1. **State the problem:** Solve the system of equations using Gaussian elimination: $$\begin{cases} x + y - z = -2 \\ 2x - y + z = 5 \\ -x + 2y + 2z = 1 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \\ -1 & 2 & 2 & 1 \end{array}\right]$$ 3. **Use row operations to get upper triangular form:** - Eliminate $x$ from rows 2 and 3. Row 2: $R_2 - 2R_1 \to R_2$ $$\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 2 & -1 & 1 & 5 \end{array}\right] - 2 \times \left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \end{array}\right] = \left[\begin{array}{ccc|c} 0 & -3 & 3 & 9 \end{array}\right]$$ Row 3: $R_3 + R_1 \to R_3$ $$\left[\begin{array}{ccc|c} -1 & 2 & 2 & 1 \end{array}\right] + \left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \end{array}\right] = \left[\begin{array}{ccc|c} 0 & 3 & 1 & -1 \end{array}\right]$$ New matrix: $$\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 0 & -3 & 3 & 9 \\ 0 & 3 & 1 & -1 \end{array}\right]$$ 4. **Eliminate $y$ from row 3:** Add $R_2$ and $R_3$ after scaling: Multiply $R_2$ by $\frac{1}{-3}$ to simplify: $$R_2 = \frac{1}{-3} R_2 = \left[0, 1, -1, -3\right]$$ Now add $R_2$ to $R_3$: $$R_3 + R_2 = \left[0, 3, 1, -1\right] + \left[0, 1, -1, -3\right] = \left[0, 4, 0, -4\right]$$ But this is incorrect because $R_2$ was scaled, so better to do: Multiply $R_2$ by $-1$ and add to $R_3$: $$R_3 + R_2 = \left[0, 3, 1, -1\right] + \left[0, 3, -3, -9\right] = \left[0, 6, -2, -10\right]$$ This is inconsistent with previous step, so correct approach: Instead, do $R_3 + R_2$ after multiplying $R_2$ by $1$: $$R_3 + R_2 = \left[0, 3, 1, -1\right] + \left[0, -3, 3, 9\right] = \left[0, 0, 4, 8\right]$$ So new $R_3$ is: $$\left[0, 0, 4, 8\right]$$ New matrix: $$\left[\begin{array}{ccc|c} 1 & 1 & -1 & -2 \\ 0 & -3 & 3 & 9 \\ 0 & 0 & 4 & 8 \end{array}\right]$$ 5. **Back substitution:** From $R_3$: $$4z = 8 \implies z = \frac{8}{4} = 2$$ From $R_2$: $$-3y + 3z = 9 \implies -3y + 3(2) = 9 \implies -3y + 6 = 9$$ $$-3y = 3 \implies y = -1$$ From $R_1$: $$x + y - z = -2 \implies x + (-1) - 2 = -2$$ $$x - 3 = -2 \implies x = 1$$ 6. **Final solution:** $$\boxed{(x, y, z) = (1, -1, 2)}$$