2021-M2-11

Hence, we have

$\begin{array}{cl}& PA{P}^{-1}\\ =& \left(\begin{array}{cc}\sin \theta & \cos \theta \\ -\cos \theta & \sin \theta \end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \beta & -\alpha \end{array}\right)\left(\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right)\\ =& \left(\begin{array}{cc}\alpha \sin \theta +\beta \cos \theta & \beta \sin \theta -\alpha \cos \theta \\ -\alpha \cos \theta +\beta \sin \theta & -\beta \cos \theta -\alpha \sin \theta \end{array}\right)\left(\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right)\\ =& \left(\begin{array}{cc}\alpha {\sin }^2\theta +\beta \sin \theta \cos \theta +\beta \sin \theta \cos \theta -\alpha {\cos }^2\theta & -\alpha \sin \theta \cos \theta +-\beta \cos {\theta }^2+\beta {\sin }^2\theta -\alpha \sin \theta \cos \theta \\ -\alpha \sin \theta \cos \theta +\beta {\sin }^2\theta -\beta {\cos }^2\theta -\alpha \sin \theta \cos \theta & \alpha {\cos }^2\theta -\beta \sin \theta \cos \theta -\beta \sin \theta \cos \theta -\alpha {\sin }^2\theta \end{array}\right)\\ =& \left(\begin{array}{cc}\alpha ({\sin }^2\theta -{\cos }^2\theta )+2\beta \sin \theta \cos \theta & \beta ({\sin }^2\theta -{\cos }^2\theta )-2\alpha \sin \theta \cos \theta \\ \beta ({\sin }^2\theta -{\cos }^2\theta )-2\alpha \sin \theta \cos \theta & \alpha ({\cos }^2\theta -{\sin }^2\theta )-2\beta \sin \theta \cos \theta \end{array}\right)\\ =& \left(\begin{array}{cc}-\alpha \cos 2\theta +\beta \sin 2\theta & -\beta \cos 2\theta -\alpha \sin 2\theta \\ -\beta \cos 2\theta -\alpha \sin 2\theta & \alpha \cos 2\theta -\beta \sin 2\theta \end{array}\right)\end{array}$

1. Sub. $\alpha =1$ and $\beta =\sqrt{3}$ into the result of (a), we have

   $\begin{array}{rcl}PB{P}^{-1}& =& \left(\begin{array}{cc}-\cos 2\theta +\sqrt{3}\sin 2\theta & -\sqrt{3}\cos 2\theta -\sin 2\theta \\ -\sqrt{3}\cos 2\theta -\sin 2\theta & \cos 2\theta -\sqrt{3}\sin 2\theta \end{array}\right)\\ \left(\begin{array}{cc}\lambda & 0\\ 0& \mu \end{array}\right)& =& \left(\begin{array}{cc}-\cos 2\theta +\sqrt{3}\sin 2\theta & -\sqrt{3}\cos 2\theta -\sin 2\theta \\ -\sqrt{3}\cos 2\theta -\sin 2\theta & \cos 2\theta -\sqrt{3}\sin 2\theta \end{array}\right)\end{array}$

   By comparing the elements of both sides, we have

   $\begin{array}{rcl}-\sqrt{3}\cos 2\theta -\sin 2\theta & =& 0\\ -\sin 2\theta & =& \sqrt{3}\cos 2\theta \\ \tan 2\theta & =& -\sqrt{3}\end{array}$

   $\therefore 2\theta =\pi -{\displaystyle \frac{\pi }{3}}$ or $2\theta =2\pi -{\displaystyle \frac{\pi }{3}}$.

   $\theta ={\displaystyle \frac{\pi }{3}}$ (rejected) or $\theta ={\displaystyle \frac{5\pi }{6}}$.

2. By the result of (b)(i), we have

   $\begin{array}{rcl}\lambda & =& -\cos 2\left({\displaystyle \frac{5\pi }{6}}\right)+\sqrt{3}\sin 2\left({\displaystyle \frac{5\pi }{6}}\right)\\ \lambda & =& -{\displaystyle \frac{1}{2}}+\sqrt{3}\times {\displaystyle \frac{-\sqrt{3}}{2}}\\ \lambda & =& -2\end{array}$

   Also, we have

   $\begin{array}{rcl}\mu & =& \cos 2\left({\displaystyle \frac{5\pi }{6}}\right)-\sqrt{3}\sin 2\left({\displaystyle \frac{5\pi }{6}}\right)\\ \mu & =& {\displaystyle \frac{1}{2}}-\sqrt{3}\times {\displaystyle \frac{-\sqrt{3}}{2}}\\ \mu & =& 2\end{array}$

   Hence, we have

   $\begin{array}{rcl}PB{P}^{-1}& =& \left(\begin{array}{cc}-2& 0\\ 0& 2\end{array}\right)\\ (PB{P}^{-1}{)}^n& =& {\left(\begin{array}{cc}-2& 0\\ 0& 2\end{array}\right)}^n\\ P{B}^n{P}^{-1}& =& \left(\begin{array}{cc}(-2{)}^n& 0\\ 0& 2^n\end{array}\right)\\ B^n& =& P^{-1}\left(\begin{array}{cc}(-2{)}^n& 0\\ 0& 2^n\end{array}\right)P\\ B^n& =& \left(\begin{array}{cc}\sin {\displaystyle \frac{5\pi }{6}}& -\cos {\displaystyle \frac{5\pi }{6}}\\ \cos {\displaystyle \frac{5\pi }{6}}& \sin {\displaystyle \frac{5\pi }{6}}\end{array}\right)\left(\begin{array}{cc}(-2{)}^n& 0\\ 0& 2^n\end{array}\right)\left(\begin{array}{cc}\sin {\displaystyle \frac{5\pi }{6}}& \cos {\displaystyle \frac{5\pi }{6}}\\ -\cos {\displaystyle \frac{5\pi }{6}}& \sin {\displaystyle \frac{5\pi }{6}}\end{array}\right)\\ B^n& =& \left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& -{\displaystyle \frac{-\sqrt{3}}{2}}\\ {\displaystyle \frac{-\sqrt{3}}{2}}& {\displaystyle \frac{1}{2}}\end{array}\right)\left(\begin{array}{cc}(-2{)}^n& 0\\ 0& 2^n\end{array}\right)\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& {\displaystyle \frac{-\sqrt{3}}{2}}\\ -{\displaystyle \frac{-\sqrt{3}}{2}}& {\displaystyle \frac{1}{2}}\end{array}\right)\\ B^n& =& {\displaystyle \frac{1}{2}}\left(\begin{array}{cc}1& \sqrt{3}\\ -\sqrt{3}& 1\end{array}\right)\times 2^n\left(\begin{array}{cc}(-1{)}^n& 0\\ 0& 1\end{array}\right)\times {\displaystyle \frac{1}{2}}\left(\begin{array}{cc}1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)\\ B^n& =& 2^{n-2}\left(\begin{array}{cc}(-1{)}^n& \sqrt{3}\\ \sqrt{3}(-1{)}^{n+1}& 1\end{array}\right)\left(\begin{array}{cc}1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)\\ B^n& =& 2^{n-2}\left(\begin{array}{cc}(-1{)}^n+3& \sqrt{3}(-1{)}^{n+1}+\sqrt{3}\\ \sqrt{3}(-1{)}^{n+1}+\sqrt{3}& 3(-1{)}^{n+2}+1\end{array}\right)\\ B^n& =& 2^{n-2}\left(\begin{array}{cc}(-1{)}^n+3& \sqrt{3}(-1{)}^{n+1}+\sqrt{3}\\ \sqrt{3}(-1{)}^{n+1}+\sqrt{3}& 3(-1{)}^n+1\end{array}\right)\end{array}$

3. $\begin{array}{cl}& (B^{-1}{)}^{555}\\ =& (B^{555}{)}^{-1}\\ =& {\left(2^{555-2}\left(\begin{array}{cc}(-1{)}^{555}+3& \sqrt{3}(-1{)}^{555+1}+\sqrt{3}\\ \sqrt{3}(-1{)}^{555+1}+\sqrt{3}& 3(-1{)}^{555}+1\end{array}\right)\right)}^{-1}\\ =& {\displaystyle \frac{1}{2^{553}}}{\left(\begin{array}{cc}2& 2\sqrt{3}\\ 2\sqrt{3}& -2\end{array}\right)}^{-1}\\ =& {\displaystyle \frac{1}{2^{553}}}\times {\displaystyle \frac{1}{\left|\begin{array}{cc}2& 2\sqrt{3}\\ 2\sqrt{3}& -2\end{array}\right|}}{\left(\begin{array}{cc}-2& -2\sqrt{3}\\ -2\sqrt{3}& 2\end{array}\right)}^T\\ =& {\displaystyle \frac{1}{2^{553}}}\times {\displaystyle \frac{1}{-16}}\left(\begin{array}{cc}-2& -2\sqrt{3}\\ -2\sqrt{3}& 2\end{array}\right)\\ =& {\displaystyle \frac{1}{2^{553}}}\times {\displaystyle \frac{1}{-2^4}}\times (-2)\left(\begin{array}{cc}1& \sqrt{3}\\ \sqrt{3}& -1\end{array}\right)\\ =& {\displaystyle \frac{-1}{2^{556}}}\left(\begin{array}{cc}1& \sqrt{3}\\ \sqrt{3}& -1\end{array}\right)\end{array}$