## College Algebra (6th Edition)

Solution set: $\{(4,-3,2)\}$
Begin with the system's augmented matrix. Use matrix row operations to get ls down the main diagonal from upper left to lower right, and Os below the $1\mathrm{s}$ (row-echelon form). $\left[\begin{array}{lllll} 1 & 0 & -3 & | & -2\\ 2 & 2 & 1 & | & 4\\ 3 & 1 & -2 & | & 5 \end{array}\right]\left\{\begin{array}{l} .\\ -2R1.+R2\rightarrow R2\\ -3R1+R3\rightarrow R3. \end{array}\right.$ $\left[\begin{array}{lllll} 1 & 0 & -3 & | & -2\\ 0 & 2 & 7 & | & 8\\ 0 & 1 & 7 & | & 11 \end{array}\right]\left\{\begin{array}{l} .\\ R3\leftrightarrow R2\\ . \end{array}\right.$ $\left[\begin{array}{lllll} 1 & 0 & -3 & | & -2\\ 0 & 1 & 7 & | & 11 \\ 0 & 2 & 7 & | & 8 \end{array}\right]\left\{\begin{array}{l} .\\ .\\ -2R2+R3\rightarrow R3. \end{array}\right.$ $\left[\begin{array}{lllll} 1 & 0 & -3 & | & -2\\ 0 & 1 & 7 & | & 11 \\ 0 & 0 & -7 & | & -14 \end{array}\right]\left\{\begin{array}{l} .\\ .\\ \Rightarrow -7z=-14. \end{array}\right.$ $z=2$ Back substitute $z=2$ into row/equation 2, $y+7(2)=11$ $y=11-14=-3$ Back substitute into row/equation 1: $x-3(2)=-2$ $x=-2+6=4$ Solution set: $\{(4,-3,2)\}$