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 22: 16


The solution to the system is $$x_1=\frac{1}{3}\qquad x_2=1\qquad x_3=-\frac{1}{3}.$$

Work Step by Step

The system associated to this matrix is: \begin{align*} 2x_1+x_2+x_3=&0\\ x_1-2x_2+x_3=&-2\\ x_1\qquad+x_3=&0 \end{align*} To solve it follow the steps below: Step 1: Subtract from the first equation the second one multiplied by $2$ to eliminate $x_1$: $$2x_1-2x_1+x_2-2(-2x_2) + x_3-2x_3 = 0-2(-2)$$ which becomes $$5x_2+3x_3=4$$ Now our system reads \begin{align*} 5x_2+3x_3=&4\\ x_1-2x_2+x_3=&-2\\ x_1\qquad+x_3=0 \end{align*} Step 2: Subtract from the second equation the third one to eliminate $x_1$ (and $x_3$ for free): $$x_1-x_1-2x_2+x_3-x_3 = -2-0$$ which becomes $$-2x_2=-2$$ and this gives $$x_2=1.$$ Step 3: Put this into the first equation to find $x_3$: $$5\times 1+ 3x_3=4\Rightarrow 3x_3 = -1$$ and this gives $$x_3=-\frac{1}{3}$$ Step 4: Put $x_2=1$ and $x_3=-\frac{1}{3}$ into the second equation and find $x_1$: $$x_1-2\times1-\frac{1}{3} = -2\Rightarrow x_1-\frac{7}{3}=-2$$ and this will give $$x_1 = \frac{1}{3}.$$
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