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

Options

A. `1/2 abs(a-b)`
B. `1/2 abs(a+b)`
C. ` abs(a-b)`
D. ` abs(a+b)`

Answer

A

Explanation

`because Z = (-1+ sqrt(3)i)/2 = omega`
` rArr omega = 1 and 1 + omega + omega ^(2) = 0 `
Now, `P = [[(-omega)^(r),omega^(2s)],[omega^(2s), omega^(r)]]`
`therefore P^(2) = [[(-omega)^(r),omega^(2s)],[omega^(2s), omega^(r)]] [[(-omega)^(r),omega^(2s)],[omega^(2s), omega^(r)]]`
` =[[omega^(2r)+omega^(4s),omega^(2s)((-omega)^(r)+omega^(r))],[omega^(2s)((-omega)^(r)+omega^(r)), omega^(4s)+omega^(2r)]]`
` =[[omega^(2r)+omega^(4s),omega^(2s)((-omega)^(r)+omega^(r))],[omega^(2s)((-omega)^(r)+omega^(s)), omega^(s)+omega^(2r)]] " " (because omega^(3) = r)`
`because P^(2) = -I= [[-1,0],[0,-1]]` ...(ii)
Form Eqs. (i) and (ii), we get
`omega^(2r) +omega^(s)=-1`
and ` omega^(2s) ((-omega)^(r)+omega^(r))=0`
`rArr r` is odd and `s = r` but not a multiple of 3, Which is possible
`therefore ` only one pair is there.