 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/18h/a/solution

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

x

A x = b, x \geq 0,

and the columns

A_{i}

for which

x_{i} > 0

are linearly independent. Equivalently, after extending those columns to a basis of the column space of

A

, one sets all nonbasic variables to zero and solves for the basic variables, obtaining a nonnegative vector.

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