2020-M2-08

By comparing the corresponding elements of both sides, we have

$\left\{ \begin{array}{ll}
-5-2b=1 & \ldots \unicode{x2460} \\
-5a-2c=0 & \ldots \unicode{x2461} \\
16+6b=b & \ldots \unicode{x2462} \\
15a+6c=0 & \ldots \unicode{x2463}
\end{array}\right.$

From $\unicode{x2460}$, we have

$\begin{array}{rcl}
-5-2b & = & 1 \\
b & = & -3
\end{array}$

From $\unicode{x2461}$, we have

$\begin{array}{rcl}
-5a-2c & = & 0 \\
a & = & \dfrac{-2}{5}c \ \ldots \unicode{x2464}
\end{array}$

It is given that $|M|=1$. Hence, we have

$\begin{array}{rcl}
|M| & = & 1 \\
\begin{vmatrix} 1 & a \\ b & c \end{vmatrix} & = & 1 \\
c-ab & = & 1 \ \ldots \unicode{x2465}
\end{array}$

Sub. $b=-3$ and $\unicode{x2464}$ into $\unicode{x2465}$, we have

$\begin{array}{rcl}
c-\dfrac{-2}{5}c \times (-3) & = & 1 \\
\dfrac{-1}{5}c & = & 1 \\
c & = & -5
\end{array}$

Sub. $c=-5$ into $\unicode{x2464}$, we have

$\begin{array}{rcl}
a & = & \dfrac{-2}{5} (-5) \\
a & = & 2
\end{array}$

1. By the result of (a), $M=\begin{pmatrix} 1 & 2 \\ -3 & -5 \end{pmatrix}$. Hence, we have

   $\begin{array}{cl}
   & M^{-1} \\
   = & \dfrac{1}{|M|} \begin{pmatrix} -5 & -(-3) \\ -2 & 1 \end{pmatrix}^T \\
   = & \dfrac{1}{1} \begin{pmatrix} -5 & -2 \\ 3 & 1 \end{pmatrix} \\
   = & \begin{pmatrix} -5 & -2 \\ 3 & 1 \end{pmatrix}
   \end{array}$

   Therefore, we have

   $\begin{array}{cl}
   & M^{-1}RM \\
   = & \begin{pmatrix} -5 & -2 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 6 & 2 \\ -15 & -5 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ -3 & -5 \end{pmatrix} \\
   = & \begin{pmatrix} 0 & 0 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ -3 & -5 \end{pmatrix} \\
   = & \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
   \end{array}$

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

   $\begin{array}{rcl}
   M^{-1}RM & = & \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\
   R & = & M\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}M^{-1}
   \end{array}$

   Also, we have

   $\begin{array}{rcl}
   PM & = MQ \\
   P & = & M\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}M^{-1}
   \end{array}$

   Hence, we have

   $\begin{array}{cl}
   & (\alpha P+\beta R)^{99} \\
   = & \left(\alpha M\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}M^{-1}+\beta M\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}M^{-1} \right)^{99} \\
   = & \left( M\begin{pmatrix} \alpha & 0 \\ 0 & 0 \end{pmatrix}M^{-1}+ M\begin{pmatrix} 0 & 0 \\ 0 & \beta \end{pmatrix}M^{-1} \right)^{99} \\
   = & \left( M\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}M^{-1}\right)^{99} \\
   = & M\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}M^{-1} \times M\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}M^{-1} \times \cdots \times M\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}M^{-1} \\
   = & M\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}^{99}M^{-1} \\
   = & M\begin{pmatrix} \alpha^{99} & 0 \\ 0 & \beta^{99} \end{pmatrix}M^{-1} \\
   = & M\begin{pmatrix} \alpha^{99} & 0 \\ 0 & 0 \end{pmatrix}M^{-1}+M\begin{pmatrix} 0 & 0 \\ 0 & \beta^{99} \end{pmatrix}M^{-1} \\
   = & \alpha^{99} M\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}M^{-1}+\beta^{99}M\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}M^{-1} \\
   = & \alpha^{99}P+\beta^{99}R
   \end{array}$