College Algebra (6th Edition)

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

Chapter 6 - Matrices and Determinants - Cumulative Review Exercises (Chapters 1-6) - Page 658: 7

Answer

$x=7$ $y=-4$ $z=6$

Work Step by Step

Build the augmented matrix of the system of equations; $\begin{bmatrix}1&-1&1&|&17\\2&3&1&|&8\\-4&1&5&|&-2\end{bmatrix}$ Use Gauss-Jordan elimination: Add $-2R_1$ to $R_2$ and $4R_1$ to $R_3$: $\begin{bmatrix}1&-1&1&|&17\\0&5&-1&|&-26\\0&-3&9&|&66\end{bmatrix}$ Multiply $R_3$ by $-\dfrac{1}{3}$: $\begin{bmatrix}1&-1&1&|&17\\0&5&-1&|&-26\\0&1&-3&|&-22\end{bmatrix}$ Add $R_3$ to $R_1$ and $-4R_3$ to $R_2$: $\begin{bmatrix}1&0&-2&|&-5\\0&1&11&|&62\\0&1&-3&|&-22\end{bmatrix}$ Add $-R_2$ to $R_3$: $\begin{bmatrix}1&0&-2&|&-5\\0&1&11&|&62\\0&0&-14&|&-84\end{bmatrix}$ Multiply $R_3$ by $-\dfrac{1}{14}$: $\begin{bmatrix}1&0&-2&|&-5\\0&1&11&|&62\\0&0&1&|&6\end{bmatrix}$ Add $2R_3$ to $R_1$ and $-11R_3$ to $R_2$: $\begin{bmatrix}1&0&0&|&7\\0&1&0&|&-4\\0&0&1&|&6\end{bmatrix}$ The solution is: $x=7$ $y=-4$ $z=6$
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