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: 31

Answer

$(1,0,1)$

Work Step by Step

Start with the augmented matrix, row reduce to reduced row echelon form (Gauss-Jordan.) $\left[\begin{array}{llll} 1 & 1 & 1 & 2\\ 2 & -3 & 2 & 4\\ 4 & 1 & -3 & 1 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ R_{2}\leftarrow R_{2}-2R_{1}.\\ R_{3}\leftarrow R_{3}-4R_{1} \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 1 & 1 & 2\\ 0 & -5 & 0 & 0\\ 0 & -3 & -7 & -7 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ R_{2}\leftarrow-\frac{1}{5}R_{2}.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 1 & 1 & 2\\ 0 & 1 & 0 & 0\\ 0 & -3 & -7 & -7 \end{array}\right]\rightarrow\left(\begin{array}{l} R_{1}\leftarrow R_{1}- R_{2}.\\ .\\ R_{3}\leftarrow R_{3}+3R_{2} \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 0 & 1 & 2\\ 0 & 1 & 0 & 0\\ 0 & 0 & -7 & -7 \end{array}\right] \rightarrow\left(\begin{array}{l} .\\ .\\ R_{3}\leftarrow-\frac{1}{7}R_{3} \end{array}\right) $ $\rightarrow\left[\begin{array}{llll} 1 & 0 & 1 & 2\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 \end{array}\right] \rightarrow\left(\begin{array}{l} R_{1}\leftarrow R_{1}-R_{3}.\\ .\\ \end{array}\right) $ $\rightarrow\left[\begin{array}{llll} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 1 \end{array}\right]$ The solution is $(1,0,1)$
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