Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 11 - Systems of Equations and Inequalities - 11.2 Systems of Linear Equations: Matrices - 11.2 Assess Your Understanding - Page 731: 51


Consistent Solution: $\left\{\left(2,-1,1\right)\right\}$

Work Step by Step

We are given the system of equations: $\begin{cases} x-2y+3z=7\\ 2x+y+z=4\\ -3x+2y-2z=-10 \end{cases}$ Write the augmented matrix: $\begin{bmatrix} 1&-2&3&|&7\\2&1&1&|&4\\-3&2&-2&|&-10\end{bmatrix}$ Perform row operations to bring the matrix to the reduced row echelon form: $R_2=-2r_1+r_2$ $\begin{bmatrix} 1&-2&3&|&7\\0&5&-5&|&-10\\-3&2&-2&|&-10\end{bmatrix}$ $R_3=3r_1+R_3$ $\begin{bmatrix}1&-2&3&|&7\\0&5&-5&|&-10\\0&-4&7&|&11\end{bmatrix}$ $R_2=\dfrac{1}{5}r_2$ $\begin{bmatrix}1&-2&3&|&7\\0&1&-1&|&-2\\0&-4&7&|&11\end{bmatrix}$ $R_1=2r_2+r_1$ $\begin{bmatrix}1&0&1&|&3\\0&1&-1&|&-2\\0&-4&7&|&11\end{bmatrix}$ $R_3=4r_2+r_3$ $\begin{bmatrix}1&0&1&|&3\\0&1&-1&|&-2\\0&0&3&|&3\end{bmatrix}$ $R_3=\dfrac{1}{3}r_3$ $\begin{bmatrix}1&0&1&|&3\\0&1&-1&|&-2\\0&0&1&|&1\end{bmatrix}$ $R_1=-r_3+r_1$ $\begin{bmatrix}1&0&0&|&2\\0&1&-1&|&-2\\0&0&1&|&1\end{bmatrix}$ $R_2=-r_3+r_2$ $\begin{bmatrix}1&0&0&|&2\\0&1&0&|&-1\\0&0&1&|&1\end{bmatrix}$ Write the corresponding system of equations: $\begin{cases} 1x+0y+0z=2\\ 0x+1y+0z=-1\\ 0x+0y+1z=1 \end{cases}$ $\begin{cases} x=2\\ y=-1\\ z=1 \end{cases}$ The system is consistent. The solution set is: $\left\{\left(2,-1,1\right)\right\}$
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