Question

Let $\omega$ be a complex number such that $2\omega +1=z$ where $z=\sqrt{-3}$. If $\left|\begin{array}{ccc}1& 1& 1\\ 1& -{\omega }^2-1& {\omega }^2\\ 1& {\omega }^2& {\omega }^7\end{array}\right|=3k$, then $k$ is equal to:

• A. $-z$
• B. $z$
• C. $-1$
• D. $1$

Answer 1

The correct option is A $-z$

From the given conditions we find that
$\omega =\frac{-1+i\sqrt{3}}{2}\Rightarrow 1+\omega +{\omega }^2=0,{\omega }^3=1\Rightarrow$ $\left|\begin{array}{ccc}1& 1& 1\\ 1& -{\omega }^2-1& {\omega }^2\\ 1& {\omega }^2& {\omega }^7\end{array}\right|=3k$ can be reduced to
$\left|\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|=3k\Rightarrow \left|\begin{array}{ccc}3& 1& 1\\ 0& \omega & {\omega }^2\\ 0& {\omega }^2& \omega \end{array}\right|=3kk={\omega }^2-\omega =-z$