Question

If $S_k=\left[\begin{array}{cc}1& k\\ 0& 1\end{array}\right],k\in N$, where $N$ is the set of all natural numbers, then ${\left(S_2\right)}^n{\left(S_k\right)}^{-1}$, for $n\in N$, is

• A. $S_{2n-k}$
• B. $S_{2n+k}$
• C. $S_{2n+k-1}$
• D. $S_{2n+k+1}$

Answer 1

The correct option is A $S_{2n-k}$

as given

$\Rightarrow$ $S_k=\left[\begin{array}{cc}1& k\\ 0& 1\end{array}\right]$

first we find

$\Rightarrow$ $adj\left(S_k\right)$=$\left[\begin{array}{cc}1& 0\\ -k& 1\end{array}\right]$,

then we take its transpose and divide by $det\left(S_k\right)$ which is equal to inverse of $S_k^{-1}$

$\Rightarrow$ $(S_k{)}^{-1}$=$\left[\begin{array}{cc}1& -k\\ 0& 1\end{array}\right]$

and on solving, we got easily

$\Rightarrow$ $(S_2{)}^n$=$\left[\begin{array}{cc}1& 2n\\ 0& 1\end{array}\right]$

now we multiply $(S_2{)}^n$ and $(S_k{)}^{-1}$

we got

$\Rightarrow$ $(S_2{)}^n\times (S_k{)}^{-1}=\left[\begin{array}{cc}1& 2n-k\\ 0& 1\end{array}\right]$

this is equal to $S_{2n-k}$

ANSWER- [A]