Let $\omega $ be a complex cube root of unity with $\omega \ne 1 $ and $P=[p_{ij}]$ be a $n \times n $ matrix with $p_{ij}=\omega^{i+j}$ . Then $P^2 \ne0$ when $n=$
(A) 57
(B) 55
(C) 58
(D) 56
Official answer is $B,C,D$. Considering the values are greater than $50$ if we see the answer, based on my assumption, the value of $n$ is not a multiple of $3$.
We need to find the square matrix whose square is not equal to zero matrix. How do we find the value of $n$?
$P^2$=$\begin{bmatrix}w^2 &w^3 &w^4& .&.&.\\ w^3 \\ w^4 \\ .\\ .\end{bmatrix}$ $\begin{bmatrix}w^2 &w^3 &w^4& .&.&.\\ w^3 \\ w^4 \\ .\\ .\end{bmatrix}$
=$\begin{bmatrix}w^4+w^6+w^8+...& .&.&.\\ . \\ . \\ .\\ .\end{bmatrix}$
=$\begin{bmatrix}w+1+w^2+...& .&.&.\\ . \\ . \\ .\\ .\end{bmatrix}$=$\begin{bmatrix}0+...& .&.&.\\ . \\ . \\ .\\ .\end{bmatrix}$
We deduce that $n$ should not be a multiple of $3$. Doubt?