1. **Problem Statement:** Find the eigenvalues and eigenvectors of matrix \( A = \begin{pmatrix} 11 & 9 & -2 \\ -8 & -6 & 2 \\ 4 & 4 & 1 \end{pmatrix} \). 2. **Characteristic Equation:** The eigenvalues \( \lambda \) satisfy \( \det(A - \lambda I) = 0 \). 3. **Calculate \( A - \lambda I \):** $$ A - \lambda I = \begin{pmatrix} 11 - \lambda & 9 & -2 \\ -8 & -6 - \lambda & 2 \\ 4 & 4 & 1 - \lambda \end{pmatrix} $$ 4. **Characteristic Polynomial:** $$ p(\lambda) = \det(A - \lambda I) = 6 - 11\lambda + 6\lambda^2 - \lambda^3 $$ 5. **Factor the polynomial:** $$ -(\lambda - 3)(\lambda - 2)(\lambda - 1) = 0 $$ Eigenvalues are \( \lambda = 3, 2, 1 \). 6. **Find eigenvectors for each eigenvalue by solving \( (A - \lambda I)v = 0 \):** - For \( \lambda = 3 \): $$ (A - 3I) = \begin{pmatrix} 8 & 9 & -2 \\ -8 & -9 & 2 \\ 4 & 4 & -2 \end{pmatrix} $$ Row reduce to find relations: $$ x = z, \quad y = 2z, \quad z \neq 0 $$ Eigenvector: $$ v = z \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} $$ - For \( \lambda = 2 \): $$ (A - 2I) = \begin{pmatrix} 9 & 9 & -2 \\ -8 & -8 & 2 \\ 4 & 4 & -1 \end{pmatrix} $$ Row reduce to find: $$ x = y, \quad z = 0, \quad y \neq 0 $$ Eigenvector: $$ v = y \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} $$ - For \( \lambda = 1 \): $$ (A - I) = \begin{pmatrix} 10 & 9 & -2 \\ -8 & -7 & 2 \\ 4 & 4 & 0 \end{pmatrix} $$ Row reduce to find: $$ x = -2z, \quad y = z, \quad z \neq 0 $$ Eigenvector: $$ v = z \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} $$ 7. **Summary:** - Eigenvalues: \( 3, 2, 1 \) - Corresponding eigenvectors: - \( \lambda=3: \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \) - \( \lambda=2: \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \) - \( \lambda=1: \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} \) 8. **Diagonalization:** Matrix \( P \) formed by eigenvectors: $$ P = \begin{pmatrix} 1 & 1 & -2 \\ 2 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix} $$ Diagonal matrix \( D = \operatorname{diag}(3, 2, 1) \). 9. **Cayley-Hamilton Theorem:** Matrix \( A \) satisfies its characteristic polynomial: $$ A^3 - 6A^2 + 11A - 6I = 0 $$ 10. **Minimal Polynomial:** Since eigenvalues are distinct, minimal polynomial is: $$ M(\lambda) = (\lambda - 3)(\lambda - 2)(\lambda - 1) $$ 11. **Final answer:** Eigenvalues are \( 3, 2, 1 \) with eigenvectors as above. Matrix \( A \) is diagonalizable with diagonal matrix \( D \) and invertible matrix \( P \) formed by eigenvectors.