Question

Let $\omega$ be a complex cube root of unity with $\omega \ne 1$ and $P=\left[p_{ij}\right]$ be a $n\times n$ matrix with $p_{ij}={\omega }^{i+j}$. Then $P^2\ne 0$, when $n=$

• A. $57$
• B. $55$
• C. $58$
• D. $56$

Answer 1

Answer: (B), (C), (D)

$56$

Assume, $\omega$ be a complex cube root of unity and $P$ be a $n\times n$ matrix with $p_{ij}={\omega }^{i+j}$
$P=\left[\begin{array}{ccccc}{\omega }^2& 1& \omega & {\omega }^2& ...\\ 1& \omega & {\omega }^2& 1& ...\\ \omega & {\omega }^2& 1& \omega & ...\\ ...& ...& ...& ...& ...\end{array}\right]$
$\Rightarrow P^2=\left[\begin{array}{ccccc}{\omega }^2& 1& \omega & {\omega }^2& ...\\ 1& \omega & {\omega }^2& 1& ...\\ \omega & {\omega }^2& 1& \omega & ...\\ ...& ...& ...& ...& ...\end{array}\right]\left[\begin{array}{ccccc}{\omega }^2& 1& \omega & {\omega }^2& ...\\ 1& \omega & {\omega }^2& 1& ...\\ \omega & {\omega }^2& 1& \omega & ...\\ ...& ...& ...& ...& ...\end{array}\right]$
$\Rightarrow ({\omega }^4+1+{\omega }^2)+({\omega }^4+1+{\omega }^2)+...=0$
Since it is possible, if $n$ is a multiple of $3$. But it is given that $P^2\ne 0$.
Therefore, the possible values of $n$ can be $55$,$58$,$56$.