Question

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) or which `P^2=-I` is
A. 1
B. 2
C. 3
D. 5

Answer 1

Correct Answer - A
We have, `z=(-1+isqrt3)/2=w and {:P=[((-z)^r,z^(2s)),(z^(2s),z^r)]:}`
`:.{:P=[((-w)^r,w^(2s)),(w^(2s),w^r)]:}`
`rArr{:P^2=[((-w)^r,w^(2s)),(w^(2s),w^r)][((-w)^r,w^(2s)),(w^(2s),w^r)]:}`
`rArr{:P^2=[(w^(2r)+w^(4s),w^(r+2s)|(-1)^2+|),(w^(r+2s)|(-1)^2+|,w^(4s)+w^(2r))]:}`
`:. P^2=-I`
`rArr w^(2r)+w^(4s)=-1 and w^r+2s){(-1)^r+1}=0`
`rArr w^(2r)+w^(4s)=-1 and (-1)^r+1=0`
`rArr w^(2r) +w^(4s)=-1 and r=1,3`
when r = 1
`w^(2r)+w^(4s)=-1rArrw^2+(w^3)^sw^s=-1rArr w^2+w^s=-1rArrs=-1`
When r = 3:
`w^(2r)+w^(4s)=-1rArr w^6 +(w^3)^sw^s=-1rArr w^s=-2`
Clearly, no value of s `in{1,2,3}" satisfies " w^s=-2`
Hence, r=1, s=1. Consequently, there is only one pair (r,s) = (1,1).
So, option (a) is correct.