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Let omega be a complex cube root of unit...

Let `omega` be a complex cube root of unity with `omega!=1a n dP=[p_(i j)]` be a `nxxn` matrix withe `p_(i j)=omega^(i+j)dot` Then `p^2!=O ,when=`
a. 55
b. 56
c. 57
d. 58

A

55

B

56

C

57

D

58

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
`P = [p_(ij)]_(nxxn)=[omega^(i+j)]_(nxxn) = [[omega^(2), omega^(3),omega^(4),...omega^(1+n)],[omega^(3), omega^(4),omega^(5) ,...omega^(2+n)],[omega^(4),omega^(5),omega^(6),...omega^(3+n)],[vdots,vdots,vdots,vdots],[omega^(n+1),omega^(n+2),omega^(n+3),...omega^(2n)]]`
`therefore P^(2) = [[0, 0,0,...0],[0, 0, 0,...0],[0,0,0,...0],[vdots,vdots,vdots,vdots],[0,0,0,...0]]=0`
If n is multiple of 3, so for `P^(2) ne 0, n` should not be a multiple of
3, i.e., n can take values `55, 56` and 58.`
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