Question

If $ \omega $ is a complex cube root of unity and $ A=\left[ \begin{matrix} \omega & 0 \\ 0 & \omega \\ \end{matrix} \right], $ then $ {{A}^{50}} $ is

• A. $ {{\omega }^{2}}A $
• B. $ \omega A $
• C. $ A $
• D. $ 0 $

Answer 1

Answer: (B)

Given, $ A=\left[ \begin{matrix} \omega & 0 \\ 0 & \omega \\ \end{matrix} \right] $
$ {{A}^{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] $
$ {{A}^{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] $
Similarly, $ {{A}^{50}}=\left[ \begin{matrix} {{\omega }^{50}} & 0 \\ 0 & {{\omega }^{50}} \\ \end{matrix} \right] $
$ =\left[ \begin{matrix} {{({{\omega }^{3}})}^{16}}{{\omega }^{2}} & 0 \\ 0 & {{({{\omega }^{3}})}^{16}}{{\omega }^{2}} \\ \end{matrix} \right] $
$ =\left[ \begin{matrix} {{\omega }^{2}} & 0 \\ 0 & {{\omega }^{2}} \\ \end{matrix} \right]=\omega A $