1. **State the problem:** Solve the system of equations using diagonal methods (diagonalization or matrix methods): $$\begin{cases} x + y + z = 0 \\ x - 2y + 2z = 4 \\ x + 2y - z = 2 \end{cases}$$ 2. **Write the system in matrix form:** $$A\mathbf{x} = \mathbf{b}$$ where $$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & 2 \\ 1 & 2 & -1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 0 \\ 4 \\ 2 \end{bmatrix}$$ 3. **Find the eigenvalues of matrix $A$ by solving $\det(A - \lambda I) = 0$:** $$\det\begin{bmatrix} 1-\lambda & 1 & 1 \\ 1 & -2-\lambda & 2 \\ 1 & 2 & -1-\lambda \end{bmatrix} = 0$$ Calculate the determinant: $$ (1-\lambda) \left((-2-\lambda)(-1-\lambda) - 4\right) - 1 \left(1(-1-\lambda) - 2 \cdot 1\right) + 1 \left(1 \cdot 2 - (-2-\lambda) \cdot 1\right) = 0 $$ Simplify step-by-step: $$ (1-\lambda)((\lambda^2 + 3\lambda + 2) - 4) - 1(-1 - \lambda - 2) + 1(2 + 2 + \lambda) = 0 $$ $$ (1-\lambda)(\lambda^2 + 3\lambda - 2) + (3 + \lambda) + (4 + \lambda) = 0 $$ $$ (1-\lambda)(\lambda^2 + 3\lambda - 2) + 7 + 2\lambda = 0 $$ Expand: $$ (\lambda^2 + 3\lambda - 2) - \lambda(\lambda^2 + 3\lambda - 2) + 7 + 2\lambda = 0 $$ $$ \lambda^2 + 3\lambda - 2 - \lambda^3 - 3\lambda^2 + 2\lambda + 7 + 2\lambda = 0 $$ Combine like terms: $$ -\lambda^3 - 2\lambda^2 + 7\lambda + 5 = 0 $$ Multiply both sides by -1: $$ \lambda^3 + 2\lambda^2 - 7\lambda - 5 = 0 $$ 4. **Find roots of the cubic:** Try rational roots $\pm1, \pm5$. Test $\lambda=1$: $$1 + 2 - 7 - 5 = -9 \neq 0$$ Test $\lambda=-1$: $$-1 + 2 + 7 - 5 = 3 \neq 0$$ Test $\lambda=5$: $$125 + 50 - 35 - 5 = 135 \neq 0$$ Test $\lambda=-5$: $$-125 + 50 + 35 - 5 = -45 \neq 0$$ No rational roots; eigenvalues are irrational or complex. For diagonal methods, we can use numerical approximation or proceed with matrix inversion. 5. **Alternative: Solve by matrix inversion (since diagonalization is complicated here):** Calculate $\mathbf{x} = A^{-1} \mathbf{b}$. 6. **Calculate determinant of $A$:** $$\det(A) = 1((-2)(-1) - 2 \cdot 2) - 1(1(-1) - 2 \cdot 1) + 1(1 \cdot 2 - (-2) \cdot 1) = 1(2 - 4) - 1(-1 - 2) + 1(2 + 2) = -2 + 3 + 4 = 5$$ 7. **Find inverse $A^{-1} = \frac{1}{5} \text{adj}(A)$:** Calculate cofactors and adjugate matrix: $$\text{adj}(A) = \begin{bmatrix} (-2)(-1) - 2 \cdot 2 & -(1(-1) - 2 \cdot 1) & 1 \cdot 2 - (-2) \cdot 1 \\ -(1(-1) - 2 \cdot 1) & 1( -1) - 1 \cdot 1 & -(1 \cdot 2 - 1 \cdot 1) \\ 1 \cdot 2 - 2 \cdot 1 & -(1 \cdot 2 - 1 \cdot 1) & 1(-2) - 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 2 - 4 & -(-1 - 2) & 2 + 2 \\ -(-1 - 2) & -1 - 1 & -(2 - 1) \\ 2 - 2 & -(2 - 1) & -2 - 1 \end{bmatrix} = \begin{bmatrix} -2 & 3 & 4 \\ 3 & -2 & -1 \\ 0 & -1 & -3 \end{bmatrix}$$ 8. **Multiply $A^{-1} \mathbf{b}$:** $$\mathbf{x} = \frac{1}{5} \begin{bmatrix} -2 & 3 & 4 \\ 3 & -2 & -1 \\ 0 & -1 & -3 \end{bmatrix} \begin{bmatrix} 0 \\ 4 \\ 2 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} -2 \cdot 0 + 3 \cdot 4 + 4 \cdot 2 \\ 3 \cdot 0 - 2 \cdot 4 - 1 \cdot 2 \\ 0 \cdot 0 - 1 \cdot 4 - 3 \cdot 2 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 0 + 12 + 8 \\ 0 - 8 - 2 \\ 0 - 4 - 6 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 20 \\ -10 \\ -10 \end{bmatrix} = \begin{bmatrix} 4 \\ -2 \\ -2 \end{bmatrix}$$ 9. **Final solution:** $$x = 4, \quad y = -2, \quad z = -2$$