 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-1/8a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let

J_{-} = (01 - 1 0), J_{+} = (0110) .

Since

J_{-}^{2} = - I

and

J_{+}^{2} = I

, separating the even and odd terms of the matrix exponential gives

R = e^{θ J_{-}} = (cos θ sin θ - sin θ cos θ),

and

S = e^{θ J_{+}} = (cosh θ sinh θ sinh θ cosh θ) .

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