Question

Let $\omega =-\frac{1}{2}=i\frac{\sqrt{3}}{2}$, then value of the determinant $△=\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-{\omega }^2& {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|$ 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 $3\omega (\omega -1)$

Applying $R_2\to R_2-R_1;R_3\to R_3-R_1$, the given determinant becomes
$\left|\begin{array}{ccc}1& 1& 1\\ 0& -2-{\omega }^2& {\omega }^2-1\\ 0& {\omega }^2-1& \omega -1\end{array}\right|$
$=1\left[\right(\omega -1\left)\right(-2-{\omega }^2)-\left\{\right({\omega }^2-1\left)\right({\omega }^2-1\left)\right\}]=1\left(0\right)+1\left(0\right)$
$=(-2-{\omega }^2)(\omega -1)-({\omega }^2-1{)}^2$
$=-2\omega +2-{\omega }^3+{\omega }^2-({\omega }^4+1-2{\omega }^2)$
$=-2\omega +1{\omega }^2-(\omega +1-2{\omega }^2)$
$=3{\omega }^2-3\omega$
$=3\omega (\omega -1)$.