Let's solve this step-by-step: --- ### **Step 1: Find the eigenvalues of \( A \).** Given: \( A = \begin{bmatrix} 2 & 2 \\ -1 & 5 \end{bmatrix} \) The eigenvalues \( \lambda \) satisfy \[ \det(A - \lambda I) = 0 \] \[ \left| \begin{matrix} 2-\lambda & 2 \\ -1 & 5-\lambda \end{matrix} \right| = 0 \] \[ (2-\lambda)(5-\lambda) + 2 = 0 \] \[ (2-\lambda)(5-\lambda) + 2 = 0 \] \[ (2-\lambda)(5-\lambda) + 2 = (10 - 2\lambda - 5\lambda + \lambda^2) + 2 = \lambda^2 - 7\lambda + 12 = 0 \] \[ \lambda^2 - 7\lambda + 12 = 0 \] Solve for \( \lambda \): \[ (\lambda - 3)(\lambda - 4) = 0 \] So, \( \lambda_1 = 3 \), \( \lambda_2 = 4 \). **Explanation:** We calculated the eigenvalues of \( A \) by solving the characteristic equation. --- ### **Step 2: Find the eigenvectors and construct \( P \) and \( D \).** #### For \( \lambda = 3 \): \[ A - 3I = \begin{bmatrix} -1 & 2 \\ -1 & 2 \end{bmatrix} \] Row reduces to: \[ -1x + 2y = 0 \implies x = 2y \] Take \( y = 1 \), \( x = 2 \): Eigenvector is \( \begin{bmatrix} 2 \\ 1 \end{bmatrix} \) #### For \( \lambda = 4 \): \[ A - 4I = \begin{bmatrix} -2 & 2 \\ -1 & 1 \end{bmatrix} \] \[ -2x + 2y = 0 \implies x = y \] Take \( x = 1 \), \( y = 1 \): Eigenvector is \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) So, \[ P = \left[ \begin{array}{cc} 1 & 2 \\ 1 & 1 \end{array} \right] \] \[ D = \left[ \begin{array}{cc} 4 & 0 \\ 0 & 3 \end{array} \right] \] **Explanation:** We found the eigenvectors for each eigenvalue and constructed \( P \) from the eigenvectors and \( D \) from the eigenvalues. --- ## **Final Answers:** \[ D = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix} \] \[ P = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} \]

Let's proceed step-by-step to find the matrix \( P \) that diagonalizes \( A \) and the resulting diagonal matrix \( D \): --- ### **Step 1: Find the eigenvalues of \( A \).** Given matrix: \[ A = \begin{bmatrix} 2 & 2 \\ -1 & 5 \end{bmatrix} \] Eigenvalues \( \lambda \) satisfy: \[ \det(A - \lambda I) = 0 \] Calculate the determinant: \[ \det \begin{bmatrix} 2 - \lambda & 2 \\ -1 & 5 - \lambda \end{bmatrix} = (2 - \lambda)(5 - \lambda) - (-1)(2) = (2 - \lambda)(5 - \lambda) + 2 \] Expand: \[ (2 - \lambda)(5 - \lambda) + 2 = [2 \times 5 - 2 \lambda - 5 \lambda + \lambda^2] + 2 = (10 - 7 \lambda + \lambda^2) + 2 \] Simplify: \[ \lambda^2 - 7 \lambda + 12 = 0 \] Solve quadratic: \[ \lambda^2 - 7 \lambda + 12 = 0 \] Factor: \[ (\lambda - 3)(\lambda - 4) = 0 \] So, the eigenvalues are: \[ \boxed{\lambda_1 = 3, \quad \lambda_2 = 4} \] **Explanation:** Eigenvalues are found by solving the characteristic polynomial. The roots indicate the values for which \( A - \lambda I \) is singular. --- ### **Step 2: Find the eigenvectors and form \( P \) and \( D \).** #### For \( \lambda = 3 \): \[ A - 3I = \begin{bmatrix} 2 - 3 & 2 \\ -1 & 5 - 3 \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ -1 & 2 \end{bmatrix} \] Solve \( (A - 3I) \mathbf{x} = 0 \): \[ -1 x + 2 y = 0 \implies x = 2 y \] Choose \( y = 1 \), then \( x = 2 \). Eigenvector: \[ \mathbf{v}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \] #### For \( \lambda = 4 \): \[ A - 4I = \begin{bmatrix} 2 - 4 & 2 \\ -1 & 5 - 4 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ -1 & 1 \end{bmatrix} \] Solve \( (A - 4I) \mathbf{x} = 0 \): \[ -2 x + 2 y = 0 \implies x = y \] Choose \( y = 1 \), then \( x = 1 \). Eigenvector: \[ \mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] Construct matrix \( P \) with eigenvectors as columns: \[ P = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \] The diagonal matrix \( D \) with eigenvalues in corresponding order: \[ D = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} \] --- ### **Final answer:** \[ \boxed{ P = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \quad \text{and} \quad D = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} } \] This matrix \( P \) diagonalizes \( A \) such that \( P^{-1} A P = D \).