Let θ = π/5 and A = [ cos θ, sin θ; - sin θ, cos θ ].
If B = A + A⁴ , then det (B) :

Solution

Maths sir · 4 steps

A is a rotation matrix, so A⁴ rotates by 4θ and det B becomes a sum of two squares.

1. Powers of A. Multiplying two matrices of this form adds their angles. So Aⁿ has the same form with θ replaced by nθ, and A⁴ uses 4θ.
2. Write B. Let p = cos θ + cos 4θ and q = sin θ + sin 4θ. Then B = A + A⁴ has p on the diagonal, q at the top right and -q at the bottom left.

   det B = p² + q² = 2 + 2(cos θ cos 4θ + sin θ sin 4θ) = 2 + 2 cos 3θ

3. Put in the angle. 3θ = 3π/5 = 108°, so cos 3θ = -sin 18° = -(√5-1)/4 ≈ -0.309. This gives det B = (5-√5)/2 ≈ 1.38.
4. So. det B ≈ 1.38, which lies in (1, 2). The answer is (a).

The idea to keep. For rotation-type matrices, Aⁿ just multiplies the angle by n. Then det(Aᵐ + Aⁿ) = 2 + 2 cos((m-n)θ), so only the angle difference matters.

Answer from the NTA final answer key. Solution by PrepWiser.