행렬 거듭제곱 집합

$\displaystyle A ^{n} = {\begin{pmatrix} 1 & & n \\ 0 & & 1 \end{pmatrix}} , B ^{n} = {\begin{pmatrix} 1 & & 0 \\ n & & 1 \end{pmatrix}}$이므로 $\displaystyle A ^{n} {\begin{pmatrix} 1 \\ 1 \end{pmatrix}} = {\begin{pmatrix} n + 1 \\ 1 \end{pmatrix}} , B ^{n} {\begin{pmatrix} 1 \\ 1 \end{pmatrix}} = {\begin{pmatrix} 1 \\ n + 1 \end{pmatrix}}$ $\displaystyle \therefore \left. \begin{array}{l} A = \left\{ \begin{array}{l} {\begin{pmatrix} n + 1 \\ 1 \end{pmatrix}} \end{array} \right| n \text{은} \text{자연수} \end{array} \right\}$, $\displaystyle \left. \begin{array}{l} B = \left\{ \begin{array}{l} {\begin{pmatrix} 1 \\ n + 1 \end{pmatrix}} \end{array} \right| n \text{은} \text{자연수} \end{array} \right\}$ 따라서 자연수 $n , m$에 대하여 ㄱ. $\displaystyle \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} n + 1 \\ 1 \end{pmatrix} \in S$이면 $\displaystyle \begin{pmatrix} b \\ a \end{pmatrix} = \begin{pmatrix} 1 \\ n + 1 \end{pmatrix} \in T$ (참) ㄴ. $\displaystyle \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} n + 1 \\ 1 \end{pmatrix} \in S$, $\displaystyle \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} m + 1 \\ 1 \end{pmatrix} \in S$이면 $b + d \neq 1$이므로 $\displaystyle \begin{pmatrix} a + c \\ b + d \end{pmatrix} = \begin{pmatrix} n + m + 1 \\ 2 \end{pmatrix} \notin S$ (거짓) ㄷ. $\displaystyle \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} n + 1 \\ 1 \end{pmatrix} \in S$, $\displaystyle \begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} 1 \\ m + 1 \end{pmatrix} \in T$이면 $\displaystyle {\begin{pmatrix} a & & p \\ b & & q \end{pmatrix}} = {\begin{pmatrix} n + 1 & & 1 \\ 1 & & m + 1 \end{pmatrix}}$이다. 이 때, $( n + 1 ) ( m + 1 ) - 1 \neq 0$이므로 역행렬을 갖는다. (참) 이상에서 옳은 것은 ㄱ, ㄷ이다.