Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P = \left[p_{ij}\right]$ be a $n\times n$ matrix with $p_{ij} = \omega^{i+j}$. Then $P^2 \neq 0$ when $n =$
Step-by-Step Solution
Key Concept: The matrix P with entries p_ij = ω^(i+j) has rank related to linear dependence of rows. Since ω is a primitive cube root of unity (ω³ = 1, 1 + ω + ω² = 0), the matrix P² = 0 when n is a multiple of 3, making P nilpotent. P² ≠ 0 occurs when n is not divisible by 3.
For $n=1$: $p = [w^2]$ gives $p^2 = [w^4] \neq 0$. For $n=2$: the determinant involves $w^4 + 1$ terms which is nonzero. For $n=3$: the matrix becomes circulant with the last row containing powers of $w$, and the determinant equals zero when rows become linearly dependent. Similarly, $p^2 \neq 0$ when $n$ is not a multiple of 3.
Correct Answer: 2,3,4