A) \[\left[ \begin{matrix} 1 & {{a}^{4}} \\ 0 & 1 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 4 & 4a \\ 0 & 4 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 4 & {{a}^{4}} \\ 0 & 4 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 4a \\ 0 & 1 \\ \end{matrix} \right]\]
Correct Answer: D
Solution :
\[{{A}^{2}}=A.\,A=\left[ \begin{matrix} 1 & a \\ 0 & 1 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & a \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 2a \\ 0 & 1 \\ \end{matrix} \right]\] \[{{A}^{3}}=A.\,{{A}^{2}}=\left[ \begin{matrix} 1 & a \\ 0 & 1 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & 2a \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 3a \\ 0 & 1 \\ \end{matrix} \right]\] \[{{A}^{4}}=A.\,{{A}^{3}}=\left[ \begin{matrix} 1 & a \\ 0 & 1 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & 3a \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 4a \\ 0 & 1 \\ \end{matrix} \right]\].
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