If \(A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\), then \(A^{n}\) for any natural number is
Step-by-step Solution:
\(A^{2}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\) \[\] \(A^{4}=A^{2}-A^{2}=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\) \[\] By the same pattern \(A^{n}=\left[\begin{array}{cc}1 & n \\ 0 & 1\end{array}\right]\) \[\] Choice (B)\[\]