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

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