Question

Let z = (-1 + √3i)/2, where i = √-1 and r, s ∈ { 1,  2, 3}. Let P = [((-z)^r, z^2s), (z^2s, z^r)]  and I be the identity matrix of order 2. Then, the total number of ordered pairs (r, s) for which P² = −I is _____.

Answer 1

It is given that

(ω is cube root of unity), where r, s ∈ {1, 2, 3}.

where r is the odd number and hence r = 1, 3.

When r = 1 (−ω)² + ω⁴s = −1

⇒ ω⁴s = −1 - ω² = + ω

Now, s can be 1 (since s ≠ 3).

That is, (r, s) = (1, 1), that is, the total number of ordered pair (r, s) is one (single) for which P² = -I.