1. **State the problem:** Solve the system of equations using the Gauss-Jordan method: $$\begin{cases}-x + y + 2z = 2 \\ 3x - y + z = 6 \\ -x + 3y + 4z = 4\end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} -1 & 1 & 2 & 2 \\ 3 & -1 & 1 & 6 \\ -1 & 3 & 4 & 4 \end{array}\right]$$ 3. **Make the pivot in row 1, column 1 equal to 1:** Multiply row 1 by $-1$: $$R1 \to -1 \times R1 = [1, -1, -2, -2]$$ Matrix becomes: $$\left[\begin{array}{ccc|c} 1 & -1 & -2 & -2 \\ 3 & -1 & 1 & 6 \\ -1 & 3 & 4 & 4 \end{array}\right]$$ 4. **Eliminate the $x$-terms in rows 2 and 3:** - Row 2: $R2 - 3 \times R1 \to R2$ $$[3, -1, 1, 6] - 3 \times [1, -1, -2, -2] = [0, 2, 7, 12]$$ - Row 3: $R3 + R1 \to R3$ $$[-1, 3, 4, 4] + [1, -1, -2, -2] = [0, 2, 2, 2]$$ Matrix now: $$\left[\begin{array}{ccc|c} 1 & -1 & -2 & -2 \\ 0 & 2 & 7 & 12 \\ 0 & 2 & 2 & 2 \end{array}\right]$$ 5. **Make the pivot in row 2, column 2 equal to 1:** Divide row 2 by 2: $$R2 \to \frac{1}{2} R2 = [0, 1, \frac{7}{2}, 6]$$ Matrix: $$\left[\begin{array}{ccc|c} 1 & -1 & -2 & -2 \\ 0 & 1 & \frac{7}{2} & 6 \\ 0 & 2 & 2 & 2 \end{array}\right]$$ 6. **Eliminate the $y$-terms in rows 1 and 3:** - Row 1: $R1 + R2 \to R1$ $$[1, -1, -2, -2] + [0, 1, \frac{7}{2}, 6] = [1, 0, \frac{3}{2}, 4]$$ - Row 3: $R3 - 2 \times R2 \to R3$ $$[0, 2, 2, 2] - 2 \times [0, 1, \frac{7}{2}, 6] = [0, 0, -5, -10]$$ Matrix: $$\left[\begin{array}{ccc|c} 1 & 0 & \frac{3}{2} & 4 \\ 0 & 1 & \frac{7}{2} & 6 \\ 0 & 0 & -5 & -10 \end{array}\right]$$ 7. **Make the pivot in row 3, column 3 equal to 1:** Divide row 3 by $-5$: $$R3 \to -\frac{1}{5} R3 = [0, 0, 1, 2]$$ Matrix: $$\left[\begin{array}{ccc|c} 1 & 0 & \frac{3}{2} & 4 \\ 0 & 1 & \frac{7}{2} & 6 \\ 0 & 0 & 1 & 2 \end{array}\right]$$ 8. **Eliminate the $z$-terms in rows 1 and 2:** - Row 1: $R1 - \frac{3}{2} \times R3 \to R1$ $$[1, 0, \frac{3}{2}, 4] - \frac{3}{2} \times [0, 0, 1, 2] = [1, 0, 0, 1]$$ - Row 2: $R2 - \frac{7}{2} \times R3 \to R2$ $$[0, 1, \frac{7}{2}, 6] - \frac{7}{2} \times [0, 0, 1, 2] = [0, 1, 0, -1]$$ Final matrix: $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right]$$ 9. **Read the solution:** $$x = 1, \quad y = -1, \quad z = 2$$ **Answer:** The solution to the system is $\boxed{(1, -1, 2)}$.