 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/5/c/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 5 / c / Solution by

Codex  0 2026-10-18

Use the lower antisymmetric block shown in the original PDF. Write

M = diag (a K, b J)

with

K = (0110), J = (0 - 1 10), K^{2} = I, J^{2} = - I .

(1)

Powers preserve the block structure. Splitting the matrix exponential series into even and odd powers gives

e^{a K} = I cosh a + K sinh a

. For the other block, the matrix exponential when the square is minus the identity gives

e^{b J} = I cos b + J sin b

. Therefore

e^{M} = cosh a sinh a 00 sinh a cosh a 00 00 cos b - sin b 00 sin b cos b .

(2)

The upper block is hyperbolic and the lower block is a rotation. The transcription's altered last row and hyperbolic functions in the lower block would describe a different problem.

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