Elementary Linear Algebra 7th Edition

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

Chapter 1 - Systems of Linear Equations - 1.2 Gaussian Elimination and Gauss-Jordan Elimination - 1.2 Exercises - Page 23: 40

Answer

$x_1=1,$ $x_2=-1,$ $x_3=2,$ $x_4=0,$ $x_5=-2,$ $x_6=1$

Work Step by Step

The augmented matrix is $\left[ {\begin{array}{*{20}{c}} 1&2&{ - 2}&2&{ - 1}&3\\ 2&{ - 1}&3&1&{ - 3}&2\\ 1&3&{ - 2}&1&{ - 2}&{ - 3}\\ 3&{ - 2}&1&{ - 1}&3&{ - 2}\\ { - 1}&{ - 2}&1&2&{ - 2}&3\\ 1&{ - 3}&1&3&{ - 2}&1 \end{array}\left| {\begin{array}{*{20}{c}} 0\\ {17}\\ { - 5}\\ { - 1}\\ {10}\\ {11} \end{array}} \right.} \right]$ By using MATLAB software, the row reduced echelon form is $\left[ {\begin{array}{*{20}{c}} 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 2\\ 0\\ { - 2}\\ 1 \end{array}} \right.} \right]$ Hence, the solution is given by $x_1=1,$ $x_2=-1,$ $x_3=2,$ $x_4=0,$ $x_5=-2,$ $x_6=1$
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