## College Algebra (10th Edition)

Solution set = $\{( 2 ,\ -1 ,\ 1 )\}$
Reduce the augmented matrix $[A|B]$ to reduced row echelon form and interpret the result $\left[\begin{array}{lllll} 1 & -2 & 3 & | & 7\\ 2 & 1 & 1 & | & 4\\ -3 & 2 & -2 & | & -10 \end{array}\right]\rightarrow\left(\begin{array}{l} .\\ R_{2}=r_{2}-2r_{1}.\\ R_{3}=r_{3}+3r_{1}. \end{array}\right)$ $\rightarrow\left[\begin{array}{lllll} 1 & -2 & 3 & | & 7\\ 0 & 5 & -5 & | & -10\\ 0 & -4 & 7 & | & 11 \end{array}\right]\rightarrow\left(\begin{array}{l} \\ R_{2}=\frac{1}{5}r_{2}.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{lllll} 1 & -2 & 3 & | & 7\\ 0 & 1 & -1 & | & -2\\ 0 & -4 & 7 & | & 11 \end{array}\right]\rightarrow\left(\begin{array}{l} R_{1}=r_{1}+2r_{2}.\\ .\\ R_{3}=r_{3}+4r_{2}. \end{array}\right)$ $\rightarrow\left[\begin{array}{lllll} 1 & 0 & 1 & | & 3\\ 0 & 1 & -1 & | & -2\\ 0 & 0 & 3 & | & 3 \end{array}\right] \rightarrow\left(\begin{array}{l} .\\ .\\ R_{3}=\frac{1}{3}r_{3}. \end{array}\right)$ $\rightarrow\left[\begin{array}{lllll} 1 & 0 & 1 & | & 3\\ 0 & 1 & -1 & | & -2\\ 0 & 0 & 1 & | & 1 \end{array}\right] \rightarrow\left(\begin{array}{l} R_{1}=r_{1}-r_{3}.\\ R_{2}=r_{2}+r_{3}.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{lllll} 1 & 0 & 0 & | & 2\\ 0 & 1 & 0 & | & -1\\ 0 & 0 & 1 & | & 1 \end{array}\right]$ The system is consistent and has a single solution. $x=2,$ $y=-1$ $z=1$ Solution set = $\{( 2 ,\ -1 ,\ 1 )\}$