Let A = ⟨ cosα -sinα; sinα cosα ⟩, (α ∈ R) such that A^32 = ⟨ 0 -1; 1 0 ⟩. Then a value of α is

Correct option is D. π/64
A = ⟨ cos α -sin α; sin α cos α ⟩
A^2 = ⟨ cos α -sin α; sin α cos α ⟩ ⟨ cos α -sin α; sin α cos α ⟩ = ⟨ cos2 α -sin2 α; sin 2α cos2 α ⟩
A^3 = ⟨ cos2 α -sin2 α; sin 2α cos2 α ⟩ ⟨ cos α -sin α; sin α cos α ⟩ = ⟨ cos 3α -sin 3α; sin 3α cos 3α ⟩
Similarly
A^32 = ⟨ cos 32α -sin 32α; sin 32α cos 32α ⟩ = ⟨ 0 -1; 1 0 ⟩
⇒ cos 32α = 0 and sin 32 α = 1
⇒ 32α = (4n + 1)π/2, n ∈ I
α = (4n + 1)π/64, n ∈ I
α = π/64 for n = 0