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}} $ .
$ \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}} $ .
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