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: $$\left[\begin{array}{cccccccccc}3& -2& 1& 12& 1& -3& -11& 1& 1& 3\end{array}\right]$$

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: $$\left[\begin{array}{cccccccccc}1& 1& 1& 32& 1& -3& -13& -2& 1& 1\end{array}\right]$$

Step 3 ：Next, we will subtract twice the first row from the second row and thrice the first row from the third row: $$\left[\begin{array}{cccccccccc}1& 1& 1& 30& -1& -5& -70& -5& -2& -8\end{array}\right]$$

Step 4 ：Next, multiply the second row by -1 and add 5 times the second row to the third row: $$\left[\begin{array}{cccccccccc}1& 1& 1& 30& 1& 5& 70& 0& -23& -27\end{array}\right]$$

Step 5 ：Next, multiply the third row by -1/23: $$\left[\begin{array}{cccccccccc}1& 1& 1& 30& 1& 5& 70& 0& 1& 27/23\end{array}\right]$$

Step 6 ：Next, subtract the third row from the first row and 5 times the third row from the second row: $$\left[\begin{array}{cccccccccc}1& 1& 0& 3-27/230& 1& 0& 7-5(27/23)0& 0& 1& 27/23\end{array}\right]$$

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