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: 39

Answer

$$x_1=2, \quad x_2=-2, \quad x_3=3, \quad x_4=-5, \quad x_5=1.$$

Work Step by Step

The augmented matrix is given by $$\left[ \begin {array}{cccccc} 1&-1&2&2&6&6\\ 3&-2&4 &4&12&14\\ 0&1&-1&-1&-3&-3\\ 2&-2& 4&5&15&10\\ 2&-2&4&4&13&13\end {array} \right] . $$ Using Maple program, we get $$ \left[ \begin {array}{cccccc} 1&-1&2&2&6&6\\ 0&1&-2 &-2&-6&-4\\ 0&0&1&1&3&1\\ 0&0&0&1& 3&-2\\ 0&0&0&0&1&1\end {array} \right] . $$ Now, the crossposting system is given by $$ \begin{align*} x_1-x_2+2x_3+2x_4+6x_5&=6 \\ x_2-2x_3-2x_4-6x_5&=-4\\ x_3+x_4+3x_5&=1\\ x_4+3x_5&=-2\\ x_5&=1. \end{align*} $$ The above system has the solution $$x_1=2, \quad x_2=-2, \quad x_3=3, \quad x_4=-5, \quad x_5=1.$$
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