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 deficiency when n is divisible by 3, since ω³ = 1 creates periodic repetition in the exponents. Computing P² requires analyzing when the sum Σₖ ω^(i+k) · ω^(k+j) = ω^(i+j) · Σₖ ω^(2k) vanishes, which occurs when n ≡ 0 (mod 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