Let θ = π/5 and A = [cosθ sinθ; -sinθ cosθ]. If B = A + A?, then det(B) :
Let θ = π/5 and A = [cosθ sinθ; -sinθ cosθ]. If B = A + A?, then det(B) :
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) + (
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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.
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