행렬 거듭제곱 연립방정식

$\displaystyle A = \begin{pmatrix} {1} & {0} \\ {1} & {1} \end{pmatrix}$ $\displaystyle A ^{2} = \begin{pmatrix} {1} & {0} \\ {1} & {1} \end{pmatrix} \begin{pmatrix} {1} & {0} \\ {1} & {1} \end{pmatrix} = \begin{pmatrix} {1} & {0} \\ {2} & {1} \end{pmatrix}$ $\displaystyle A ^{3} = \begin{pmatrix} {1} & {0} \\ {2} & {1} \end{pmatrix} \begin{pmatrix} {1} & {0} \\ {1} & {1} \end{pmatrix} = \begin{pmatrix} {1} & {0} \\ {3} & {1} \end{pmatrix}$ …… 이므로 $\displaystyle A ^{n} = \begin{pmatrix} {1} & {0} \\ {n} & {1} \end{pmatrix}$ $\displaystyle A ^{n} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 3 \\ 8 \end{pmatrix}$의 양변에 $( A ^{n} ) ^{-1}$을 곱하면 $\displaystyle \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ - n & 1 \end{pmatrix} \begin{pmatrix} 3 \\ 8 \end{pmatrix} = \begin{pmatrix} 3 \\ - 3 n + 8 \end{pmatrix}$ $\alpha + \beta = 2$에서 $3 + ( - 3 n + 8 ) = 2$ $\therefore n = 3$