1. **State the problem:** Find the eigenvalues and corresponding eigenvectors of the matrix $$A = \begin{pmatrix}-2 & 2 & -3 \\ 2 & 1 & -6 \\ -1 & -2 & 0\end{pmatrix}$$ 2. **Recall the formula:** Eigenvalues $\lambda$ satisfy the characteristic equation $$\det(A - \lambda I) = 0$$ where $I$ is the identity matrix. 3. **Form the matrix $A - \lambda I$:** $$A - \lambda I = \begin{pmatrix}-2 - \lambda & 2 & -3 \\ 2 & 1 - \lambda & -6 \\ -1 & -2 & -\lambda \end{pmatrix}$$ 4. **Calculate the determinant:** $$\det(A - \lambda I) = (-2 - \lambda) \begin{vmatrix}1 - \lambda & -6 \\ -2 & -\lambda \end{vmatrix} - 2 \begin{vmatrix}2 & -6 \\ -1 & -\lambda \end{vmatrix} - 3 \begin{vmatrix}2 & 1 - \lambda \\ -1 & -2 \end{vmatrix}$$ Calculate each minor: - First minor: $$ (1 - \lambda)(-\lambda) - (-6)(-2) = -\lambda + \lambda^2 - 12 = \lambda^2 - \lambda - 12 $$ - Second minor: $$ 2(-\lambda) - (-6)(-1) = -2\lambda - 6 $$ - Third minor: $$ 2(-2) - (1 - \lambda)(-1) = -4 + 1 - \lambda = -3 - \lambda $$ 5. **Substitute back:** $$\det(A - \lambda I) = (-2 - \lambda)(\lambda^2 - \lambda - 12) - 2(-2\lambda - 6) - 3(-3 - \lambda)$$ 6. **Expand and simplify:** $$(-2 - \lambda)(\lambda^2 - \lambda - 12) = -2\lambda^2 + 2\lambda + 24 - \lambda^3 + \lambda^2 + 12\lambda = -\lambda^3 - \lambda^2 + 14\lambda + 24$$ Add the other terms: $$- 2(-2\lambda - 6) = 4\lambda + 12$$ $$- 3(-3 - \lambda) = 9 + 3\lambda$$ Sum all: $$-\lambda^3 - \lambda^2 + 14\lambda + 24 + 4\lambda + 12 + 9 + 3\lambda = -\lambda^3 - \lambda^2 + (14 + 4 + 3)\lambda + (24 + 12 + 9)$$ $$= -\lambda^3 - \lambda^2 + 21\lambda + 45$$ 7. **Characteristic polynomial:** $$-\lambda^3 - \lambda^2 + 21\lambda + 45 = 0$$ Multiply both sides by $-1$: $$\lambda^3 + \lambda^2 - 21\lambda - 45 = 0$$ 8. **Find roots (eigenvalues):** Try rational roots using factors of 45: $\pm1, \pm3, \pm5, \pm9, \pm15, \pm45$ Test $\lambda = 3$: $$3^3 + 3^2 - 21(3) - 45 = 27 + 9 - 63 - 45 = -72 \neq 0$$ Test $\lambda = -3$: $$(-3)^3 + (-3)^2 - 21(-3) - 45 = -27 + 9 + 63 - 45 = 0$$ So $\lambda = -3$ is a root. 9. **Divide polynomial by $(\lambda + 3)$:** $$\lambda^3 + \lambda^2 - 21\lambda - 45 = (\lambda + 3)(\lambda^2 - 2\lambda - 15)$$ 10. **Solve quadratic:** $$\lambda^2 - 2\lambda - 15 = 0$$ Use quadratic formula: $$\lambda = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2} = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm 8}{2}$$ So, $$\lambda = 5 \quad \text{or} \quad \lambda = -3$$ Eigenvalues are: $$\lambda_1 = -3, \quad \lambda_2 = 5, \quad \lambda_3 = -3$$ 11. **Find eigenvectors:** For each eigenvalue solve $$ (A - \lambda I)\mathbf{v} = 0 $$ - For $\lambda = -3$: $$A + 3I = \begin{pmatrix}1 & 2 & -3 \\ 2 & 4 & -6 \\ -1 & -2 & 3\end{pmatrix}$$ Row reduce or solve system: From first row: $$v_1 + 2v_2 - 3v_3 = 0$$ From second row: $$2v_1 + 4v_2 - 6v_3 = 0$$ (which is twice the first, so no new info) From third row: $$-v_1 - 2v_2 + 3v_3 = 0$$ (negative of first row, no new info) Set free variables: Let $v_3 = t$, then $$v_1 = 3t - 2v_2$$ Choose $v_2 = s$, then eigenvector is $$\mathbf{v} = \begin{pmatrix}3t - 2s \\ s \\ t \end{pmatrix}$$ For simplicity, choose $s=0$, $t=1$: $$\mathbf{v} = \begin{pmatrix}3 \\ 0 \\ 1 \end{pmatrix}$$ - For $\lambda = 5$: $$A - 5I = \begin{pmatrix}-7 & 2 & -3 \\ 2 & -4 & -6 \\ -1 & -2 & -5\end{pmatrix}$$ Solve: $$-7v_1 + 2v_2 - 3v_3 = 0$$ $$2v_1 - 4v_2 - 6v_3 = 0$$ $$-v_1 - 2v_2 - 5v_3 = 0$$ From second equation: $$2v_1 = 4v_2 + 6v_3 \Rightarrow v_1 = 2v_2 + 3v_3$$ Substitute into first: $$-7(2v_2 + 3v_3) + 2v_2 - 3v_3 = 0 \Rightarrow -14v_2 - 21v_3 + 2v_2 - 3v_3 = 0$$ $$-12v_2 - 24v_3 = 0 \Rightarrow -12v_2 = 24v_3 \Rightarrow v_2 = -2v_3$$ Substitute $v_2$ back into $v_1$: $$v_1 = 2(-2v_3) + 3v_3 = -4v_3 + 3v_3 = -v_3$$ Eigenvector: $$\mathbf{v} = \begin{pmatrix} -v_3 \\ -2v_3 \\ v_3 \end{pmatrix} = v_3 \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}$$ Choose $v_3 = 1$: $$\mathbf{v} = \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}$$ **Final answer:** Eigenvalues: $$\lambda_1 = -3, \quad \lambda_2 = 5, \quad \lambda_3 = -3$$ Corresponding eigenvectors: $$\lambda = -3: \quad \mathbf{v} = \begin{pmatrix}3 \\ 0 \\ 1 \end{pmatrix} \text{ and any vector of form } \begin{pmatrix}3t - 2s \\ s \\ t \end{pmatrix}$$ $$\lambda = 5: \quad \mathbf{v} = \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}$$