1. **Problem Statement:** Find the eigenvalues and eigenvectors of the matrix $$A = \begin{bmatrix} 2 & 1 & 1 \\ 2 & 1 & -2 \\ -1 & 0 & -2 \end{bmatrix}$$. 2. **Formula and Explanation:** Eigenvalues $\lambda$ satisfy the characteristic equation $$\det(A - \lambda I) = 0,$$ where $I$ is the identity matrix. Eigenvectors $\mathbf{v}$ satisfy $$ (A - \lambda I)\mathbf{v} = \mathbf{0}.$$ 3. **Calculate the characteristic polynomial:** $$A - \lambda I = \begin{bmatrix} 2-\lambda & 1 & 1 \\ 2 & 1-\lambda & -2 \\ -1 & 0 & -2-\lambda \end{bmatrix}$$ Calculate the determinant: $$\det(A - \lambda I) = (2-\lambda) \begin{vmatrix} 1-\lambda & -2 \\ 0 & -2-\lambda \end{vmatrix} - 1 \begin{vmatrix} 2 & -2 \\ -1 & -2-\lambda \end{vmatrix} + 1 \begin{vmatrix} 2 & 1-\lambda \\ -1 & 0 \end{vmatrix}$$ 4. **Evaluate minors:** $$= (2-\lambda)((1-\lambda)(-2-\lambda) - 0) - 1(2(-2-\lambda) - (-1)(-2)) + 1(2 \cdot 0 - (-1)(1-\lambda))$$ Simplify each term: $$= (2-\lambda)((1-\lambda)(-2-\lambda)) - 1(-4 - 2\lambda - 2) + 1(0 + 1 - \lambda)$$ $$= (2-\lambda)((1-\lambda)(-2-\lambda)) + 6 + 2\lambda + 1 - \lambda$$ 5. **Expand $(1-\lambda)(-2-\lambda)$:** $$= 1 \cdot (-2-\lambda) - \lambda(-2-\lambda) = -2 - \lambda + 2\lambda + \lambda^2 = -2 + \lambda + \lambda^2$$ 6. **Substitute back:** $$= (2-\lambda)(-2 + \lambda + \lambda^2) + 7 + \lambda$$ Expand: $$= (2)(-2 + \lambda + \lambda^2) - \lambda(-2 + \lambda + \lambda^2) + 7 + \lambda$$ $$= -4 + 2\lambda + 2\lambda^2 + 2\lambda - \lambda^2 - \lambda^3 + 7 + \lambda$$ Combine like terms: $$= (-4 + 7) + (2\lambda + 2\lambda + \lambda) + (2\lambda^2 - \lambda^2) - \lambda^3$$ $$= 3 + 5\lambda + \lambda^2 - \lambda^3$$ Rewrite: $$-\lambda^3 + \lambda^2 + 5\lambda + 3 = 0$$ 7. **Solve the cubic equation:** Try rational roots $\pm1, \pm3$. Test $\lambda = -1$: $$-(-1)^3 + (-1)^2 + 5(-1) + 3 = 1 + 1 - 5 + 3 = 0$$ So $\lambda = -1$ is a root. 8. **Factor polynomial:** Divide by $(\lambda + 1)$: $$-\lambda^3 + \lambda^2 + 5\lambda + 3 = -(\lambda + 1)(\lambda^2 - 2\lambda - 3)$$ 9. **Solve quadratic:** $$\lambda^2 - 2\lambda - 3 = 0$$ Use quadratic formula: $$\lambda = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm 4}{2}$$ So, $$\lambda = 3 \text{ or } \lambda = -1$$ 10. **Eigenvalues:** $$\lambda_1 = -1, \quad \lambda_2 = -1, \quad \lambda_3 = 3$$ 11. **Find eigenvectors:** For $\lambda = 3$: Solve $(A - 3I)\mathbf{v} = 0$: $$\begin{bmatrix} -1 & 1 & 1 \\ 2 & -2 & -2 \\ -1 & 0 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ From first row: $-x + y + z = 0 \Rightarrow y = x - z$ From second row: $2x - 2y - 2z = 0 \Rightarrow x - y - z = 0$ Substitute $y$: $$x - (x - z) - z = 0 \Rightarrow x - x + z - z = 0$$ True for all $x,z$. From third row: $-x - 5z = 0 \Rightarrow x = -5z$ Then $y = x - z = -5z - z = -6z$ Eigenvector: $$\mathbf{v}_3 = z \begin{bmatrix} -5 \\ -6 \\ 1 \end{bmatrix}$$ 12. For $\lambda = -1$: Solve $(A + I)\mathbf{v} = 0$: $$\begin{bmatrix} 3 & 1 & 1 \\ 2 & 2 & -2 \\ -1 & 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$$ From third row: $-x - z = 0 \Rightarrow x = -z$ From first row: $3x + y + z = 0 \Rightarrow 3(-z) + y + z = 0 \Rightarrow y - 2z = 0 \Rightarrow y = 2z$ From second row: $2x + 2y - 2z = 0 \Rightarrow 2(-z) + 2(2z) - 2z = 0 \Rightarrow -2z + 4z - 2z = 0$ True for all $z$. Eigenvector: $$\mathbf{v}_{-1} = z \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$$ 13. **Summary:** Eigenvalues: $$\lambda_1 = 3, \quad \lambda_2 = -1, \quad \lambda_3 = -1$$ Eigenvectors: $$\mathbf{v}_1 = \begin{bmatrix} -5 \\ -6 \\ 1 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$$ The eigenvalue $-1$ has multiplicity 2 with eigenvector above. Final answer: Eigenvalues are $3, -1, -1$ with corresponding eigenvectors as shown.