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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/digital-systems-linear-algebra-introduction-to-formal-proof-resit/6/e/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Digital systems, linear algebra, introduction to formal proof (resit) / 6 / e / Solution by

Codex  0 2026-10-31

The second, third and fourth linear equations give

x_{1} + x_{2} = 2, x_{2} + x_{3} = 0, x_{3} + x_{4} = - 1.

Setting

x_{4} = t

yields

x_{3} = - 1 - t

,

x_{2} = 1 + t

and

x_{1} = 1 - t

. The first linear equation then gives

(1 - t) + (1 + t) + (- 1 - t) + t = 1

, and the fifth gives

(1 - t) + t = 1

, so both remaining conditions hold for every real

t

. Consequently

x = 11 - 1 0 + t - 1 1 - 1 1, t \in R .

This is a particular solution plus the whole null space, an affine line in a vector space of solutions. The derivation both produces solutions and shows that none have been omitted.

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