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In $ \omega \ne 1 $ is the complex cube root of unity and matrix $ H=\left[ \begin{matrix}
   \omega & 0 \\
   0 & \omega \\
\end{matrix} \right], $ then $ {{H}^{70}} $ is equal to
A) $ H $
B) $ 0 $
C) $ -H $
D) $ {{H}^{2}} $

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Last updated date: 14th Jun 2024
Total views: 402k
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Answer
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Hint: We will find the value of $ \left| H \right| $ and power up the equation with $ 70 $ to get the value of $ {{H}^{70}} $ . After getting the value of $ {{H}^{70}} $ as $ {{\omega }^{140}} $ we use the formula $ {{\omega }^{3n}}=1 $ from the given data that $ \omega $ is the cube root of unity i.e. when we raised to the power f $ 3 $ we get the value as $ 1 $ . Mathematically $ {{\omega }^{3}}=1 $ for our convenience we can write
 $ \begin{align}
  & {{\left( {{\omega }^{3}} \right)}^{n}}={{1}^{n}} \\
 & {{\omega }^{3n}}=1
\end{align} $
by using the property $ {{1}^{n}}=1 $ and simplify the equation to get the result.

Complete step by step answer:
Given that, $ H=\left[ \begin{matrix}
   \omega & 0 \\
   0 & \omega \\
\end{matrix} \right], $ $ \omega \ne 1 $
$\Rightarrow$ The value of $ \left| H \right| $ is
 $ \begin{align}
  & \left| H \right|=\left| \begin{matrix}
   \omega & 0 \\
   0 & \omega \\
\end{matrix} \right| \\
 & ={{\omega }^{2}}
\end{align} $
$\Rightarrow$ Add power of $ 70 $ to the above equation, then
 $ \begin{align}
  & {{\left| H \right|}^{70}}={{\left( {{\omega }^{2}} \right)}^{70}} \\
 & {{H}^{70}}={{\omega }^{140}}....\left( \text{i} \right)
\end{align} $
$\Rightarrow$ Given that, $ \omega $ is the cube root of the unity i.e.
 $ \begin{align}
  & \omega =\sqrt[3]{1} \\
 & {{\omega }^{3n}}=1 \\
\end{align} $
$\Rightarrow$ Now from the equation $ \left( \text{i} \right) $
 $ {{H}^{70}}={{\omega }^{140}} $
$\Rightarrow$ We are going to write $ 140 $ as multiple of $ 3 $ , then
 $ {{H}^{70}}={{\omega }^{3\times 46+2}} $
Using the formula $ {{a}^{b+c}}={{a}^{b}}.{{a}^{c}} $ in the above equation then we have
 $ {{H}^{70}}={{\omega }^{3\times 46}}.{{\omega }^{2}} $
$\Rightarrow$ We have $ {{\omega }^{3n}}=1 $ , so we can write $ {{\omega }^{3\times 46}}=1 $ in the above equation, then
 $ \begin{align}
  & {{H}^{70}}=1\left( {{\omega }^{2}} \right) \\
 & {{H}^{70}}={{\omega }^{2}} \\
\end{align} $
$\Rightarrow$ But we have $ {{\omega }^{2}}=H $ so
 $ {{H}^{70}}=H $

Note:
We can solve the problem by finding the values of $ {{H}^{2}},{{H}^{3}},{{H}^{4}},{{H}^{5}},{{H}^{6}} $
 $ \begin{align}
  & {{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 \left( \omega \right) & 0 \\
   0 & \omega \left( \omega \right) \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   {{\omega }^{2}} & 0 \\
   0 & {{\omega }^{2}} \\
\end{matrix} \right]
\end{align} $
 $ \begin{align}
  & {{H}^{3}}={{H}^{2}}H \\
 & =\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 }^{2}}\left( \omega \right) & 0 \\
   0 & {{\omega }^{2}}\left( \omega \right) \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right] \\
 & =I
\end{align} $
 $ \begin{align}
  & {{H}^{4}}={{H}^{3}}.H \\
 & =I.H \\
 & =H
\end{align} $
 $ \begin{align}
  & {{H}^{5}}={{H}^{4}}.H \\
 & =H.H \\
 & ={{H}^{2}}
\end{align} $
 $ \begin{align}
  & {{H}^{6}}={{H}^{5}}.H \\
 & ={{H}^{2}}.H \\
 & ={{H}^{3}} \\
 & =I
\end{align} $
From the above sequence we can say that
 $ \begin{align}
  & {{H}^{69}}={{\left( {{H}^{3}} \right)}^{13}} \\
 & ={{I}^{13}} \\
 & =I
\end{align} $
Now the value of $ {{H}^{70}} $ is
 $ \begin{align}
  & {{H}^{70}}={{H}^{69}}.H \\
 & =I.H \\
 & =H
\end{align} $
Don’t use the value of $ \omega $ as $ {{1}^{\dfrac{1}{3}}} $ in the equation $ {{H}^{70}}={{\omega }^{140}} $ as we can’t get the proper value of $ {{H}^{70}} $ . You only use the formula $ {{\omega }^{3n}}=1 $ and simplify $ {{\omega }^{140}} $ .