Let's solve the problem **step by step**. ### **Step 1: Write the Matrix** Let \[ A = \begin{bmatrix} -20 & -36 \\ 4 & 20 \end{bmatrix} \] We are asked to find \( A^n \) for positive integer \( n \) by **diagonalizing** \( A \). --- ### **Step 2: Find Eigenvalues of A** The characteristic polynomial is: \[ \det(A - \lambda I) = 0 \] So, \[ \begin{vmatrix} -20 - \lambda & -36 \\ 4 & 20 - \lambda \end{vmatrix} = 0 \] \[ (-20 - \lambda)(20 - \lambda) - (4)(-36) = 0 \] \[ (-20 - \lambda)(20 - \lambda) + 144 = 0 \] Let's expand \((-20 - \lambda)(20 - \lambda)\): \[ = (-20)(20 - \lambda) - \lambda(20 - \lambda) \] \[ = -400 + 20\lambda - 20\lambda + \lambda^2 \] \[ = -400 + \lambda^2 \] So, \[ -400 + \lambda^2 + 144 = 0 \] \[ \lambda^2 - 256 = 0 \] \[ \lambda^2 = 256 \] \[ \lambda = 16, \quad \lambda = -16 \] --- ### **Step 3: Find Eigenvectors** #### For \(\lambda_1 = 16\): \(A - 16I\): \[ \begin{bmatrix} -20 - 16 & -36 \\ 4 & 20 - 16 \end{bmatrix} = \begin{bmatrix} -36 & -36 \\ 4 & 4 \end{bmatrix} \] Let eigenvector be \(\vec{v}_1 = \begin{bmatrix} x \\ y \end{bmatrix}\): \[ \begin{cases} -36x - 36y = 0 \\ 4x + 4y = 0 \end{cases} \implies x + y = 0 \implies y = -x \] So, eigenvector: \(\vec{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\) --- #### For \(\lambda_2 = -16\): \(A + 16I\): \[ \begin{bmatrix} -20 + 16 & -36 \\ 4 & 20 + 16 \end{bmatrix} = \begin{bmatrix} -4 & -36 \\ 4 & 36 \end{bmatrix} \] Let eigenvector be \(\vec{v}_2 = \begin{bmatrix} x \\ y \end{bmatrix}\): \[ \begin{cases} -4x - 36y = 0 \\ 4x + 36y = 0 \end{cases} \implies x + 9y = 0 \implies y = -\frac{1}{9}x \] So, eigenvector: \(\vec{v}_2 = \begin{bmatrix} 1 \\ -\frac{1}{9} \end{bmatrix}\) --- ### **Step 4: Diagonalize \(A\)** Let \[ P = \begin{bmatrix} 1 & 1 \\ -1 & -\frac{1}{9} \end{bmatrix}, \quad D = \begin{bmatrix} 16 & 0 \\ 0 & -16 \end{bmatrix} \] Then \[ A = P D P^{-1} \] So, \[ A^n = P D^n P^{-1} \] where \[ D^n = \begin{bmatrix} 16^n & 0 \\ 0 & (-16)^n \end{bmatrix} \] --- ### **Step 5: Compute \(P^{-1}\)** First, find \(\det P\): \[ \det P = (1) \left(-\frac{1}{9}\right) - (1)(-1) = -\frac{1}{9} + 1 = \frac{8}{9} \] So, \[ P^{-1} = \frac{1}{\det P} \begin{bmatrix} -\frac{1}{9} & -1 \\ 1 & 1 \end{bmatrix} = \frac{9}{8} \begin{bmatrix} -\frac{1}{9} & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{8} & -\frac{9}{8} \\ \frac{9}{8} & \frac{9}{8} \end{bmatrix} \] --- ### **Step 6: Write the Final Formula** \[ A^n = P D^n P^{-1} \] Plug in the matrices: \[ A^n = \begin{bmatrix} 1 & 1 \\ -1 & -\frac{1}{9} \end{bmatrix} \begin{bmatrix} 16^n & 0 \\ 0 & (-16)^n \end{bmatrix} \begin{bmatrix} -\frac{1}{8} & -\frac{9}{8} \\ \frac{9}{8} & \frac{9}{8} \end{bmatrix} \] --- ## **Final Answer (in boxed form):** \[ \boxed{ \begin{bmatrix} -20 & -36 \\ 4 & 20 \end{bmatrix}^n = \begin{bmatrix} 1 & 1 \\ -1 & -\frac{1}{9} \end{bmatrix} \begin{bmatrix} 16^n & 0 \\ 0 & (-16)^n \end{bmatrix} \begin{bmatrix} -\frac{1}{8} & -\frac{9}{8} \\ \frac{9}{8} & \frac{9}{8} \end{bmatrix} } \] This formula provides \(A^n\) for any positive integer \(n\).