Science:Math Exam Resources/Courses/MATH152/April 2012/Question 04 (d)/Solution 1

From part (c), we have that $\displaystyle A=MDM^{-1}$ where

$M={\begin{bmatrix}1&1\\-1&1\end{bmatrix}}$

$M^{-1}={\begin{bmatrix}1/2&-1/2\\1/2&1/2\end{bmatrix}}$

and

$D={\begin{bmatrix}1&0\\0&5\end{bmatrix}}$

Taking powers yields

${\begin{aligned}A^{k}&=(MDM^{-1})^{k}\\&=MDM^{-1}\cdot MDM^{-1}\cdot \ldots \cdot MDM^{-1}\\&=MD^{k}M^{-1}\\&={\begin{bmatrix}1&1\\-1&1\end{bmatrix}}{\begin{bmatrix}1&0\\0&5^{k}\end{bmatrix}}{\begin{bmatrix}1/2&-1/2\\1/2&1/2\end{bmatrix}}\\&=\left({\frac {1}{2}}\right){\begin{bmatrix}1+5^{k}&-1+5^{k}\\-1+5^{k}&1+5^{k}\end{bmatrix}}\end{aligned}}$