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

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Complex Numbers

Let

ω

be a complex cube root of unity with

ω \neq = 1

and

P = [p_{ij}]

be a

n \times n

matrix with

p_{ij} = ω^{i + j}

. Then

P^{2} \neq = 0

, when

n =

Answer: (B), (C), (D)

The matrix

P

has entries

p_{ij} = ω^{i + j}

. We can factor:

P = ω^{2} ω^{3} \dots ω^{n + 1} 11 ⋮ 1 ω ω ⋮ ω \dots \dots \dots ω^{n - 1} ω^{n - 1} ⋮ ω^{n - 1}

The matrix has all rows identical. For

P^{2} \neq = 0

, we need the row sum

1 + ω + ω^{2} + \dots + ω^{n - 1} \neq = 0

.

Since

ω

is a cube root of unity,

1 + ω + ω^{2} = 0

. The sum

1 + ω + \dots + ω^{n - 1} = \frac{1 - ω ^{n}}{1 - ω}

.

This is non-zero when

ω^{n} \neq = 1

, i.e., when

n

is not a multiple of 3.

Among the options: 57 = 3×19 (multiple of 3), 55, 58, and 56 are not multiples of 3.

Therefore,

n = 55, 56, 58

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · One or more correct options