Question

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

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

Answer 1

Answer: (A)

$-3\omega$

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

$\Rightarrow R_1=R_1-R_3$

$\Rightarrow \left|\begin{array}{ccc}0& 1-{\omega }^2& 1-{\omega }^2\\ 1& -1-{\omega }^2& {\omega }^2\\ 1& {\omega }^2& {\omega }^2\end{array}\right|$

$\Rightarrow C_2=C_2-C_3$

$\Rightarrow \left|\begin{array}{ccc}0& 0& 1-{\omega }^2\\ 1& -1-2{\omega }^2& {\omega }^2\\ 1& 0& {\omega }^2\end{array}\right|$

$Det=(1-{\omega }^2)\times [1\times 0+(1+2{\omega }^2\left)\right]=(1-{\omega }^2)\left[\right(1+{\omega }^2)+{\omega }^2]=(1-{\omega }^4)+{\omega }^2(1-{\omega }^2)=1-\omega +{\omega }^2-{\omega }^4=1-\omega +{\omega }^2-\omega =1+{\omega }^2-2\omega =-3\omega$