 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-35/1/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Direct substitution into the three equations gives

\frac{4 \cdot 13 + 13 + 12}{11} = 7, \frac{13 + 9 \cdot 13 + 2 \cdot 12}{11} = 14, \frac{13 + 2 \cdot 13 + 16 \cdot 12}{11} = 21.

Since the covariance matrix is invertible, this identifies its unique solution:

z = \frac{1}{11} (13, 13, 12)^{T} .

The original PDF has all three components; the converted TeX's damaged binomial notation should not be used.

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