Question

If $\omega (\ne 1)$ is a cube root of unity and

$A=\left[\begin{array}{ccc}1+2{\omega }^{100}+{\omega }^{200}& {\omega }^2& 1\\ 1& 1+2{\omega }^{100}+{\omega }^{200}& \omega \\ \omega & {\omega }^2& 2+{\omega }^{100}+2{\omega }^{200}\end{array}\right]$ then

• A. $|A|=0$
• B. $\left|A\right|\ne 0$
• C. $\text{A is symmetric}$
• D. $\text{A is skew-symmetric}$

Answer 1

The correct option is A $|A|=0$

$\left|\begin{array}{ccc}1+2\omega +{\omega }^2& {\omega }^2& 1\\ 1& 1+{\omega }^2+2\omega & \omega \\ \omega & {\omega }^2& 2+\omega +2{\omega }^2\end{array}\right|$
We know $1+\omega +{\omega }^2=0$ & ${\omega }^3=1$
$\Rightarrow \left|\begin{array}{ccc}\omega & {\omega }^2& 1\\ 1& \omega & \omega \\ \omega & {\omega }^2& 1+{\omega }^2\end{array}\right|=0$$(\because$ after taking $\omega$ common from $C2$ we get $C1$ and $C2$ same$)$
Clearly $A$ is neither symmetric nor skew-symmetric.