A) \[\left[ \begin{matrix} 1 & 2n \\ 0 & 1 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 2 & n \\ 0 & 1 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & n \\ 0 & -1 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & n \\ 0 & 1 \\ \end{matrix} \right]\]
E) \[\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
\[\because \]\[A=\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]\] \[{{A}^{2}}=\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & 4 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 2\times 2 \\ 0 & 1 \\ \end{matrix} \right]\] \[{{A}^{3}}=\left[ \begin{matrix} 1 & 4 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 2\times 3 \\ 0 & 1 \\ \end{matrix} \right]\] Similarly, \[{{A}^{n}}=\left[ \begin{matrix} 1 & 2n \\ 0 & 1 \\ \end{matrix} \right]\]
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