91Ó°ÊÓ

To determine if matrix \( A \) is diagonalizable, we first need to find its eigenvalues. The eigenvalues \( \lambda \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). Compute:\[A - \lambda I = \begin{bmatrix} 5-\lambda & 2 \ 2 & 5-\lambda \end{bmatrix}\]The determinant is:\[\det(A - \lambda I) = (5-\lambda)(5-\lambda) - 2 \times 2 = (5-\lambda)^2 - 4\]Set the determinant equal to zero:\[(5-\lambda)^2 - 4 = 0\]\[(5-\lambda)^2 = 4\]\[5-\lambda = \pm 2\]\[\lambda = 5 \pm 2\]Thus, the eigenvalues are \( \lambda_1 = 7 \) and \( \lambda_2 = 3 \).

For \( \lambda_1 = 7 \), substitute into \( A - \lambda I \) and solve for the eigenvector:\[A - 7I = \begin{bmatrix} 5-7 & 2 \ 2 & 5-7 \end{bmatrix} = \begin{bmatrix} -2 & 2 \ 2 & -2 \end{bmatrix}\]Solving \( \begin{bmatrix} -2 & 2 \ 2 & -2 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \):From the first row, \(-2x + 2y = 0\) gives \(y = x\). An eigenvector for \( \lambda_1 = 7 \) can be \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \).For \( \lambda_2 = 3 \), substitute into \( A - \lambda I \) and solve similarly:\[A - 3I = \begin{bmatrix} 5-3 & 2 \ 2 & 5-3 \end{bmatrix} = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix}\]Solving \( \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \):From the first row, \(2x + 2y = 0\) gives \(y = -x\). An eigenvector for \( \lambda_2 = 3 \) is \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \).

Since we found distinct eigenvectors for each eigenvalue, matrix \( A \) is diagonalizable. Construct the matrix \( P \) using the eigenvectors as columns:\[P = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix}\]Now, form the diagonal matrix \( D \) using the eigenvalues:\[D = \begin{bmatrix} 7 & 0 \ 0 & 3 \end{bmatrix}\]

Verify the diagonalization by checking if \( P^{-1}AP = D \). First, find \( P^{-1} \). The determinant of \( P \) is \(-2 \) and the inverse is:\[P^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & -1 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} 0.5 & 0.5 \ 0.5 & -0.5 \end{bmatrix}\]Multiply \( P^{-1} \), \( A \), and \( P \):\[P^{-1}AP = \begin{bmatrix} 0.5 & 0.5 \ 0.5 & -0.5 \end{bmatrix} \begin{bmatrix} 5 & 2 \ 2 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \ 0 & 3 \end{bmatrix} = D\]This confirms the diagonalization is correct.