A) \[{{2}^{n-1}}X\]
B) \[{{n}^{2}}X\]
C) \[nX\]
D) \[{{2}^{n+1}}X\]
E) \[{{2}^{n}}X\]
Correct Answer: A
Solution :
\[\because \]\[X=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[{{X}^{2}}=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\] \[=2\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\] and \[{{X}^{3}}=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & 2 \\ 2 & 2 \\ \end{matrix} \right]\] \[={{2}^{2}}\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\] Similarly, \[{{X}^{n}}={{2}^{n-1}}\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]={{2}^{n-1}}X\]
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