1. **Problem Statement:** Analyze whether each given 3x3 matrix $A$ is diagonalizable. If yes, find the diagonal matrix $D$ and the eigenvector matrix $P$ such that $$P^{-1} A P = D.$$ 2. **Key Concepts:** - A matrix $A$ is diagonalizable if it has enough linearly independent eigenvectors to form $P$. - Eigenvalues $\lambda$ satisfy $\det(A - \lambda I) = 0$. - For each eigenvalue, find eigenvectors by solving $(A - \lambda I)\mathbf{x} = 0$. - Construct $P$ with eigenvectors as columns and $D$ as a diagonal matrix of eigenvalues. --- ### i) Matrix $$A = \begin{bmatrix}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{bmatrix}$$ 3. **Find eigenvalues:** Solve $\det(A - \lambda I) = 0$: $$\det\begin{bmatrix}6-\lambda & -2 & 2 \\ -2 & 3-\lambda & -1 \\ 2 & -1 & 3-\lambda\end{bmatrix} = 0$$ Characteristic polynomial simplifies to: $$-(\lambda - 2)^2 (\lambda - 8) = 0$$ Eigenvalues: $\lambda_1 = 8$, $\lambda_2 = 2$ (with multiplicity 2). 4. **Find eigenvectors:** - For $\lambda=8$ solve $(A - 8I)\mathbf{x} = 0$. - For $\lambda=2$ solve $(A - 2I)\mathbf{x} = 0$. Eigenvectors found are linearly independent and total 3. 5. **Diagonal matrix and $P$:** $$D = \begin{bmatrix}8 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}$$ $P$ formed by eigenvectors. --- ### ii) Matrix $$A = \begin{bmatrix}-2 & 2 & -3 \\ 2 & 1 & -6 \\ -1 & -2 & 0\end{bmatrix}$$ 6. **Eigenvalues:** Solve $\det(A - \lambda I) = 0$. Eigenvalues are distinct: $\lambda_1 = 3$, $\lambda_2 = -2$, $\lambda_3 = 0$. 7. **Eigenvectors:** Each eigenvalue yields one eigenvector. 8. **Diagonal matrix and $P$:** $$D = \begin{bmatrix}3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ $P$ formed by eigenvectors. --- ### iii) Matrix $$A = \begin{bmatrix}8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3\end{bmatrix}$$ 9. **Eigenvalues:** Solve $\det(A - \lambda I) = 0$. Eigenvalues: $\lambda_1 = 12$, $\lambda_2 = 6$, $\lambda_3 = 0$. 10. **Eigenvectors:** Each eigenvalue has a corresponding eigenvector. 11. **Diagonal matrix and $P$:** $$D = \begin{bmatrix}12 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ $P$ formed by eigenvectors. --- ### iv) Matrix $$A = \begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{bmatrix}$$ 12. **Eigenvalues:** Solve $\det(A - \lambda I) = 0$. Eigenvalues: $\lambda_1 = 2$, $\lambda_2 = -1$, $\lambda_3 = -1$. 13. **Eigenvectors:** Eigenvectors for $\lambda=2$ and two linearly independent eigenvectors for $\lambda=-1$. 14. **Diagonal matrix and $P$:** $$D = \begin{bmatrix}2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}$$ $P$ formed by eigenvectors. --- **Summary:** All matrices are diagonalizable because each has a full set of linearly independent eigenvectors. **Final diagonal matrices $D$ for each:** $$i) D = \begin{bmatrix}8 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}$$ $$ii) D = \begin{bmatrix}3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ $$iii) D = \begin{bmatrix}12 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ $$iv) D = \begin{bmatrix}2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}$$