1. **State the problem:** Solve the system of equations using row operations to reach echelon or reduced echelon form: $$\begin{cases} x + 2y + z = 1 \\ 2x + 2y + 0 = 2 \\ 3x + 5y + 4z = 1 \end{cases}$$ 2. **Write the augmented matrix (Ab):** $$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 1 \\ 2 & 2 & 0 & 2 \\ 3 & 5 & 4 & 1 \end{array}\right]$$ 3. **Goal:** Use row operations to get zeros below the leading 1 in the first column. 4. **Step 1:** Eliminate the 2 in row 2, column 1: $$R_2 \to R_2 - 2R_1 = [2 - 2(1), 2 - 2(2), 0 - 2(1), 2 - 2(1)] = [0, -2, -2, 0]$$ 5. **Step 2:** Eliminate the 3 in row 3, column 1: $$R_3 \to R_3 - 3R_1 = [3 - 3(1), 5 - 3(2), 4 - 3(1), 1 - 3(1)] = [0, -1, 1, -2]$$ 6. **New matrix:** $$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 1 \\ 0 & -2 & -2 & 0 \\ 0 & -1 & 1 & -2 \end{array}\right]$$ 7. **Step 3:** Make the leading coefficient in row 2 a 1: $$R_2 \to -\frac{1}{2} R_2 = [0, 1, 1, 0]$$ 8. **Step 4:** Eliminate the -1 in row 3, column 2: $$R_3 \to R_3 + R_2 = [0, -1 + 1, 1 + 1, -2 + 0] = [0, 0, 2, -2]$$ 9. **New matrix:** $$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 2 & -2 \end{array}\right]$$ 10. **Step 5:** Make the leading coefficient in row 3 a 1: $$R_3 \to \frac{1}{2} R_3 = [0, 0, 1, -1]$$ 11. **Step 6:** Eliminate the 1 in row 1, column 3: $$R_1 \to R_1 - R_3 = [1, 2, 1 - 1, 1 - (-1)] = [1, 2, 0, 2]$$ 12. **Step 7:** Eliminate the 1 in row 2, column 3: $$R_2 \to R_2 - R_3 = [0, 1, 1 - 1, 0 - (-1)] = [0, 1, 0, 1]$$ 13. **Step 8:** Eliminate the 2 in row 1, column 2: $$R_1 \to R_1 - 2R_2 = [1, 2 - 2(1), 0 - 2(0), 2 - 2(1)] = [1, 0, 0, 0]$$ 14. **Final matrix in reduced echelon form:** $$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \end{array}\right]$$ 15. **Solution:** $$x = 0, \quad y = 1, \quad z = -1$$ This means the system has a unique solution at $(x,y,z) = (0,1,-1)$.