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. $3(-2\omega -1)$

Answer 1

Answer: (B), (D)

$3(-2\omega -1)$
$3\omega (\omega -1)$

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

$\omega =$cube root of unity

$\Rightarrow {\omega }^3=1$

$1+\omega +{\omega }^2=0$

$\Rightarrow -1-{\omega }^2=\omega$

${\omega }^4=\left({\omega }^3\right)\omega =\omega$

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

$=1({\omega }^2-{\omega }^4)-1(\omega -{\omega }^2)+1({\omega }^2-\omega )$

$={\omega }^2-\omega -+{\omega }^2+{\omega }^2-\omega$

$=3({\omega }^2-\omega )$

$=3\omega (\omega -1)=3(-2\omega -1)$

Option B and D