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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 toA) $H$ B) $0$ C) $-H$ D) ${{H}^{2}}$

Last updated date: 14th Jun 2024
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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.

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 ${{\omega }^{140}}$ .