Let `z= (-1+sqrt(3i))/2, where i=sqrt(-1) and r,s epsilon P1,2,3}. Let P= [((-z)^r, z^(2s)),(z^(2s), z^r)]` and I be the idenfity matrix or order 2. Then the total number of ordered pairs (r,s) for which `P^2=-I` is

Answer

1

Explanation

`Z=(-1+isqrt(3))/2= omega`
`P=[((-omega)^(r),omega^(2s)),(omega^(2s),omega^(r))]`
`P^(2)=[((-omega)^(r),omega^(2s)),(omega^(2s),omega^(r))][((-omega)^(r),omega^(2s)),(omega^(2s),omega^(r))]`
`=[((-omega)^(2r)+(omega^(2s))^(2),omega^(2s) (-omega)^(r)+omega^(r) omega^(2s)),(omega^(2s) (-omega)^(r)+omega^(r) omega^(2s),omega^(4s)+omega^(2s))]`
`=[(omega^(4s)+omega^(2r),omega^(2s)(omega^(r)+(-omega)^(r))),(omega^(2s) (omega^(r)+(-omega)^(r)),omega^(4s)+omega^(2r))]`
`=-I` (Given)
`:. Omega^(4S)+omega^(2r)=-1` and
`omega^(2s) (omega^(t) + (-omega)^(r))=0`
`:. omega^(r)+(-omega)^(r)=0`
`:. r=1` and `s=1`
So only one pair