A) \[\left[ \begin{matrix} 1 & 2n \\ 0 & 1 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 2 & n \\ 0 & 1 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & 2n \\ 0 & -1 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 2n \\ 0 & 1 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
\[{{A}^{2}}=\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 4 \\ 0 & 1 \\ \end{matrix} \right]\] and\[{{A}^{3}}={{A}^{2}}A\]. = \[\left[ \,\begin{matrix} 1 & 4 \\ 0 & 1 \\ \end{matrix}\, \right]\,\,\left[ \,\begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix}\, \right]=\left[ \,\begin{matrix} 1 & 6 \\ 0 & 1 \\ \end{matrix}\, \right]\] and so on. \[\therefore \] \[{{A}^{n}}=\left[ \,\begin{matrix} 1 & 2n \\ 0 & 1 \\ \end{matrix}\, \right]\].
You need to login to perform this action.
You will be redirected in
3 sec
