Mathematics - Complex Number Question with Solution | TestHub

Let $A=\left[\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right]$ Now, $A^2=\left[\begin{array}{lll}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$ and $\mathrm{Tr}(A)=0,|A|=0$ $\therefore A^3=0$ $\Rightarrow \left|\begin{array}{ccc}z+1& \omega & {\omega }^2\\ \omega & z+{\omega }^2& 1\\ {\omega }^2& 1& z+\omega \end{array}\right|=|A+zI|=0$ $\Rightarrow z^3=0$ $\Rightarrow z=0$, the number of $z$ satisfying the given equation is 1 .