Let ω be a complex cube root of unity with ω ≠ 1 and P = p_ij be a n x n matrix with p_ij = ω^(i + j) . Then P^2 ≠ 0 , when n =

Correct answer: (B) 55; (C) 58; (D) 56. n = 1 n = 2 P = ω^2 P = ω^2 ω^3 ω^3 ω^4 = ω^2 1 1 ω P^2 = ω^4 ≠ 0 P^2 = ω^4 + 1 ....... ....... ....... ≠ 0 n = 3 P = ω^2 1 ω 1 ω ω^2 ω ω^2 1 ω^2 1 ω 1 ω ω^2 ω ω^2 1 = 0 0 0 0 0 0 0 0 0 Similarly P^2 ≠ 0 when n is not multiple of 3.

Q23 / 60 · JEE Advanced 2013 (Paper 2) · Mathematics

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

A
57
Correct answer
B
55
Correct answer
C
58
Correct answer
D
56

Solution

n = 1 n = 2

P = [ω^{2}]

P = [\begin{matrix} ω^{2} & ω^{3} \\ ω^{3} & ω^{4} \end{matrix}] = [\begin{matrix} ω^{2} & 1 \\ 1 & ω \end{matrix}]

P^{2} = [ω^{4}] \neq 0

P^{2} = [\begin{matrix} ω^{4} + 1 & ....... \\ ....... & ....... \end{matrix}] \neq 0

n = 3

P = [\begin{matrix} ω^{2} & 1 & ω \\ 1 & ω & ω^{2} \\ ω & ω^{2} & 1 \end{matrix}] [\begin{matrix} ω^{2} & 1 & ω \\ 1 & ω & ω^{2} \\ ω & ω^{2} & 1 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]

Similarly

P^{2} \neq 0

when n is not multiple of 3.

Hence, the correct answer is options: 'B', 'C', 'D'.

Let ω be a complex cube root of unity with ω ≠ 1 and P = p ij be a n x n matrix with p ij = ω i + j . Then P 2 ≠ 0 , when n =

Solution

More Matrices Questions

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