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Let A = [[cos θ, -sin θ], [sin θ, cos θ]]. If A^T + A = I₂, then the value of θ is

1. 2nπ + π/3, n ∈ Z
2. 2nπ + π/2, n ∈ Z
3. 2nπ + π, n ∈ Z
4. 2nπ + π/6, n ∈ Z

Correct answer: 2nπ + π/3, n ∈ Z

Solution

For A=[[cos,-sin],[sin,cos]], A^T+A has diagonal 2cos(theta) and off-diagonals cancel, giving 2cos(theta)*I = I. Hence cos(theta)=1/2, so theta = 2n*pi +/- pi/3; the matching option is 2n*pi + pi/3.

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