Alternatively, we can also show this by straightforward computation, using the breakdown in part b):
A 1995 = ( B [ 1 0 0 − 1 ] B − 1 ) 1995 = B [ 1 0 0 − 1 ] 1995 B − 1 = B [ 1 1995 0 0 ( − 1 ) 1995 ] B − 1 = B [ 1 0 0 − 1 ] B − 1 = A {\displaystyle {\begin{aligned}A^{1995}&=\left(B\left[{\begin{array}{cc}1&0\\0&-1\end{array}}\right]B^{-1}\right)^{1995}\\&=B\left[{\begin{array}{cc}1&0\\0&-1\end{array}}\right]^{1995}B^{-1}\\&=B\left[{\begin{array}{cc}1^{1995}&0\\0&(-1)^{1995}\end{array}}\right]B^{-1}\\&=B\left[{\begin{array}{cc}1&0\\0&-1\end{array}}\right]B^{-1}\\&=A\end{aligned}}}
![{\displaystyle {\begin{aligned}A^{1995}&=\left(B\left[{\begin{array}{cc}1&0\\0&-1\end{array}}\right]B^{-1}\right)^{1995}\\&=B\left[{\begin{array}{cc}1&0\\0&-1\end{array}}\right]^{1995}B^{-1}\\&=B\left[{\begin{array}{cc}1^{1995}&0\\0&(-1)^{1995}\end{array}}\right]B^{-1}\\&=B\left[{\begin{array}{cc}1&0\\0&-1\end{array}}\right]B^{-1}\\&=A\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/25b82a53ee2c4cc6bd31805df5b552d73eb5165b)
