1. **Problem Statement:** Solve the system of linear equations using the Gauss-Jordan elimination method: $$\begin{cases} 2x + y - z = 8 \\ -3x - y + 2z = -11 \\ -2x + y + 2z = -3 \end{cases}$$ 2. **Method Overview:** Gauss-Jordan elimination transforms the augmented matrix of the system into reduced row echelon form (RREF) to find the solution directly. 3. **Step 1: Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array}\right]$$ 4. **Step 2: Make the pivot of the first row 1 by dividing row 1 by 2:** $$R1 \to \frac{1}{2}R1 = \left[1 \quad \frac{1}{2} \quad -\frac{1}{2} \quad 4\right]$$ 5. **Step 3: Eliminate the first column entries in rows 2 and 3:** - Row 2: $R2 + 3 \times R1 \to R2$ $$-3 + 3 \times 1 = 0$$ $$-1 + 3 \times \frac{1}{2} = \frac{1}{2}$$ $$2 + 3 \times -\frac{1}{2} = \frac{1}{2}$$ $$-11 + 3 \times 4 = 1$$ - Row 3: $R3 + 2 \times R1 \to R3$ $$-2 + 2 \times 1 = 0$$ $$1 + 2 \times \frac{1}{2} = 2$$ $$2 + 2 \times -\frac{1}{2} = 1$$ $$-3 + 2 \times 4 = 5$$ 6. **New matrix:** $$\left[\begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ 0 & 2 & 1 & 5 \end{array}\right]$$ 7. **Step 4: Make the pivot of row 2 equal to 1 by multiplying row 2 by 2:** $$R2 \to 2R2 = \left[0 \quad 1 \quad 1 \quad 2\right]$$ 8. **Step 5: Eliminate the second column entries in rows 1 and 3:** - Row 1: $R1 - \frac{1}{2} \times R2 \to R1$ $$\frac{1}{2} - \frac{1}{2} \times 1 = 0$$ $$-\frac{1}{2} - \frac{1}{2} \times 1 = -1$$ $$4 - \frac{1}{2} \times 2 = 3$$ - Row 3: $R3 - 2 \times R2 \to R3$ $$2 - 2 \times 1 = 0$$ $$1 - 2 \times 1 = -1$$ $$5 - 2 \times 2 = 1$$ 9. **New matrix:** $$\left[\begin{array}{ccc|c} 1 & 0 & -1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & -1 & 1 \end{array}\right]$$ 10. **Step 6: Make the pivot of row 3 equal to 1 by multiplying row 3 by -1:** $$R3 \to -1 \times R3 = \left[0 \quad 0 \quad 1 \quad -1\right]$$ 11. **Step 7: Eliminate the third column entries in rows 1 and 2:** - Row 1: $R1 + R3 \to R1$ $$-1 + 1 = 0$$ $$3 + (-1) = 2$$ - Row 2: $R2 - R3 \to R2$ $$1 - 1 = 0$$ $$2 - (-1) = 3$$ 12. **Final matrix in RREF:** $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array}\right]$$ 13. **Solution:** $$x = 2, \quad y = 3, \quad z = -1$$ This completes the Gauss-Jordan elimination method to solve the system.