 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/p/p3/a/solution

Q

Q^{T} Q = I

, equivalently

Q^{- 1} = Q^{T}

. For

c = cos θ

and

s = sin θ

M (θ)^{T} M (θ) = (c s - s c) (c - s s c) = (c^{2} + s^{2} 0 0 c^{2} + s^{2}) = I .

Thus $M (θ)$ is orthogonal for every allowed angle, and indeed for every real angle.

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