Browse Questions

# Let $\omega$ be a complex cube root of unity with $\omega\neq 1$ and $p=[P_{ij}]$ be a $n\times n$ matrix with $P_{ij}=\omega^{i+j}$.Then $P^2\neq 0$ when $n$=

$(a)\;57\qquad(b)\;58\qquad(c)\;60\qquad(d)\;61$

Can you answer this question?

For $n=3$
P=$\begin{bmatrix}\omega^2&\omega^3&\omega^4\\\omega^3&\omega^4&\omega^5\\\omega^4&\omega^5&\omega^6\end{bmatrix}$
and $P^2=\begin{bmatrix}0 &0&0\\0&0&0\\0&0&0\end{bmatrix}$
It shows $P^2=0$ if $n$ is a multiple of 3.
So for $P^2\neq 0$ n should not be a multiple of 3 (i.e) n can take values 55,58,56.
Hence (b) is the correct answer.
answered Nov 21, 2013