Let's solve the problem **step by step**, as instructed: Given matrix: \[ A = \begin{bmatrix} -20 & -36 \\ 4 & 20 \end{bmatrix} \] We are **diagonalize** this matrix and then compute \(A^n\). --- ### **Step 1: Find the Eigenvalues** The characteristic polynomial is: \[ \det(A - \lambda I) = 0 \] \[ \begin{vmatrix} -20 - \lambda & -36 \\ 4 & 20 - \lambda \end{vmatrix} = 0 \] \[ (-20-\lambda)(20-\lambda) - (-36)(4) = 0 \] \[ (-20-\lambda)(20-\lambda) + 144 = 0 \] \[ [-400 + 20\lambda - 20\lambda + \lambda^2] + 144 = 0 \] \[ -400 + \lambda^2 + 144 = 0 \] \[ \lambda^2 - 256 = 0 \] \[ \lambda^2 = 256 \implies \lambda = 16, -16 \] --- ### **Step 2: Find the Eigenvectors** #### For \(\lambda_1 = 16\): \[ A - 16I = \begin{bmatrix}-20-16 & -36 \\ 4 & 20-16\end{bmatrix} = \begin{bmatrix}-36 & -36 \\ 4 & 4\end{bmatrix} \] Solve: \([-36, -36] \begin{bmatrix}x_1 \\ x_2\end{bmatrix} = 0\) \[ -36x_1 - 36x_2 = 0 \implies x_1 + x_2 = 0 \] So, one eigenvector: \(\begin{bmatrix}1 \\ -1\end{bmatrix}\) --- #### For \(\lambda_2 = -16\): \[ A + 16I = \begin{bmatrix}-20+16 & -36 \\ 4 & 20+16\end{bmatrix} = \begin{bmatrix}-4 & -36 \\ 4 & 36\end{bmatrix} \] Solve: \([-4, -36] \begin{bmatrix}x_1 \\ x_2\end{bmatrix} = 0\) \[ -4x_1 - 36x_2 = 0 \implies x_1 = -9x_2 \] So, one eigenvector: \(\begin{bmatrix}-9 \\ 1\end{bmatrix}\) --- ### **Step 3: Diagonalize** Let \[ P = \begin{bmatrix}1 & -9 \\ -1 & 1\end{bmatrix},\quad D = \begin{bmatrix}16 & 0 \\ 0 & -16\end{bmatrix} \] \(A = PDP^{-1}\) --- ### **Step 4: Compute \(A^n\)** \[ A^n = P D^n P^{-1} \] Where \(D^n = \begin{bmatrix}16^n & 0 \\ 0 & (-16)^n\end{bmatrix}\) Now, \(P^{-1}\): \[ \det(P) = (1)(1) - (-9)(-1) = 1 - 9 = -8 \] \[ P^{-1} = \frac{1}{-8} \begin{bmatrix} 1 & 9 \\ 1 & 1 \end{bmatrix} \] --- ### **Step 5: Write the Formula** \[ A^n = P D^n P^{-1} \] \[ = \begin{bmatrix}1 & -9 \\ -1 & 1\end{bmatrix} \begin{bmatrix}16^n & 0 \\ 0 & (-16)^n\end{bmatrix} \frac{1}{-8} \begin{bmatrix}1 & 9 \\ 1 & 1\end{bmatrix} \] Or, more explicitly: \[ A^n = \frac{1}{-8} \begin{bmatrix}1 & -9 \\ -1 & 1\end{bmatrix} \begin{bmatrix}16^n & 0 \\ 0 & (-16)^n\end{bmatrix} \begin{bmatrix}1 & 9 \\ 1 & 1\end{bmatrix} \] You can leave your answer **in this form**, or if needed, multiply out the matrices for the full formula. --- ## **Final Answer** \[ \boxed{ A^n = \frac{1}{-8} \begin{bmatrix} 1 & -9 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 16^n & 0 \\ 0 & (-16)^n \end{bmatrix} \begin{bmatrix} 1 & 9 \\ 1 & 1 \end{bmatrix} } \] Or, filling in the answer boxes: \[ \left[ \begin{array}{cc} \frac{1}{-8}\left(16^n + 9(-16)^n\right) & \frac{1}{-8}\left(9 \cdot 16^n + 9(-16)^n\right) \\ \frac{1}{-8}\left(-16^n - (-16)^n\right) & \frac{1}{-8}\left(-9 \cdot 16^n (-16)^n\right) \end{array} \right] \]