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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.`57` b. `55` c. `58` d. `56`

A

57

B

55

C

58

D

56

Text Solution

Verified by Experts

The correct Answer is:
B, C, D
`P=[(omega^(2),omega^(3),omega^(4),...,omega^(n+1)),(omega^(3),omega^(4),omega^(5),...,omega^(n+2)),(vdots,vdots,vdots,vdots,vdots),(omega^(n+1),omega^(n+2),...,...,omega^(2n))]`
or
`P^(2)=[(omega^(4)+omega^(6)...,omega^(5)+omega^(7)+...,...,...),(omega^(5)+omega^(7)+...,...,...,...),(vdots,vdots,vdots,...),(omega^(n+2)+omega^(n+4)...,...,...,omega^(2n+2)+omega^(2n+4)...)]`
= Null matrix if n is a multiple of 3.
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