Question

If $\omega \ne 1$ is the complex cube root of unity and matrix $H=\left[\begin{array}{cc}\omega & 0\\ 0& \omega \end{array}\right]$, then $H^{70}$ is equal to

• A. $H$
• B. $0$
• C. $-H$
• D. $H^2$

Answer 1

The correct option is A $H$

$H=\left|\begin{array}{cc}\omega & 0\\ 0& \omega \end{array}\right|$

$H={\omega }^2$

$H^{70}={\left({\omega }^2\right)}^{70}$

$H^{70}={\omega }^{140}$

As we know that, if $\omega$ is a cube root of unity, then

${\omega }^{3n}=1$

$\therefore H^{70}={\omega }^{140}={\omega }^{(3\times 46)+2}$

$\Rightarrow H^{70}={\omega }^{3\times 46}.{\omega }^2$

$\Rightarrow H^{70}=1.{\omega }^2$

As $H={\omega }^2$

$\therefore H^{70}=H$