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)

The correct options are
B $55$
C $58$
D $56$

We know $C=A\times B=[\sum a_{ik}\times b_{kj}]$

So, $P^2=[{\omega }^{i+j}\times {\sum }_{k=1}^n{\omega }^{2k}]$

$\Rightarrow P^2=[{\omega }^{i+j}\times {\omega }^2\times \frac{{\omega }^{2n}-1}{w^2-1}]$

Since $P^2\ne 0$ every element in the matrix is non-zero

So, ${\omega }^{2n}-1$ should be non-zero $\Rightarrow 2n$ must not be a multiple of 3

S0, Option (B),(C) and (D)