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: (A)

$z$

Using $1+\omega +{\omega }^2=0$ & ${\omega }^3=1$

$\to \left|\begin{array}{ccc}1& 1& 1\\ 1& -{\omega }^2-1& {\omega }^2\\ 1& {\omega }^2& {\omega }^7\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|$

Now, $R_1\to R_1+R_2+R_3$

$\to \left|\begin{array}{ccc}3& 0& 0\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|$

Now, expand along $R_1$, we get

$\Rightarrow 3\left|\begin{array}{cc}\omega & {\omega }^2\\ {\omega }^2& \omega \end{array}\right|=3({\omega }^2-{\omega }^4)=3({\omega }^2-\omega )$

$\Rightarrow 3(-1-\omega -\omega )=-3(2\omega +1)=-3(-z)=3z$

Compare $3z=3k$

$z=k$