College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 6 - Summary, Review, and Test - Review Exercises - Page 655: 4

Answer

$x=-2$ $y=-1$ $z=0$

Work Step by Step

Build the augmented matrix of the system of equations; $\begin{bmatrix}1&-2&1&|&0\\0&1&-3&|&-1\\0&2&5&|&-2\end{bmatrix}$ Bring the matrix to the row reduced echelon form: Add $R_3$ to $R_1$: $\begin{bmatrix}1&0&6&|&-2\\0&1&-3&|&-1\\0&2&5&|&-2\end{bmatrix}$ Add $-2R_2$ to $R_3$: $\begin{bmatrix}1&0&6&|&-2\\0&1&-3&|&-1\\0&0&11&|&0\end{bmatrix}$ Multiply $R_3$ by $\dfrac{1}{11}$: $\begin{bmatrix}1&0&6&|&-2\\0&1&-3&|&-1\\0&0&1&|&0\end{bmatrix}$ Add $3R_3$ to $R_2$: $\begin{bmatrix}1&0&6&|&-2\\0&1&0&|&-1\\0&0&1&|&0\end{bmatrix}$ Add $-6R_3$ to $R_1$: $\begin{bmatrix}1&0&0&|&-2\\0&1&0&|&-1\\0&0&1&|&0\end{bmatrix}$ The solution is: $x=-2$ $y=-1$ $z=0$
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