 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-1/7b/c/solution

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

Here

A^{†} A = \frac{1}{2} (5 - 3 - 3 5),

4, 1

and corresponding normalized vectors

(1, - 1)^{T} / 2

and

(1, 1)^{T} / 2

. A compatible choice is

U = \frac{1}{2} (1 - 1 11), V = (10 0 - 1), D = (2001) .

V^{†} A U = D

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