1. **State the problem:** Solve the system of equations using the Gauss Elimination Method: $$\begin{cases} 2y + z = -8 \\ x - 2y - 3z = 0 \\ -x + y + 2z = 3 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 0 & 2 & 1 & -8 \\ 1 & -2 & -3 & 0 \\ -1 & 1 & 2 & 3 \end{array}\right]$$ 3. **Swap rows to get a leading 1 in the first row:** Swap row 1 and row 2: $$\left[\begin{array}{ccc|c} 1 & -2 & -3 & 0 \\ 0 & 2 & 1 & -8 \\ -1 & 1 & 2 & 3 \end{array}\right]$$ 4. **Eliminate the first variable from row 3:** Add row 1 to row 3: $$R_3 = R_3 + R_1 \Rightarrow \left[\begin{array}{ccc|c} 1 & -2 & -3 & 0 \\ 0 & 2 & 1 & -8 \\ 0 & -1 & -1 & 3 \end{array}\right]$$ 5. **Make the leading coefficient of row 2 equal to 1:** Divide row 2 by 2: $$R_2 = \frac{1}{2} R_2 \Rightarrow \left[\begin{array}{ccc|c} 1 & -2 & -3 & 0 \\ 0 & 1 & \frac{1}{2} & -4 \\ 0 & -1 & -1 & 3 \end{array}\right]$$ 6. **Eliminate the second variable from row 3:** Add row 2 to row 3: $$R_3 = R_3 + R_2 \Rightarrow \left[\begin{array}{ccc|c} 1 & -2 & -3 & 0 \\ 0 & 1 & \frac{1}{2} & -4 \\ 0 & 0 & -\frac{1}{2} & -1 \end{array}\right]$$ 7. **Make the leading coefficient of row 3 equal to 1:** Multiply row 3 by -2: $$R_3 = -2 R_3 \Rightarrow \left[\begin{array}{ccc|c} 1 & -2 & -3 & 0 \\ 0 & 1 & \frac{1}{2} & -4 \\ 0 & 0 & 1 & 2 \end{array}\right]$$ 8. **Back substitution:** - From row 3: $z = 2$ - Substitute $z$ into row 2: $y + \frac{1}{2} \times 2 = -4 \Rightarrow y + 1 = -4 \Rightarrow y = -5$ - Substitute $y$ and $z$ into row 1: $x - 2(-5) - 3(2) = 0 \Rightarrow x + 10 - 6 = 0 \Rightarrow x + 4 = 0 \Rightarrow x = -4$ **Final solution:** $$\boxed{x = -4, y = -5, z = 2}$$