Let z = -1 + √3i / 2, where i = √-1 and r, s ∈ {1, 2, 3}. Let P = [(-z)r z²s; z²s zr] and I be the identity matrix of order 2. Then the total number of ordered pairs (r, s) for which P² = -I is

z = ω (where ω is cube root of unity)
P = [(-ω)r ω²s; ω²s ωr]
P² = [(-ω)r ω²s; ω²s ωr] [(-ω)r ω²s; ω²s ωr] = [(-ω)²r² + ω⁴s² (-ω)rω²s + rω²s + ωrω²s; (-ω)rω²s + rω²s + ωrω²s (-ω)²s² + ω²r²] = [-ω²r² + ω⁴s² (-ω)rω²s + rω²s + ω³rs; (-ω)rω²s + rω²s + ω³rs -ω²s² + ω²r²] = [-1 0; 0 -1] ⇒ ω²r² + ω⁴s² = -1, ((-ω)r + r + ω)ω²s = 0; r, s ∈ {1, 2, 3} ⇒ second equation represents r = 1, 3
Case - 1: r = 1
ω⁴s² = -1 - ω² = -ω
ωs² = -1 ⇒ s² = 1/ω = ω² ⇒ s = 1
Case - 2: r = 3
ω⁴s² = -1 - 9ω² = -1 - 9ω² which has no solution for s
⇒ Total number of ordered pairs (r, s) = 1