1. **State the problem:** Solve the system of equations using the Gauss-Jordan method: $$\begin{cases} 2x + y - z = 8 \\ -3x - y + 2z = -11 \\ -2x + y + 2z = -3 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[ \begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array} \right]$$ 3. **Make the leading coefficient of the first row 1:** Divide row 1 by 2: $$R1 \to \frac{1}{2}R1 = \left[1 \quad \frac{1}{2} \quad -\frac{1}{2} \quad 4\right]$$ Matrix now: $$\left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array} \right]$$ 4. **Eliminate the first column entries below the pivot:** - For row 2: $R2 + 3 \times R1$: $$-3 + 3 \times 1 = 0$$ $$-1 + 3 \times \frac{1}{2} = -1 + \frac{3}{2} = \frac{1}{2}$$ $$2 + 3 \times (-\frac{1}{2}) = 2 - \frac{3}{2} = \frac{1}{2}$$ $$-11 + 3 \times 4 = -11 + 12 = 1$$ - For row 3: $R3 + 2 \times R1$: $$-2 + 2 \times 1 = 0$$ $$1 + 2 \times \frac{1}{2} = 1 + 1 = 2$$ $$2 + 2 \times (-\frac{1}{2}) = 2 - 1 = 1$$ $$-3 + 2 \times 4 = -3 + 8 = 5$$ Matrix now: $$\left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ 0 & 2 & 1 & 5 \end{array} \right]$$ 5. **Make the leading coefficient of row 2 equal to 1:** Multiply row 2 by 2: $$R2 \to 2 \times R2 = \left[0 \quad 1 \quad 1 \quad 2\right]$$ Matrix now: $$\left[ \begin{array}{ccc|c} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & 1 & 1 & 2 \\ 0 & 2 & 1 & 5 \end{array} \right]$$ 6. **Eliminate the second column entries above and below the pivot in row 2:** - For row 1: $R1 - \frac{1}{2} \times R2$: $$\frac{1}{2} - \frac{1}{2} \times 1 = 0$$ $$-\frac{1}{2} - \frac{1}{2} \times 1 = -\frac{1}{2} - \frac{1}{2} = -1$$ $$4 - \frac{1}{2} \times 2 = 4 - 1 = 3$$ - For row 3: $R3 - 2 \times R2$: $$2 - 2 \times 1 = 0$$ $$1 - 2 \times 1 = 1 - 2 = -1$$ $$5 - 2 \times 2 = 5 - 4 = 1$$ Matrix now: $$\left[ \begin{array}{ccc|c} 1 & 0 & -1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & -1 & 1 \end{array} \right]$$ 7. **Make the leading coefficient of row 3 equal to 1:** Multiply row 3 by -1: $$R3 \to -1 \times R3 = \left[0 \quad 0 \quad 1 \quad -1\right]$$ Matrix now: $$\left[ \begin{array}{ccc|c} 1 & 0 & -1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & -1 \end{array} \right]$$ 8. **Eliminate the third column entries above the pivot in row 3:** - For row 1: $R1 + R3$: $$-1 + 1 = 0$$ $$3 + (-1) = 2$$ - For row 2: $R2 - R3$: $$1 - 1 = 0$$ $$2 - (-1) = 3$$ Final matrix: $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array} \right]$$ 9. **Read the solution:** $$x = 2, \quad y = 3, \quad z = -1$$ **Final answer:** $$(x, y, z) = (2, 3, -1)$$