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: 52

Answer

Consistent Solutions set: $\left\{\left(\frac{56}{13},-\frac{7}{13},\frac{35}{13}\right)\right\}$

Work Step by Step

We are given the system of equations: $\begin{cases} 2x+y-3z=0\\ -2x+2y+z=-7\\ 3x-4y-3z=7 \end{cases}$ Write the augmented matrix: $\begin{bmatrix} 2&1&-3&|&0\\-2&2&1&|&-7\\3&-4&-3&|&7\end{bmatrix}$ Perform row operations to bring the matrix to the reduced row echelon form: $R_2=r_1+r_2$ $\begin{bmatrix} 2&1&-3&|&0\\0&3&-2&|&-7\\3&-4&-3&|&7\end{bmatrix}$ $R_1=r_3-r_1$ $\begin{bmatrix}1&-5&0&|&7\\0&3&-2&|&-7\\3&-4&-3&|&7\end{bmatrix}$ $R_3=-3r_1+r_3$ $\begin{bmatrix}1&-5&0&|&7\\0&3&-2&|&-7\\0&11&-3&|&-14\end{bmatrix}$ $R_2=\dfrac{1}{3}r_2$ $\begin{bmatrix}1&-5&0&|&7\\0&1&-\frac{2}{3}&|&-\frac{7}{3}\\0&11&-3&|&-14\end{bmatrix}$ $R_1=5r_2+r_1$ $\begin{bmatrix}1&0&-\frac{10}{3}&|&-\frac{14}{3}\\0&1&-\frac{2}{3}&|&-\frac{7}{3}\\0&11&-3&|&-14\end{bmatrix}$ $R_3=-11r_2+r_3$ $\begin{bmatrix}1&0&-\frac{10}{3}&|&-\frac{14}{3}\\0&1&-\frac{2}{3}&|&-\frac{7}{3}\\0&0&\frac{13}{3}&|&\frac{35}{3}\end{bmatrix}$ $R_3=\dfrac{3}{13}r_3$ $\begin{bmatrix}1&0&-\frac{10}{3}&|&-\frac{14}{3}\\0&1&-\frac{2}{3}&|&-\frac{7}{3}\\0&0&1&|&\frac{35}{13}\end{bmatrix}$ $R_1=\dfrac{10}{3}r_3+r_1$ $\begin{bmatrix}1&0&0&|&\frac{56}{13}\\0&1&-\frac{2}{3}&|&-\frac{7}{3}\\0&0&1&|&\frac{35}{13}\end{bmatrix}$ $R_2=\dfrac{2}{3}r_3+r_2$ $\begin{bmatrix}1&0&0&|&\frac{56}{13}\\0&1&0&|&-\frac{7}{13}\\0&0&1&|&\frac{35}{13}\end{bmatrix}$ Write the corresponding system of equations: $\begin{cases} 1x+0y+0z=\frac{56}{13}\\ 0x+1y+0z=-\frac{7}{13}\\ 0x+0y+1z=\frac{35}{13} \end{cases}$ $\begin{cases} x=\frac{56}{13}\\ y=-\frac{7}{13}\\ z=\frac{35}{13} \end{cases}$ The system is consistent. The solution set is: $\left\{\left(\frac{56}{13},-\frac{7}{13},\frac{35}{13}\right)\right\}$
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