Binomial TheoremHard
Question
Let ω be a complex cube root of unity with ω ≠ 1 and P = [pij] be a n × n matrix with pij = ωi+j Then P2 ≠ 0 when n =
Options
A.57
B.55
C.58
D.56
Solution
P2 = [αij ] n × n
If n is a multiple of 3 then P2 = 0
⇒ n is not a multiple of 3
⇒ n can be 55, 58, 56
If n is a multiple of 3 then P2 = 0
⇒ n is not a multiple of 3
⇒ n can be 55, 58, 56
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