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

Given, $2\omega$ + 1 = z
$\Rightarrow 2\omega +1=\sqrt{-3}[\because z=\sqrt{-3}]\Rightarrow \omega =\frac{-1+\sqrt{3}i}{2}$
Since, $\omega$ is cube root of unity.
$\therefore {\omega }^2=\frac{-1-\sqrt{3}i}{2}and{\omega }^{3n}=1$
Now, $\left|\begin{array}{ccc}1& 1& 1\\ 1& -{\omega }^2-1& {\omega }^2\\ 1& {\omega }^2& {\omega }^7\end{array}\right|=3k$
$\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|=3k$
$[\because 1+\omega +{\omega }^2=0and{\omega }^7=({\omega }^3{)}^2.\omega =\omega ]$
On applying $R_1\to R_1+R_2+R_3$, we get
$\left|\begin{array}{ccc}3& 1+\omega +{\omega }^2& 1+\omega +{\omega }^2\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|=3k\Rightarrow \left|\begin{array}{ccc}3& 0& 0\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|$
$\Rightarrow 3({\omega }^2-{\omega }^4)=3k\Rightarrow ({\omega }^2-\omega )=k\therefore k=\left(\frac{-1-\sqrt{3}i}{2}\right)-\left(\frac{-1+\sqrt{3}i}{2}\right)=-\sqrt{3}i=-z$