## College Algebra 7th Edition

Solution set: $\{(-1,3,5)\}$
Start with the augmented matrix, row reduce to reduced row echelon form (Gauss-Jordan.) $\left[\begin{array}{llll} 2 & -3 & 5 & 14\\ 4 & -1 & -2 & -17\\ -1 & -1 & 1 & 3 \end{array}\right]\rightarrow\left(\begin{array}{l} +R_{3}.\\ -2R_{1}.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & -4 & 6 & 17\\ 0 & 5 & -12 & -45\\ -1 & -1 & 1 & 3 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ .\\ +R_{1}. \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & -4 & 6 & 17\\ 0 & 5 & -12 & -45\\ 0 & -5 & 7 & 20 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ .\\ +R_{2}. \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & -4 & 6 & 17\\ 0 & 5 & -12 & -45\\ 0 & 0 & -5 & -25 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ .\\ \div(-5). \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & -4 & 6 & 17\\ 0 & 5 & -12 & -45\\ 0 & 0 & 1 & 5 \end{array}\right]\rightarrow\left(\begin{array}{l} -6R_{3}.\\ +12R_{3}.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & -4 & 0 & -13\\ 0 & 5 & 0 & 15\\ 0 & 0 & 1 & 5 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ \div 5.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & -4 & 0 & -13\\ 0 & 1 & 0 & 3\\ 0 & 0 & 1 & 5 \end{array}\right]\rightarrow\left(\begin{array}{l} +4R_{2}.\\ .\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{llll} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & 3\\ 0 & 0 & 1 & 5 \end{array}\right]$ Solution set: $\{(-1,3,5)\}$