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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/11/a/iii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 11 / a / iii / Solution by

Codex  0 2026-10-18

No. Every vector in the plane can be written as

(x, y, z)^{T} = (- 2 y + 3 z, y, z)^{T} = y (- 2, 1, 0)^{T} + z (3, 0, 1)^{T} .

The two linearly independent vectors in part (ii) therefore form a vector space basis for this vector subspace. Its dimension is

2

, so any three vectors in it have linear dependence.

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