 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/6c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Since

P = P^{'} = 0

, it suffices to map the two edge vectors. Let the columns of the required matrix be

u_{1}, u_{2}, u_{3}

. The condition

MQ = Q^{'}

gives

u_{2} = (- \frac{1}{2}, 0, \frac{3}{2})^{T} .

The condition

MR = R^{'}

then gives

\frac{3}{2} u_{1} + \frac{1}{2} u_{2} = R^{'},

and hence

u_{1} = (\frac{3}{2}, 0, \frac{1}{2})^{T} .

These are orthonormal. Taking

u_{3} = u_{1} \times u_{2} = (0, - 1, 0)^{T}

produces the orthogonal matrix

M = 3 /2 0 1/2 - 1/2 0 3 /2 0 - 1 0 .

Its columns are orthonormal, and direct substitution verifies

MP = P^{'}

MQ = Q^{'}

, and

MR = R^{'}

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