2021-M2-11

由此，可得

$\begin{array}{cl}
& PAP^{-1} \\
= & \begin{pmatrix} \sin\theta & \cos\theta \\ -\cos\theta & \sin\theta \end{pmatrix}\begin{pmatrix} \alpha & \beta \\ \beta & -\alpha \end{pmatrix}\begin{pmatrix} \sin\theta & -\cos\theta \\ \cos \theta & \sin\theta \end{pmatrix} \\
= & \begin{pmatrix} \alpha\sin\theta+\beta\cos\theta & \beta\sin\theta-\alpha\cos\theta \\-\alpha\cos\theta+\beta\sin\theta & -\beta\cos\theta-\alpha\sin\theta \end{pmatrix}\begin{pmatrix} \sin\theta & -\cos\theta \\ \cos \theta & \sin\theta \end{pmatrix} \\
= & \begin{pmatrix} \alpha\sin^2\theta+\beta\sin\theta\cos\theta+\beta\sin\theta\cos\theta-\alpha\cos^2\theta & -\alpha\sin\theta\cos\theta+-\beta\cos\theta^2+\beta\sin^2\theta-\alpha\sin\theta\cos\theta \\ -\alpha\sin\theta\cos\theta+\beta\sin^2\theta-\beta\cos^2\theta-\alpha\sin\theta\cos\theta & \alpha\cos^2\theta-\beta\sin\theta\cos\theta-\beta\sin\theta\cos\theta-\alpha\sin^2\theta \end{pmatrix} \\
= & \begin{pmatrix} \alpha(\sin^2\theta-\cos^2\theta)+2\beta\sin\theta\cos\theta & \beta(\sin^2\theta-\cos^2\theta) -2\alpha\sin\theta\cos\theta \\ \beta(\sin^2\theta-\cos^2\theta)-2\alpha\sin\theta\cos\theta & \alpha(\cos^2\theta-\sin^2\theta)-2\beta\sin\theta\cos\theta \end{pmatrix} \\
= & \begin{pmatrix} -\alpha\cos2\theta+\beta\sin2\theta & -\beta\cos2\theta-\alpha\sin2\theta \\ -\beta\cos2\theta-\alpha\sin2\theta & \alpha\cos2\theta-\beta\sin2\theta \end{pmatrix}
\end{array}$

1. 把 $\alpha=1$ 及 $\beta=\sqrt{3}$ 代入 (a) 的結果，可得

   $\begin{array}{rcl}
   PBP^{-1} & = & \begin{pmatrix} -\cos2\theta+\sqrt{3}\sin2\theta & -\sqrt{3}\cos2\theta-\sin2\theta \\ -\sqrt{3}\cos2\theta-\sin2\theta & \cos2\theta-\sqrt{3}\sin2\theta \end{pmatrix} \\
   \begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix} & = & \begin{pmatrix} -\cos2\theta+\sqrt{3}\sin2\theta & -\sqrt{3}\cos2\theta-\sin2\theta \\ -\sqrt{3}\cos2\theta-\sin2\theta & \cos2\theta-\sqrt{3}\sin2\theta \end{pmatrix}
   \end{array}$

   透過比較兩方的元素，可得

   $\begin{array}{rcl}
   -\sqrt{3}\cos2\theta-\sin2\theta & = & 0 \\
   -\sin2\theta & = & \sqrt{3}\cos2\theta \\
   \tan2\theta & = & -\sqrt{3}
   \end{array}$

   $\therefore 2\theta =\pi-\dfrac{\pi}{3}$ 或 $2\theta = 2\pi-\dfrac{\pi}{3}$。

   $\theta =\dfrac{\pi}{3}$ (捨去) 或 $\theta=\dfrac{5\pi}{6}$。

2. 利用 (b)(i) 的結果，可得

   $\begin{array}{rcl}
   \lambda & = & -\cos2\left(\dfrac{5\pi}{6}\right)+\sqrt{3}\sin2\left(\dfrac{5\pi}{6}\right) \\
   \lambda & = & -\dfrac{1}{2}+\sqrt{3} \times \dfrac{-\sqrt{3}}{2} \\
   \lambda & = & -2
   \end{array}$

   另外，可得

   $\begin{array}{rcl}
   \mu & = & \cos 2\left(\dfrac{5\pi}{6}\right)-\sqrt{3}\sin2\left(\dfrac{5\pi}{6}\right) \\
   \mu & = & \dfrac{1}{2} -\sqrt{3}\times\dfrac{-\sqrt{3}}{2} \\
   \mu & = & 2
   \end{array}$

   由此，可得

   $\begin{array}{rcl}
   PBP^{-1} & = & \begin{pmatrix} -2 & 0 \\ 0 & 2 \end{pmatrix} \\
   (PBP^{-1})^n & = & \begin{pmatrix} -2 & 0 \\ 0 & 2 \end{pmatrix}^n \\
   PB^nP^{-1} & = & \begin{pmatrix} (-2)^n & 0 \\ 0 & 2^n \end{pmatrix} \\
   B^n & = & P^{-1}\begin{pmatrix} (-2)^n & 0 \\ 0 & 2^n \end{pmatrix}P \\
   B^n & = & \begin{pmatrix} \sin\dfrac{5\pi}{6} & -\cos\dfrac{5\pi}{6} \\ \cos\dfrac{5\pi}{6} & \sin\dfrac{5\pi}{6} \end{pmatrix} \begin{pmatrix} (-2)^n & 0 \\ 0 & 2^n \end{pmatrix} \begin{pmatrix} \sin\dfrac{5\pi}{6} & \cos\dfrac{5\pi}{6} \\ -\cos\dfrac{5\pi}{6} & \sin\dfrac{5\pi}{6} \end{pmatrix} \\
   B^n & = & \begin{pmatrix} \dfrac{1}{2} & -\dfrac{-\sqrt{3}}{2} \\ \dfrac{-\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix} \begin{pmatrix} (-2)^n & 0 \\ 0 & 2^n \end{pmatrix} \begin{pmatrix} \dfrac{1}{2} & \dfrac{-\sqrt{3}}{2} \\ -\dfrac{-\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix} \\
   B^n & = & \dfrac{1}{2} \begin{pmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{pmatrix} \times 2^n\begin{pmatrix} (-1)^n & 0 \\ 0 & 1 \end{pmatrix} \times \dfrac{1}{2}\begin{pmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{pmatrix} \\
   B^n & = & 2^{n-2} \begin{pmatrix} (-1)^n & \sqrt{3} \\ \sqrt{3}(-1)^{n+1} & 1 \end{pmatrix} \begin{pmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{pmatrix} \\
   B^n & = & 2^{n-2}\begin{pmatrix} (-1)^n+3 & \sqrt{3}(-1)^{n+1}+\sqrt{3} \\ \sqrt{3}(-1)^{n+1}+\sqrt{3} & 3(-1)^{n+2}+1 \end{pmatrix} \\
   B^n & = & 2^{n-2}\begin{pmatrix} (-1)^n+3 & \sqrt{3}(-1)^{n+1}+\sqrt{3} \\ \sqrt{3}(-1)^{n+1}+\sqrt{3} & 3(-1)^n+1 \end{pmatrix}
   \end{array}$

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