Question

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

• A. Zero
• B. $3$
• C. $-1$
• D. $1$

Answer 1

Answer: (A)

Zero

$\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|\Rightarrow 1\times ({\omega }^2\times \omega -1\times 1)-\omega (\omega \times \omega -1\times {\omega }^2)+{\omega }^2(1\times \omega -{\omega }^2\times {\omega }^2)\Rightarrow ({\omega }^3-1)-\omega ({\omega }^2-{\omega }^2)+{\omega }^2(\omega -{\omega }^4)\Rightarrow (1-1)-\omega \left(0\right)+{\omega }^2(\omega -\omega {.\omega }^3)\Rightarrow 0+0+{\omega }^2(\omega -\omega )\Rightarrow 0$

{Using the result that $\omega$ is cube root of unity so ${\omega }^3=1$}