Solve the following system of linear equations using an augmented matrix: \[\begin{align*} 3x - 2y + z &= 1,\\ 2x + y - z &= -1,\\ x + 2y + 3z &= 5.\end{align*}\]

Step 1 ：Form the augmented matrix for the system: \[\left[\begin{array}{ccc|c} 3 & -2 & 1 & 1 \\ 2 & 1 & -1 & -1 \\ 1 & 2 & 3 & 5 \end{array}\right].\]

Step 2 ：Apply row operations to bring the matrix to row-echelon form. Swap Row 1 and Row 3, then subtract 2 times Row 3 from Row 2 and 3 times Row 3 from Row 1: \[\left[\begin{array}{ccc|c} 1 & 2 & 3 & 5 \\ 0 & -3 & -7 & -11 \\ 0 & -4 & -7 & -14 \end{array}\right].\]

Step 3 ：To eliminate the negative numbers, multiply Row 2 by -1/3 and Row 3 by -1/4: \[\left[\begin{array}{ccc|c} 1 & 2 & 3 & 5 \\ 0 & 1 & 7/3 & 11/3 \\ 0 & 1 & 7/4 & 7/2 \end{array}\right].\]

Step 4 ：Subtract Row 2 from Row 3: \[\left[\begin{array}{ccc|c} 1 & 2 & 3 & 5 \\ 0 & 1 & 7/3 & 11/3 \\ 0 & 0 & -1/12 & -5/6 \end{array}\right].\]

Step 5 ：Multiply Row 3 by -12 to make the leading coefficient 1: \[\left[\begin{array}{ccc|c} 1 & 2 & 3 & 5 \\ 0 & 1 & 7/3 & 11/3 \\ 0 & 0 & 1 & 7 \end{array}\right].\]

Step 6 ：Subtract 3 times Row 3 from Row 1 and 7/3 times Row 3 from Row 2 to make the other entries in column 3 zero: \[\left[\begin{array}{ccc|c} 1 & 2 & 0 & -16 \\ 0 & 1 & 0 & -10 \\ 0 & 0 & 1 & 7 \end{array}\right].\]

Step 7 ：Finally, subtract 2 times Row 2 from Row 1 to make the other entries in column 2 zero: \[\left[\begin{array}{ccc|c} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -10 \\ 0 & 0 & 1 & 7 \end{array}\right].\]

Step 8 ：The solution to the system is (x, y, z) = (4, -10, 7).