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 questions regarding the matrix \( A=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{array}\right] \), let's get started: ### a) Characteristic Equation: The characteristic polynomial of a matrix \( A \) is given by \( \det(A - \lambda I) \) where \( I \) is the identity matrix and \( \lambda \) is the eigenvalue. For your matrix \( A \), we compute: \[ A - \lambda I = \left[\begin{array}{ccc}-1 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 1 \\ 0 & 1 & 1 - \lambda\end{array}\right] \] Now, we calculate the determinant: \[ \det(A - \lambda I) = (-1 - \lambda) \cdot \det\left(\begin{array}{cc}1 - \lambda & 1 \\ 1 & 1 - \lambda\end{array}\right) \] The determinant of the 2x2 matrix: \[ \det\left(\begin{array}{cc}1 - \lambda & 1 \\ 1 & 1 - \lambda\end{array}\right) = (1 - \lambda)(1 - \lambda) - 1 = (1 - \lambda)^2 - 1 = \lambda^2 - 2\lambda \] Putting it all together, we have: \[ \det(A - \lambda I) = (-1 - \lambda)(\lambda^2 - 2\lambda) = -(\lambda + 1)(\lambda^2 - 2\lambda) \] Thus, the characteristic equation is: \[ -(\lambda + 1)(\lambda^2 - 2\lambda) = 0 \quad \Rightarrow \quad \lambda(\lambda - 2)(\lambda + 1) = 0 \] ### b) Eigenvalues and Eigenvectors: From the characteristic equation, the eigenvalues are: \[ \lambda_1 = 0, \quad \lambda_2 = 2, \quad \lambda_3 = -1 \] Now, we'll find the eigenvectors corresponding to each eigenvalue. 1. **For \( \lambda_1 = 0 \):** \[ A - 0I = A = \left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1\end{array}\right] \] Solving \( Ax = 0 \): The system gives \( x_1 = 0 \) and \( x_2 + x_3 = 0 \), so we can choose \( x_3 = t \) (free variable): \[ \text{Eigenvector: } \begin{bmatrix} 0 \\ -t \\ t \end{bmatrix} \Rightarrow \text{One eigenvector is } \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} \] 2. **For \( \lambda_2 = 2 \):** \[ A - 2I = \left[\begin{array}{ccc}-3 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1\end{array}\right] \] The system yields \( x_1 = 0 \) and \( x_2 = x_3 \), giving: \[ \text{Eigenvector: } \begin{bmatrix} 0 \\ t \\ t \end{bmatrix} \Rightarrow \text{One eigenvector is } \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \] 3. **For \( \lambda_3 = -1 \):** \[ A + I = \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 2\end{array}\right] \] The system yields \( 2x_2 + x_3 = 0 \) leading to \( x_3 = -2x_2 \), giving: \[ \text{Eigenvector: } \begin{bmatrix} 0 \\ t \\ -2t \end{bmatrix} \Rightarrow \text{One eigenvector is } \begin{bmatrix} 0 \\