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. $3{\omega }^2(1-{\omega }^2)$

Answer 1

The correct option is D $3{\omega }^2(1-{\omega }^2)$

$\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-w^2& w^2\\ 1& w^2& w^4\end{array}\right|$

$C_2\to C_2-C_1,C_3\to C_3-C_1$

$=\left|\begin{array}{ccc}1& 0& 0\\ 1& -2-w^2& w^2-1\\ 1& w^2-1& w^4-1\end{array}\right|$

$=1\times \left|\begin{array}{cc}-2-w^2& w^2-1\\ w^2-1& w^4-1\end{array}\right|-0+0$

$=(-2-w^2)(w^4-1)-(w^2-1)(w^2-1)$

$=-2w^4+2-w^6+w^2-(w^4+1-2w^2)$

$=-2w^4+2-1+w^2-w^4-1+2w^2$

$=-3w^4+3w^2$

$=3w^2(1-w^2)$