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: 38

Answer

$(1,1,-2)$

Work Step by Step

Start with the augmented matrix, row reduce to reduced row echelon form (Gauss-Jordan.) $\left[\begin{array}{llll} 10 & 10 & -20 & 60\\ 15 & 20 & 30 & -25\\ -5 & 30 & -10 & 45 \end{array}\right]\rightarrow\left(\begin{array}{l} \div 10\\ -1.5R_{1}\\ +0.5R_{1} \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 1 & -2 & 6\\ 0 & 5 & 60 & -115\\ 0 & 35 & -20 & 75 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ \div 5\\ -7R_{2} \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 1 & -2 & 6\\ 0 & 1 & 12 & -23\\ 0 & 0 & -440 & 880 \end{array}\right]\rightarrow\left(\begin{array}{l} -R_{2}.\\ .\\ \div(-440) \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 0 & -14 & 29\\ 0 & 1 & 12 & -23\\ 0 & 0 & 1 & -2 \end{array}\right]\rightarrow\left(\begin{array}{l} +14R_{3}.\\ -12R_{3}.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & -2 \end{array}\right]$ The solution is $(1,1,-2)$
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