College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 6, Matrices and Determinants - Section 6.1 - Matrices and Systems of Linear Equations - 6.1 Exercises - Page 501: 44

Answer

The solution set is $\{(2+3t\ ,\ t,\ 4) \ |\ \ t\in \mathbb{R}\}$ (Dependent system)

Work Step by Step

Start with the augmented matrix, row reduce to reduced row echelon form (Gauss-Jordan.) $\left[\begin{array}{llll} -2 & 6 & -2 & -12\\ 1 & -3 & 2 & 10\\ -1 & 3 & 2 & 6 \end{array}\right]\rightarrow\left(\begin{array}{l} \leftrightarrow R_{2}.\\ .\\ +R_{2}. \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & -3 & 2 & 10\\ -2 & 6 & -2 & -12\\ 0 & 0 & 4 & 16 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ +2R_{1}.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & -3 & 2 & 10\\ 0 & 0 & 2 & 8\\ 0 & 0 & 4 & 16 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ \div 2.\\ -2R_{2}. \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & -3 & 2 & 10\\ 0 & 0 & 1 & 4\\ 0 & 0 & 0 & 0 \end{array}\right]$ The last row represents 0=0, which is always true. The system is consistent (has solutions) and dependent. $y$ has no leading 1 in its corresponding column. Let $y=t\in \mathbb{R}$, an arbitrary number, Row $2 \Rightarrow z=4$ Row $1 \Rightarrow x-3t+2(4)=10 \Rightarrow x=2+3t$ The solution set is $\{(2+3t\ ,\ t,\ 4) \ |\ \ t\in \mathbb{R}\}$ (Dependent system)
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