Let's solve the problem step by step. Given: \[ A = \begin{bmatrix}-20 & -36 \\ & 20\end{bmatrix} \] We are to find \(A^n\) **by diagonalizing** \(A\). ### 1. **Find Eigenvalues** The characteristic equation is: \[ \det(A - \lambda I) = 0 \] \[ \begin{vmatrix} -20 - \lambda & -36 \\ 4 & 20 - \lambda \end{vmatrix} = 0 \] \[ (-20 - \lambda)(20 - \lambda) - (4)(-36) = 0 \] \[ (-20 - \lambda)(20 - \lambda) + 144 = 0 \] \[ (-20 - \lambda)(20 - \lambda) = -144 \] Now expand: \[ (-20 - \lambda)(20 - \lambda) = -400 + 20\lambda + 20\lambda - \lambda^2 = -400 + 40\lambda - \lambda^2 \] So: \[ -400 + 40\lambda - \lambda^2 + 144 = 0 \] \[ -400 + 144 = -256 \] \[ -256 + 40\lambda - \lambda^2 = 0 \] \[ -\lambda^2 + 40\lambda - 256 = 0 \] Multiply both sides by -1: \[ \lambda^2 - 40\lambda + 256 = 0 \] Solve for \(\lambda\): \[ \lambda = \frac{40 \pm \sqrt{1600 - 1024}}{2} = \frac{40 \pm \sqrt{576}}{2} = \frac{40 \pm 24}{2} \] Thus, \[ \lambda_1 = \frac{40 + 24}{2} = 32 \] \[ \lambda_2 = \frac{40 - 24}{2} = 8 \] --- ### 2. **Find Eigenvectors** #### For \(\lambda = 32\): \[ A - 32I = \begin{bmatrix}-52 & -36 \\ 4 & -12\end{bmatrix} \] Row 1: \(-52x - 36y = 0\) \(\implies 13x + 9y = 0\) \(\implies x = -\frac{9}{13}y\) Let \(y = 13\), then \(x = -9\). So, eigenvector: \(\vec{v}_1 = \begin{bmatrix} -9 \\ 13 \end{bmatrix}\) --- #### For \(\lambda = 8\): \[ A - 8I = \begin{bmatrix}-28 & -36 \\ 4 & 12\end{bmatrix} \] Row 1: \(-28x - 36y = 0\) \(\implies 7x + 9y = 0\) \(\implies x = -\frac{9}{7}y\) Let \(y = 7\), then \(x = -9\). So, eigenvector: \(\vec{v}_2 = \begin{bmatrix} -9 \\ 7 \end{bmatrix}\) --- ### 3. **Diagonalization** Let \(P = \begin{bmatrix} -9 & -9 \\ 13 & 7 \end{bmatrix}\), and \(D = \begin{bmatrix} 32 & 0 \\ 0 & 8 \end{bmatrix}\). \[ A = PDP^{-1} \] \[ A^n = P D^n P^{-1} \] where \(D^n = \begin{bmatrix} 32^n & 0 \\ 0 & 8^n \end{bmatrix}\) --- ### 4. **Find \(P^{-1}\)** \[ \det P = (-9)(7) - (-9)(13) = -63 + 117 = 54 \] So, \[ P^{-1} = \frac{1}{54}\begin{bmatrix} 7 & 9 \\ -13 & -9 \end{bmatrix} \] --- ### 5. **Final Formula** \[ A^n = P D^n P^{-1} \] Write this explicitly: \[ A^n = \begin{bmatrix} -9 & -9 \\ 13 & 7 \end{bmatrix} \begin{bmatrix} 32^n & 0 \\ 0 & 8^n \end{bmatrix} \frac{1}{54} \begin{bmatrix} 7 & 9 \\ -13 & -9 \end{bmatrix} \] You can leave it in this **diagonalized form**, or multiply out for each entry if required. --- ## **Final Boxed Answer** \[ \boxed{ A^n = \begin{bmatrix} -9 & -9 \\ 13 & 7 \end{bmatrix} \begin{bmatrix} 32^n & 0 \\ 0 & 8^n \end{bmatrix} \frac{1}{54} \begin{bmatrix} 7 & 9 \\ -13 & -9 \end{bmatrix} } \] This is the formula for any positive integer \( n \).