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

From part (c), we have that $A=MD{M}^{-1}$ where

$M=\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]$

$M^{-1}=\left[\begin{array}{cc}1/2& -1/2\\ 1/2& 1/2\end{array}\right]$

and

$D=\left[\begin{array}{cc}1& 0\\ 0& 5\end{array}\right]$

Taking powers yields

$\begin{array}{rl}{A}^k& =(MD{M}^{-1}{)}^k\\ & =MD{M}^{-1}\cdot MD{M}^{-1}\cdot \dots \cdot MD{M}^{-1}\\ & =M{D}^k{M}^{-1}\\ & =\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 5^k\end{array}\right]\left[\begin{array}{cc}1/2& -1/2\\ 1/2& 1/2\end{array}\right]\\ & =\left(\frac{1}{2}\right)\left[\begin{array}{cc}1+5^k& -1+5^k\\ -1+5^k& 1+5^k\end{array}\right]\end{array}$