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Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P=\left[p_j\right]$ be a $n \times n$ matrix with $p_{i j}=\omega^{i+j}$. Then $P ^2 \neq 0$, when $n =$
$(A)$ $57$ $(B)$ $55$ $(C)$ $58$ $(D)$ $56$
$(A,B,C)$
$(A,B,D)$
$(A,C,D)$
$(B,C,D)$
Solution
$n=1 $
$P=\left[\omega^2\right] $
$P^2=\left[\omega^4\right] \neq 0$
$n=2 $
$P=\left[\begin{array}{ll}\omega^2 & \omega^3 \\ \omega^3 & \omega^4\end{array}\right]=\left[\begin{array}{cc}\omega^2 & 1 \\ 1 & \omega\end{array}\right]$
$P ^2=$ $\left[\begin{array}{cc}\omega^4+1 & \ldots \\ \cdots & \ldots\end{array}\right] \neq 0$
$n=3 $
$P=\left[\begin{array}{ccc}\omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1\end{array}\right]\left[\begin{array}{ccc}\omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
Similarly $P ^2 \neq 0$ when $n$ is not multiple of $3.$
