## Intermediate Algebra for College Students (7th Edition)

$\{(1,-1,2)\}$.
The given system of equations is $x+y-z=-2$ $2x-y+z=5$ $-x+2y+2z=1$ The augmented matrix is $\Rightarrow \left[\begin{array}{ccc|c} 1 & 1 & -1& -2\\ 2 & -1 & 1& 5 \\ -1&2&2&1 \end{array}\right]$ Perform $R_2\rightarrow R_2-2\times R_1$ and $R_3\rightarrow R_3+ R_1$. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 1 & -1& -2\\ 2-2\times (1) & -1-2\times (1) & 1-2\times (-1)& 5-2\times (-2) \\ -1+1&2+1&2-1&1-2 \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 1 & -1& -2\\ 0 & -3 & 3& 9 \\ 0&3&1&-1 \end{array}\right]$ Perform $R_2\rightarrow \frac{R_2}{-3}$. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 1 & -1& -2\\ 0 & -3/(-3) & 3/(-3)& 9/(-3) \\ 0&3&1&-1 \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 1 & -1& -2\\ 0 & 1 & -1& -3 \\ 0&3&1&-1 \end{array}\right]$ Perform $R_3\rightarrow R_3-3\times R_2$ and $R_1\rightarrow R_1- R_2$. $\Rightarrow \left[\begin{array}{ccc|c} 1-0 & 1-1 & -1-(-1)& -2-(-3)\\ 0 & 1 & -1& -3 \\ 0-3\times (0)&3-3\times (1)&1-3\times (-1)&-1-3\times (-3) \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 0 & 0& 1\\ 0 & 1 & -1& -3 \\ 0&0&4&8 \end{array}\right]$ Perform $R_3\rightarrow \frac{R_3}{4}$. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 0 & 0& 1\\ 0 & 1 & -1& -3 \\ 0/(4)&0/(4)&4/(4)&8/(4) \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 0 & 0& 1\\ 0 & 1 & -1& -3 \\ 0&0&1&2 \end{array}\right]$ Perform $R_2\rightarrow R_2+R_3$ $\Rightarrow \left[\begin{array}{ccc|c} 1 & 0 & 0& 1\\ 0+0 & 1+0 & -1+1& -3+2 \\ 0&0&1&2 \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 0 & 0& 1\\ 0 & 1 & 0& -1 \\ 0&0&1&2 \end{array}\right]$ Use back substitution to solve the linear system. $\Rightarrow x=1$ and $\Rightarrow y=-1$. and $\Rightarrow z=2$. The solution set is $\{(x,y,z)\}=\{(1,-1,2)\}$.