Determinant of a Matrix

A = [cosθ, sinθ], [-sinθ, cosθ]
A² = [cos2θ, sin2θ], [-sin2θ, cos2θ]
⇒ A? = [cos4θ, sin4θ], [-sin4θ, cos4θ]
B = [cos4θ, sin4θ], [-sin4θ, cos4θ] + [cosθ, sinθ], [-sinθ, cosθ]
= [cos4θ + cosθ, sin4θ + sinθ], [- (sin4θ + sinθ), cos4θ + cosθ]
det (B) = (cos4θ + cosθ)² + (sin4θ + sinθ)²
= (cos²4θ + sin²4θ) + (cos²θ + sin²θ) + 2 (cos4θcosθ + sin4θsinθ)
= 1 + 1 + 2cos (4θ - θ)
= 2 + 2cos3θ
Given 3θ = 3π/5
|B| = 2 + 2cos (3π/5)
= 2 + 2 (- (√5-1)/4) = 2 - (√5-1)/2 = (4-√5+1)/2 = (5-√5)/2 ∈ (1,2)

M = {A = [a b; c d] | a, b, c, d ∈ {±3, ±2, ±1,0}
f (A) = det (A) = ad-bc=15
Case 1: ad=9 (2 ways: 3*3, -3*-3), bc=-6 (4 ways: 3*-2, -3*2, 2*-3, -2*3). Total = 2*4=8.
Case 2: ad=6 (4 ways), bc=-9 (2 ways). Total = 4*2=8.
Total no. of possible such cases = 8+8=16.