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 $3\omega (\omega -1)$

$△=\left|\begin{array}{ccc}3& 1& 1\\ 0& -1-{\omega }^2& {\omega }^2\\ 0& {\omega }^2& \omega \end{array}\right|(C_1\to C_1+C_2+C_3)(\because 1+\omega +{\omega }^2=0)=3[\omega .\omega -{\omega }^4]=3({\omega }^2-\omega )=3\omega (\omega -1).$