A) A
B) \[{{3}^{n}}A\]
C) \[({{3}^{n-1}})A\]
D) 3A
Correct Answer: C
Solution :
[c] Given matrix is \[A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]\] So, \[{{A}^{2}}\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \\ \end{matrix} \right]=3\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]\] and \[{{A}^{3}}=3\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]\] \[=3\left[ \begin{matrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \\ \end{matrix} \right]=9\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right]={{3}^{2}}A\] Similarly \[{{A}^{4}}={{3}^{3}}A\]. Hence, \[{{A}^{n}}={{3}^{n-1}}A\]
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