A) \[{{2}^{100}}A\]
B) \[{{2}^{99}}A\]
C) \[{{2}^{101}}A\]
D) None of above
Correct Answer: B
Solution :
[b] Let \[A=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\] \[{{A}^{2}}=\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]=2\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]=2A\] \[{{A}^{3}}={{2}^{2}}\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right],{{A}^{4}}={{2}^{3}}\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\] \[{{A}^{3}}={{2}^{2}}A,\] \[{{A}^{4}}={{2}^{3}}A\therefore \,\,{{A}^{n}}={{2}^{n-1}}\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]\] \[\Rightarrow {{A}^{100}}={{2}^{100-1}}A\therefore {{A}^{100}}={{2}^{99}}A\]
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