A) \[\left( \frac{5-i\sqrt{3}}{2} \right)\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
B) \[\left( \frac{3-i\sqrt{3}}{2} \right)\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
D) \[(2+i\sqrt{3})\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
Correct Answer: D
Solution :
Thus,\[A=\left[ \begin{matrix} \frac{\omega }{i} & \frac{{{\omega }^{2}}}{i} \\ -\frac{{{\omega }^{2}}}{i} & -\frac{\omega }{i} \\ \end{matrix} \right]=\frac{\omega }{i}\left[ \begin{matrix} 1 & \omega \\ -\omega & -1 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{2}}=-{{\omega }^{2}}\left[ \begin{matrix} 1-{{\omega }^{2}} & 0 \\ 0 & 1-{{\omega }^{2}} \\ \end{matrix} \right]\] \[=-\left[ \begin{matrix} -{{\omega }^{2}}+{{\omega }^{4}} & 0 \\ 0 & -{{\omega }^{2}}+{{\omega }^{4}} \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} -{{\omega }^{2}}+\omega & 0 \\ 0 & -{{\omega }^{2}}+\omega \\ \end{matrix} \right]\]\[\because \,\,f(x)={{x}^{2}}+2\] \[\therefore \]\[f(A)={{A}^{2}}+2\left[ \begin{matrix} -{{\omega }^{2}}+\omega & 0 \\ 0 & -{{\omega }^{2}}+\omega \\ \end{matrix} \right]+\left[ \begin{matrix} 2 & 0 \\ 0 & 2 \\ \end{matrix} \right]\] \[=-[{{\omega }^{2}}+\omega +2]\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\] \[=(3+2\omega )\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=(2+i\sqrt{3})\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]\]
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