If \(S_{k}=\begin{pmatrix}1 & k \\ 0 & 1\end{pmatrix}\), \(k \in \mathrm{N}\), where \(\mathrm{N}\) is the set of natural numbers, then \((\mathrm{S}_{2})^{\mathrm{n}}(\mathrm{S}_{\mathrm{k}})^{-1}\), for \(\mathrm{n} \in \mathrm{N}\), is :
Show Hint
- \(S_{2n+k-1}\)
- \(S_{2^{n}+k-1}\)
- \(S_{2^{n}-k}\)
- \(S_{2n-k}\)
The Correct Option is D
Solution and Explanation
Therefore \((S_2)^n = S_{2n}\) (add 2 n times).
Also \(S_kS_{-k}=S_0=I\), so \((S_k)^{-1}=S_{-k}\).
Hence \((S_2)^n(S_k)^{-1}=S_{2n}S_{-k}=S_{2n-k}\).
Answer: (4) \(S_{2n-k}\)
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