Let be a complex cube root of unity with 1 and P = [ p i j ] be an n × n matrix with p i j = i + j . Then P 2 0 , when n = [2013]

Question

Let $ω$ be a complex cube root of unity with $ω \neq 1$ and $P = [p_{i j}]$ be an $n \times n$ matrix with $p_{i j} = ω^{i + j} .$ Then $P^{2} \neq 0,$ when $n =$ [2013]

Smart Achievers · Accepted Answer

(2, 3, 4)

For $ω \neq 1$ , $P = [p_{i j}]$ and $P^{2} = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

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, hence $n$ can take values 55, 56 and 58.

Solution

Q.
Let $ω$ be a complex cube root of unity with $ω \neq 1$ and $P = [p_{i j}]$ be an $n \times n$ matrix with $p_{i j} = ω^{i + j} .$ Then $P^{2} \neq 0,$ when $n =$ [2013]

1 57

2 55

3 58

4 56

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