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

Answer

1

Explanation

Here, `z = (-1 + isqrt(3))/(2) = omega`
`therefore 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^(4s), omega^(r+2s)[(-1)^(r) + 1]),(omega^(r+2s)[(-1)^(r) + 1], omega^(4s) + omega^(2r)):}]`
Given, `P^(2) = -I`
`therefore omega^(2r) + omega^(4s) = -1 " and " omega^(r+2s)[(-1)^(r) + 1]=0`
Since, `r in {1, 2, 3) " and " (-1)^(r) + 1 = 0`
`rArr r = {1, 3}`
Also, `omega^(2r) + omega^(4s) = -1`
If r = 1, then `omega^(2) + omega^(4s) = -1`
which is only possible, when s = 1
As, `omega^(2) + omega^(4) = -1`
`therefore r = 1, s = 1`
Again, if r = 3, then
`omega^(6) + omega^(4s) = -1`
`rArr omega^(4s) = -2 " " ["never possible"]`
`therefore r ne 3`
`rArr (r,s) = (1,1)` is the only solution.
Hence, the total number of ordered pairs is 1.