A) 0
B) \[-H\]
C) 2
D) \[{{H}^{2}}\]
E) 1
Correct Answer: C
Solution :
\[\because \] \[H=\left[ \begin{matrix} \omega & 0 \\ 0 & \omega \\ \end{matrix} \right]\] \[{{H}^{2}}=\left[ \begin{matrix} \omega & 0 \\ 0 & \omega \\ \end{matrix} \right]\left[ \begin{matrix} \omega & 0 \\ 0 & \omega \\ \end{matrix} \right]=\left[ \begin{matrix} {{\omega }^{2}} & 0 \\ 0 & {{\omega }^{2}} \\ \end{matrix} \right]\] \[{{H}^{3}}=\left[ \begin{matrix} {{\omega }^{2}} & 0 \\ 0 & {{\omega }^{2}} \\ \end{matrix} \right]\left[ \begin{matrix} \omega & 0 \\ 0 & \omega \\ \end{matrix} \right]=\left[ \begin{matrix} {{\omega }^{3}} & 0 \\ 0 & {{\omega }^{3}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[{{H}^{6}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] and \[{{H}^{4}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} \omega & 0 \\ 0 & \omega \\ \end{matrix} \right]=\left[ \begin{matrix} \omega & 0 \\ 0 & \omega \\ \end{matrix} \right]\] \[\therefore \] \[{{H}^{70}}={{({{H}^{6}})}^{11}}{{H}^{4}}\] \[=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} \omega & 0 \\ 0 & \omega \\ \end{matrix} \right]=\left[ \begin{matrix} \omega & 0 \\ 0 & \omega \\ \end{matrix} \right]=H\]
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