 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-1/2c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Componentwise,

[(A B)^{T}]_{ij} = (A B)_{ji} = \sum_{k} A_{jk} B_{ki} = \sum_{k} (B^{T})_{ik} (A^{T})_{kj} = (B^{T} A^{T})_{ij},

(A B)^{T} = B^{T} A^{T} .

Repeated application gives

(A^{k})^{T} = (A^{T})^{k}

. Transposing the matrix exponential term by term therefore yields

(e^{A})^{T} = e^{A^{T}} .

e^{t A} e^{t A^{T}} = I + t (A + A^{T}) + t^{2} (\frac{1}{2} A^{2} + A A^{T} + \frac{1}{2} (A^{T})^{2}) + O (t^{3}) .

Hence

Q_{0} = I, Q_{1} = A + A^{T}, Q_{2} = \frac{1}{2} A^{2} + A A^{T} + \frac{1}{2} (A^{T})^{2} .

e^{t A}

is an orthogonal matrix for every real

t

, then

e^{t A} (e^{t A})^{T} = e^{t A} e^{t A^{T}} = I .

t

must vanish, so

A^{T} = - A;

that is,

A

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