 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ia/paper-1/8c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Yes. Suppose

α_{1} a_{1} + α_{2} a_{2} + α_{3} a_{3} = 0

. Taking its dot product with

a_{2} \times a_{3}

eliminates the last two terms and gives

α_{1} a_{1} \cdot (a_{2} \times a_{3}) = 0.

α_{1} = 0

. Cyclically using

a_{3} \times a_{1}

and

a_{1} \times a_{2}

α_{2} = α_{3} = 0

as well. This is the definition of linear independence. Equivalently the three-column matrix has nonzero determinant.

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