## College Algebra (6th Edition)

Solution set: $\{ ( -1$ , $6$ , $3 ) \}$
Augmented matrix: $\left[\begin{array}{lllll} 3 & 1 & -1 & | & 0\\ 2 & 3 & -5 & | & 1\\ 1 & -2 & 3 & | & -4 \end{array}\right]\left\{\begin{array}{l} \leftrightarrow R_{3}.\\ .\\ . \end{array}\right.$ $\left[\begin{array}{lllll} 1 & -2 & 3 & | & -4\\ 2 & 3 & -5 & | & 1\\ 3 & 1 & -1 & | & 0 \end{array}\right]\left\{\begin{array}{l} .\\ \leftarrow R_{2}-2R_{1}\\ \leftarrow R_{3}-3R_{1}. \end{array}\right.$ $\left[\begin{array}{lllll} 1 & -2 & 3 & | & -4\\ 0 & 7 & -11 & | & 9\\ 0 & 7 & -10 & | & 12 \end{array}\right]\left\{\begin{array}{l} .\\ .\\ \leftarrow R_{3}-R_{2}. \end{array}\right.$ $\left[\begin{array}{lllll} 1 & -2 & 3 & | & -4\\ 0 & 7 & -11 & | & 9\\ 0 & 0 & 1 & | & 3 \end{array}\right]\left\{\begin{array}{l} .\\ .\\ \Rightarrow c=3 \end{array}\right.$ Back substitute into row/equation 2, $7b-11(3)=9$ $7b=9+33$ $7b=42$ $b=6$ Back substitute into row/equation 1: $a-2(6)+3(3)=-4$ $a=-4+12-9=-1$ Solution set: $\{ ( -1$ , $6$ , $3 ) \}$