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. -1
• C. 1
• D. -z

Answer 1

Answer: (D)

$2\omega +1=z,z=\sqrt{3}i$
$\omega =\frac{-1+\sqrt{3}i}{2}\to$ Cube root of unity.
$C_1\to C_1+C_2+C_3$
$\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-{\omega }^2& {\omega }^2\\ 1& {\omega }^2& {\omega }^7\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|=\left|\begin{array}{ccc}3& 1& 1\\ 0& \omega & {\omega }^2\\ 0& {\omega }^2& \omega \end{array}\right|$
$=3\left({\omega }^2-{\omega }^4\right)$
$=3\left[\left(\frac{-1-\sqrt{3}i}{2}\right)-\left(\frac{-1+\sqrt{3}i}{2}\right)\right]$
$=-3\sqrt{3}i$
$=-3z$
$\therefore k=-z$