Answer
$\{(-1,2,2)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& -4y &+4z&=&-1\\
2x& -y & +5z&=&6\\
-x& +3y &-z &=&5
\end{matrix}\right.$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -4 & 4& -1\\
2 & -1 & 5& 6 \\
-1&3&-1&5
\end{array}\right]$
Perform $R_2\rightarrow R_2-2\times R_1$ and $R_3\rightarrow R_3+ R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -4 & 4& -1\\
2-2(1) & -1-2(-4) & 5-2(4)& 6-2(-1) \\
-1+1&3+(-4)&-1+4&5+(-1)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -4 & 4& -1\\
0 & 7 & -3& 8 \\
0&-1&3&4
\end{array}\right]$
Perform $R_2\rightarrow \frac{R_2}{7}$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -4 & 4& -1\\
0/7 & 7/7 & -3/7& 8/7 \\
0&-1&3&4
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -4 & 4& -1\\
0 & 1 & -3/7& 8/7 \\
0&-1&3&4
\end{array}\right]$
Perform $R_1\rightarrow R_1+4\times R_2$ and $R_3\rightarrow R_3+ R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1+4(0) & -4+4(1) & 4+4(-3/7)& -1+4(8/7)\\
0 & 1 & -3/7& 8/7 \\
0+0&-1+1&3+(-3/7)&4+(8/7)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 16/7&25/7\\
0 & 1 & -3/7& 8/7 \\
0&0&18/7&36/7
\end{array}\right]$
Perform $R_3\rightarrow R_3(7/18)$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 16/7&25/7\\
0 & 1 & -3/7& 8/7 \\
0(7/18)&0(7/18)&18/7(7/18)&36/7(7/18)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 16/7&25/7\\
0 & 1 & -3/7& 8/7 \\
0&0&1&2
\end{array}\right]$
Perform $R_1\rightarrow R_1-(16/7) R_3$ and $R_2\rightarrow R_2+(3/7) R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-(16/7) (0) & 0-(16/7) (0) & 16/7-(16/7) (1)&25/7-(16/7) (2)\\
0+(3/7)(0) & 1+(3/7)(0) & -3/7+(3/7)(1)& 8/7+(3/7)(2) \\
0&0&1&2
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0&-1\\
0 & 1 & 0& 2 \\
0&0&1&2
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=-1$
and
$\Rightarrow y=2$.
and
$\Rightarrow z=2$.
The solution set is $\{(x,y,z)\}=\{(-1,2,2)\}$.
