Question

Let $\omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}$. Then the value of the determinant $\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-{\omega }^2& {\omega }^2\\ 1& {\omega }^2& {\omega }^4\end{array}\right|$ is

• A. $3\omega$
• B. $3\omega (\omega -1)$
• C. $3{\omega }^2$
• D. $3\omega (1-\omega )$

Answer 1

The correct option is B $3\omega (\omega -1)$

Let $\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-{\omega }^2& {\omega }^2\\ 1& {\omega }^2& {\omega }^4\end{array}\right|$

As $\omega$ is cube root of unity , $1+\omega +{\omega }^2=0$ and $w^3=1$

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

Applying $C_1\to C_1+C_2+C_3$

$\Delta =\left|\begin{array}{ccc}3& 1& 1\\ 1+\omega +{\omega }^2& \omega & {\omega }^2\\ 1+\omega +{\omega }^2& {\omega }^2& \omega \end{array}\right|=\left|\begin{array}{ccc}3& 1& 1\\ 0& \omega & {\omega }^2\\ 0& {\omega }^2& \omega \end{array}\right|=3\omega (\omega -1)$

Hence, option 'B' is correct.