Answer
$(-2,1,3)$
Work Step by Step
Step 1. Establish the augmented matrix of the system and use the Gauss Eliminations method:
$\begin{vmatrix} 3 & -1 & 2 & -1 \\ 4 & -2 & 1 & -7\\-1 & 3 & -2 & -1 \end{vmatrix} \begin{array}(-R_3\leftrightarrow R_1 \\.\\.\\ \end{array}$
Step 2. exchange -row3 with row 1:
$\begin{vmatrix} 1 & -3 & 2 & 1 \\ 4 & -2 & 1 & -7\\3 & -1 & 2 & -1 \end{vmatrix} \begin{array}(.\\R_2-4R_1\to R_2\\R_3-3R_1\to R_3\\ \end{array}$
Step 3. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & -3 & 2 & 1 \\ 0 & 10 & -7 & -11\\0 & 8 & -4 & -4 \end{vmatrix} \begin{array}(.\\R_3/8\leftrightarrow R_2\\.\\ \end{array}$
Step 4. divide row 3 by 8 and exchange it with row 2:
$\begin{vmatrix} 1 & -3 & 2 & 1 \\ 0 & 1 & -1/2 & -1/2\\0 & 10 & -7 & -11 \end{vmatrix} \begin{array}(.\\.\\R_3-10R_2\to R_3\\ \end{array}$
Step 5. Do the operations given on the right side of the matrix:
$\begin{vmatrix} 1 & -3 & 2 & 1 \\ 0 & 1 & -1/2 & -1/2\\0 & 0 & -2 & -6 \end{vmatrix} \begin{array}(.\\.\\R_3-10R_2\to R_3\\ \end{array}$
Step 6. From the third row we have $z=3$, back-substitute it into row 2 to get $y-3/2=-1/2$ and $y=1$, use row 1 with the known values, we have $x-3\times1+2\times3=1$ and $x=-2$, thus the solution is $(-2,1,3)$
