1. **Problem:** Solve the system using Gauss-Jordan Elimination. Given system: $$\begin{cases} 2x + y - z = 3 \\ 4x - y + 2z = 1 \\ -2x + 3y + z = 4 \end{cases}$$ 2. **Formula and rules:** Gauss-Jordan elimination transforms the augmented matrix to reduced row echelon form (RREF) to find the solution. 3. **Step 1: Write augmented matrix** $$\left[\begin{array}{ccc|c} 2 & 1 & -1 & 3 \\ 4 & -1 & 2 & 1 \\ -2 & 3 & 1 & 4 \end{array}\right]$$ 4. **Step 2: Make leading 1 in row 1** Divide row 1 by 2: $$\left[\begin{array}{ccc|c} \cancel{2}/2 & \cancel{1}/2 & -1/2 & 3/2 \\ 4 & -1 & 2 & 1 \\ -2 & 3 & 1 & 4 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & \frac{3}{2} \\ 4 & -1 & 2 & 1 \\ -2 & 3 & 1 & 4 \end{array}\right]$$ 5. **Step 3: Eliminate x in rows 2 and 3** Row 2 = Row 2 - 4*Row 1: $$4 - 4*1 = \cancel{4} - 4 = 0$$ $$-1 - 4*\frac{1}{2} = -1 - 2 = -3$$ $$2 - 4*(-\frac{1}{2}) = 2 + 2 = 4$$ $$1 - 4*\frac{3}{2} = 1 - 6 = -5$$ Row 3 = Row 3 + 2*Row 1: $$-2 + 2*1 = 0$$ $$3 + 2*\frac{1}{2} = 3 + 1 = 4$$ $$1 + 2*(-\frac{1}{2}) = 1 - 1 = 0$$ $$4 + 2*\frac{3}{2} = 4 + 3 = 7$$ Matrix now: $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & \frac{3}{2} \\ 0 & -3 & 4 & -5 \\ 0 & 4 & 0 & 7 \end{array}\right]$$ 6. **Step 4: Make leading 1 in row 2** Divide row 2 by -3: $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & \frac{3}{2} \\ 0 & \cancel{-3}/-3 & \frac{4}{-3} & \frac{-5}{-3} \\ 0 & 4 & 0 & 7 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & \frac{3}{2} \\ 0 & 1 & -\frac{4}{3} & \frac{5}{3} \\ 0 & 4 & 0 & 7 \end{array}\right]$$ 7. **Step 5: Eliminate y in rows 1 and 3** Row 1 = Row 1 - (1/2)*Row 2: $$\frac{1}{2} - \frac{1}{2}*1 = 0$$ $$-\frac{1}{2} - \frac{1}{2}*(-\frac{4}{3}) = -\frac{1}{2} + \frac{2}{3} = \frac{1}{6}$$ $$\frac{3}{2} - \frac{1}{2}*\frac{5}{3} = \frac{3}{2} - \frac{5}{6} = \frac{4}{3}$$ Row 3 = Row 3 - 4*Row 2: $$4 - 4*1 = 0$$ $$0 - 4*(-\frac{4}{3}) = 0 + \frac{16}{3} = \frac{16}{3}$$ $$7 - 4*\frac{5}{3} = 7 - \frac{20}{3} = \frac{1}{3}$$ Matrix now: $$\left[\begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{4}{3} \\ 0 & 1 & -\frac{4}{3} & \frac{5}{3} \\ 0 & 0 & \frac{16}{3} & \frac{1}{3} \end{array}\right]$$ 8. **Step 6: Make leading 1 in row 3** Multiply row 3 by $\frac{3}{16}$: $$\left[\begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{4}{3} \\ 0 & 1 & -\frac{4}{3} & \frac{5}{3} \\ 0 & 0 & \cancel{\frac{16}{3}}*\frac{3}{16} & \frac{1}{3}*\frac{3}{16} \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{4}{3} \\ 0 & 1 & -\frac{4}{3} & \frac{5}{3} \\ 0 & 0 & 1 & \frac{1}{16} \end{array}\right]$$ 9. **Step 7: Eliminate z in rows 1 and 2** Row 1 = Row 1 - $\frac{1}{6}$*Row 3: $$\frac{1}{6} - \frac{1}{6}*1 = 0$$ $$\frac{4}{3} - \frac{1}{6}*\frac{1}{16} = \frac{4}{3} - \frac{1}{96} = \frac{127}{96}$$ Row 2 = Row 2 + $\frac{4}{3}$*Row 3: $$-\frac{4}{3} + \frac{4}{3}*1 = 0$$ $$\frac{5}{3} + \frac{4}{3}*\frac{1}{16} = \frac{5}{3} + \frac{1}{12} = \frac{21}{12} = \frac{7}{4}$$ 10. **Final matrix:** $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & \frac{127}{96} \\ 0 & 1 & 0 & \frac{7}{4} \\ 0 & 0 & 1 & \frac{1}{16} \end{array}\right]$$ 11. **Solution:** $$x = \frac{127}{96}, \quad y = \frac{7}{4}, \quad z = \frac{1}{16}$$ --- **Answer:** $$\boxed{\left(x, y, z\right) = \left(\frac{127}{96}, \frac{7}{4}, \frac{1}{16}\right)}$$