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`

Options

A. 57
B. 55
C. 58
D. 56

Answer

A

Explanation

We have, `P=[P_(ij)]_(nxxx)`
`:. (P^2)_(ij)=sum_(r=1)^n p ir" " prj=sum_(r=1)^nomega^(i+r)omega^(r+j)=sum_(r-1)^nomega^(i+j+2r)`
`=omega^(i+j)sum_(r=1)^n(omega^2)^r=omega^(i+j)[omega^2{((omega^2)^n-1)/(omega^2-1)}]`
`=(omega^(i+j+2))/(omega^2-1)(omega^(2n)-1)`
Now, `P^2=O`
`rArr (P^2)_(ij)=0" for all " o,j`
`rArr (omega^(i+j+2))/(omega^2-1)(omega^(2n)-1)=0" for all "i,j`
`rArr omega^(2n)=1`
`rArr " n is a multiple of 3"`
Clearly, 57 is a multiple of 3, So, option (a) is correct.