Question
For the following matrices, check whether there exists an invertible matrix P such that P-1AP is diagonal. When such a P exists, find P.
Answer :
Word Count : 959
### Problem 1: Diagonalizability of Matrix A Given matrix \( A \): \[ A = \begin{bmatrix} 0 & 1 & -3 \\ 2 & -1 & 6 \\ 1 & -1 & 4 \end{bmatrix} \] #### Step 1: Find the eigenvalues of \( A \) To check if \( A \) is diagonalizable, we first need to find its eigenvalues by solving the characteristic equation: \[ \text{det}(A - \lambda I) = 0 \] Where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalue. \[ A - \lambda I = \begin{bmatrix} 0 - \lambda & 1 & -3 \\ 2 & -1 - \lambda & 6 \\ 1 & -1 & 4 - \lambda \end{bmatrix} \] The determinant of \( A - \lambda I \) is: \[ \text{det}(A - \lambda I) = \text{det}\begin{bmatrix} -\lambda & 1 & -3 \\ 2 & -1-\lambda & 6 \\ 1 & -1 & 4-\lambda \end{bmatrix} \] Expanding along the first row: \[ = -\lambda \text{det} \begin{bmatrix} -1-\lambda & 6 \\ -1 & 4-\lambda \end{bmatrix} - 1 \text{det} \begin{bmatrix} 2 & 6 \\ 1 & 4-\lambda \end{bmatrix} + (-3) \text{det} \begin{bmatrix} 2 ___ _______ _________ ______ _____ ______ __________ _____ ___ __________ ___ _____.
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### Problem 1: Diagonalizability of Matrix A Given matrix \( A \): \[ A = \begin{bmatrix} 0 & 1 & -3 \\ 2 & -1 & 6 \\ 1 & -1 & 4 \end{bmatrix} \] #### Step 1: Find the eigenvalues of \( A \) To check if \( A \) is diagonalizable, we first need to find its eigenvalues by solving the characteristic equation: \[ \text{det}(A - \lambda I) = 0 \] Where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalue. \[ A - \lambda I = \begin{bmatrix} 0 - \lambda & 1 & -3 \\ 2 & -1 - \lambda & 6 \\ 1 & -1 & 4 - \lambda \end{bmatrix} \] The determinant of \( A - \lambda I \) is: \[ \text{det}(A - \lambda I) = \text{det}\begin{bmatrix} -\lambda & 1 & -3 \\ 2 & -1-\lambda & 6 \\ 1 & -1 & 4-\lambda \end{bmatrix} \] Expanding along the first row: \[ = -\lambda \text{det} \begin{bmatrix} -1-\lambda & 6 \\ -1 & 4-\lambda \end{bmatrix} - 1 \text{det} \begin{bmatrix} 2 & 6 \\ 1 & 4-\lambda \end{bmatrix} + (-3) \text{det} \begin{bmatrix} 2 ___ _______ _________ ______ _____ ______ __________ _____ ___ __________ ___ _____.
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