rule to solve the linear system. Thus the linear system has the unique solution $x_1=|A_1|/|A|$ and $x_2=|A_2|/|A|$ where $A_1=\left[\begin {array}{c} -2&4\\ 4&3 \end{array}\right]$ $A_2=\left[\begin {array}{c} 3&-2\\ 5&4 \end{array}\right]$ and then $|A_1|=-22 $ and $|A_2|=22$ therefore, $x_1=|A_1|/|A|=22/11=2$ and $x_2=|A_2|/|A|=-22/11=-2$
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