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Question:

If A = [cosθ -sinθ; sinθ cosθ], then the matrix A⁵ when θ = π/12, is equal to

⎡⎢⎢⎢⎣√32𕒵212√32⎤⎥⎥⎥⎦

⎡⎢⎢⎢⎣12√32−√3212⎤⎥⎥⎥⎦

⎡⎢⎢⎢⎣12−√32√3212⎤⎥⎥⎥⎦

⎡⎢⎢⎢⎣√3212𕒵2√32⎤⎥⎥⎥⎦

Solution:

Here, AAT = I ⇒ A⁻¹ = AT = [cosθ sinθ; -sinθ cosθ]
Also, Aⁿ = [cos(nθ) sin(nθ); -sin(nθ) cos(nθ)]
∴ A⁵ = [cos(5θ) sin(5θ); -sin(5θ) cos(5θ)]
∴ A⁵ = ⎡⎢⎢⎣cos(5π/12) sin(5π/12); -sin(5π/12) cos(5π/12)⎤⎥⎥⎦
50 × π/12 = 4(360) + 60
∴ A⁵ = [cos(4(360) + 60) sin(4(360) + 60); -sin(4(360) + 60) cos(4(360) + 60)]
∴ A⁵ = [cos(60) sin(60); -sin(60) cos(60)]
⎡⎢⎢⎢⎣√3/2 1/2; -1/2 √3/2⎤⎥⎥⎥⎦