1. **State the problem:** Solve the system of linear equations using the Sarrus method (which applies to 3x3 determinants): $$\begin{cases} x - y + z = 1 \\ x + y = 4 \\ 2x + 2y + 2z = 9 \end{cases}$$ 2. **Rewrite the system in standard form:** $$\begin{cases} x - y + z = 1 \\ x + y + 0z = 4 \\ 2x + 2y + 2z = 9 \end{cases}$$ 3. **Write the coefficient matrix $A$ and constants vector $B$: ** $$A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & 0 \\ 2 & 2 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 4 \\ 9 \end{bmatrix}$$ 4. **Calculate the determinant of $A$ using Sarrus' rule:** $$\det(A) = 1 \cdot 1 \cdot 2 + (-1) \cdot 0 \cdot 2 + 1 \cdot 1 \cdot 2 - (1 \cdot 1 \cdot 2 + 1 \cdot 0 \cdot 1 + 2 \cdot (-1) \cdot 2)$$ Calculate each term: $$= 2 + 0 + 2 - (2 + 0 - 4) = 4 - (-2) = 4 + 2 = 6$$ 5. **Calculate determinants for $x$, $y$, and $z$ by replacing respective columns with $B$: ** - For $x$ (replace first column): $$A_x = \begin{bmatrix} 1 & -1 & 1 \\ 4 & 1 & 0 \\ 9 & 2 & 2 \end{bmatrix}$$ $$\det(A_x) = 1 \cdot 1 \cdot 2 + (-1) \cdot 0 \cdot 9 + 1 \cdot 4 \cdot 2 - (1 \cdot 1 \cdot 9 + 4 \cdot 0 \cdot 1 + 2 \cdot (-1) \cdot 4)$$ $$= 2 + 0 + 8 - (9 + 0 - 8) = 10 - 1 = 9$$ - For $y$ (replace second column): $$A_y = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 4 & 0 \\ 2 & 9 & 2 \end{bmatrix}$$ $$\det(A_y) = 1 \cdot 4 \cdot 2 + 1 \cdot 0 \cdot 2 + 1 \cdot 1 \cdot 9 - (1 \cdot 4 \cdot 2 + 1 \cdot 0 \cdot 1 + 2 \cdot 1 \cdot 9)$$ $$= 8 + 0 + 9 - (8 + 0 + 18) = 17 - 26 = -9$$ - For $z$ (replace third column): $$A_z = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & 4 \\ 2 & 2 & 9 \end{bmatrix}$$ $$\det(A_z) = 1 \cdot 1 \cdot 9 + (-1) \cdot 4 \cdot 2 + 1 \cdot 1 \cdot 2 - (1 \cdot 1 \cdot 2 + 1 \cdot 4 \cdot 1 + 9 \cdot (-1) \cdot 2)$$ $$= 9 - 8 + 2 - (2 + 4 - 18) = 3 - (-12) = 3 + 12 = 15$$ 6. **Calculate the variables using Cramer's rule:** $$x = \frac{\det(A_x)}{\det(A)} = \frac{9}{6} = \frac{3}{2}$$ $$y = \frac{\det(A_y)}{\det(A)} = \frac{-9}{6} = -\frac{3}{2}$$ $$z = \frac{\det(A_z)}{\det(A)} = \frac{15}{6} = \frac{5}{2}$$ **Final answer:** $$\boxed{x = \frac{3}{2}, \quad y = -\frac{3}{2}, \quad z = \frac{5}{2}}$$