A) \[\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
Correct Answer: D
Solution :
[d] We have \[A=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]\] Now, \[{{A}^{2}}=A.A=\left( \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right)\left( \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right)=\left( \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right)=-I\] where \[I=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right)\] is identity matrix \[{{({{A}^{2}})}^{8}}={{(-I)}^{8}}=I\]. Hence, \[{{A}^{16}}=I\]
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