4-) \( \left(20+20\right. \) point For given matrix \( A=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{array}\right] \) a) Find characteristic equation. b) Find eigenvalues and eigenvectors. c) If possible find an invertible matrix \( P \) such that \( D=P^{-1} A P \) is diagonal and write this diagonal matrix \( D \). d) Find \( A^{35} \). e) Find \( A^{-1} \) by using Cayley Hamilton Theorem.

To tackle the problems for the given matrix \( A \), let's go through them step by step: ### a) Find the Characteristic Equation The characteristic equation of a matrix \( A \) is found using the determinant of \( A - \lambda I = 0 \). For matrix \( A = \begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{bmatrix} \): \[ A - \lambda I = \begin{bmatrix}-1 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 1 \\ 0 & 1 & 1 - \lambda\end{bmatrix} \] Calculating the determinant, we have: \[ \text{det}(A - \lambda I) = (-1 - \lambda) \cdot \text{det} \begin{bmatrix}1 - \lambda & 1 \\ 1 & 1 - \lambda\end{bmatrix} \] \[ = (-1 - \lambda)((1 - \lambda)(1 - \lambda) - 1) = (-1 - \lambda)(\lambda^2 - 2\lambda) \] So, the characteristic equation is: \[ (-1 - \lambda)(\lambda^2 - 2 \lambda) = 0 \] ### b) Find Eigenvalues and Eigenvectors The eigenvalues are obtained from the characteristic equation: 1. \( \lambda_1 = -1 \) 2. From \( \lambda^2 - 2\lambda = 0 \), we find \( \lambda_2 = 0 \) and \( \lambda_3 = 2 \). Now let's find the eigenvectors: - **For \( \lambda_1 = -1 \)**: \[ (A + I)x = 0 \text{ leading to } \begin{bmatrix}0 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 2\end{bmatrix} \Rightarrow \text{Eigenvector } v_1 = \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} \] - **For \( \lambda_2 = 0 \)**: \[ Ax = 0 \Rightarrow \begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} = 0 \Rightarrow x_1 = 0, x_2 = -x_3 \text{, choose } x_3 = 1 \Rightarrow \text{Eigenvector } v_2 = \begin{bmatrix}0 \\ 1 \\ -1\end{bmatrix} \] - **For \( \lambda_3 = 2 \)**: \[ (A - 2I)x = 0 \Rightarrow \begin{bmatrix}-3 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1\end{bmatrix} \Rightarrow x_1 = 0, x_2 = x_3 \text{, choose } x_3 = 1 \Rightarrow \text{Eigenvector } v_3 = \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix} \] ### c) Find \( P \) such that \( D = P^{-1}AP \) Using the eigenvectors as columns of \( P \): \[ P = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1\end{bmatrix} \] We can calculate \( D = P^{-1}AP \) where \( D \) will be a diagonal matrix with eigenvalues on the diagonal: \[ D = \begin{bmatrix}-1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2\end{bmatrix} \] ### d) Find \( A^{35} \) Since \( A \) can be diagonalized as \( A = PDP^{-1} \), we can use \( D \) to simplify \( A^{35} \): \[ A^{35} = PD^{35}P^{-1} \] Calculating \( D^{35} \): \[ D^{35} = \begin{bmatrix}(-1)^{35} & 0 & 0 \\ 0