Answer
$\{(2,1,-1)\}$
Work Step by Step
Use Gauss-Jordan to perform the row operations given, we have:
$\begin{bmatrix} 2 & 4 & 4 & 4 \\ 1 & 3 & 1 & 4 \\ -1 & 3 & 2 & -1 \end{bmatrix} \begin{array} .R1/2\to R1\\.\\ .\\ \end{array}$
$\begin{bmatrix} 1 & 2 & 2 & 2 \\ 1 & 3 & 1 & 4 \\ -1 & 3 & 2 & -1 \end{bmatrix} \begin{array} ..\\R2-R1\to R2\\ R3+R1\to R3\\ \end{array}$
$\begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 1 & -1 & 2 \\ 0 & 5 & 4 & 1 \end{bmatrix} \begin{array} ..\\.\\ R3-5R2\to R3\\ \end{array}$
$\begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 9 & -9 \end{bmatrix} \begin{array} ..\\.\\ R3/9\to R3\\ \end{array}$
$\begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 1 & -1 \end{bmatrix} \begin{array} .R1-2R3\to R1\\R2+R3\to R2\\ .\\ \end{array}$
$\begin{bmatrix} 1 & 2 & 0 & 4 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \end{bmatrix} \begin{array} .R1-2R2\to R1\\.\\ .\\ \end{array}$
$\begin{bmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \end{bmatrix} $
Thus the solution set is $\{(2,1,-1)\}$
