Solve the following system of linear equations: \[3x - 2y + z = 1\] \[2x + y - 3z = -1\] \[x + y + z = 3\]

Step 1 ：Let's represent these equations as an augmented matrix: \[ \begin{bmatrix} 3 & -2 & 1 & 1 \ \ 2 & 1 & -3 & -1 \ \ 1 & 1 & 1 & 3 \end{bmatrix} \]

Step 2 ：Next, we will perform row operations to get this matrix in row-echelon form. First, we will swap the first and the third row: \[ \begin{bmatrix} 1 & 1 & 1 & 3 \ \ 2 & 1 & -3 & -1 \ \ 3 & -2 & 1 & 1 \end{bmatrix} \]

Step 3 ：Next, we will subtract twice the first row from the second row and thrice the first row from the third row: \[ \begin{bmatrix} 1 & 1 & 1 & 3 \ \ 0 & -1 & -5 & -7 \ \ 0 & -5 & -2 & -8 \end{bmatrix} \]

Step 4 ：Next, multiply the second row by -1 and add 5 times the second row to the third row: \[ \begin{bmatrix} 1 & 1 & 1 & 3 \ \ 0 & 1 & 5 & 7 \ \ 0 & 0 & -23 & -27 \end{bmatrix} \]

Step 5 ：Next, multiply the third row by -1/23: \[ \begin{bmatrix} 1 & 1 & 1 & 3 \ \ 0 & 1 & 5 & 7 \ \ 0 & 0 & 1 & 27/23 \end{bmatrix} \]

Step 6 ：Next, subtract the third row from the first row and 5 times the third row from the second row: \[ \begin{bmatrix} 1 & 1 & 0 & 3 - 27/23 \ \ 0 & 1 & 0 & 7 - 5(27/23) \ \ 0 & 0 & 1 & 27/23 \end{bmatrix} \]

Step 7 ：Finally, subtract the second row from the first row: \[ \begin{bmatrix} 1 & 0 & 0 & (3 - 27/23) - (7 - 5(27/23)) \ \ 0 & 1 & 0 & 7 - 5(27/23) \ \ 0 & 0 & 1 & 27/23 \end{bmatrix} \]