행렬 거듭제곱 방정식

$\displaystyle A ^{2} = \begin{pmatrix} - 2 & - 3 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} - 2 & - 3 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ - 1 & - 2 \end{pmatrix}$ $\displaystyle A ^{3} = A ^{2} A = \begin{pmatrix} 1 & 3 \\ - 1 & - 2 \end{pmatrix} \begin{pmatrix} - 2 & - 3 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = E$ $\therefore A ^{2005} = ( A ^{3} ) ^{668} \cdot A = A ( \because A ^{3} = E )$ $\displaystyle \therefore A ^{2005} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$에서 $\displaystyle A \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ 그런데 $\displaystyle A ^{- 1} = \begin{pmatrix} 1 & 3 \\ - 1 & - 2 \end{pmatrix}$이므로 $\displaystyle \begin{pmatrix} x \\ y \end{pmatrix} = A ^{- 1} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ - 1 & - 2 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ - 5 \end{pmatrix}$ $\therefore x - y = 12$