 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/digital-systems-linear-algebra-introduction-to-formal-proof/5/e/solution

The reduced row echelon form gives

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

x_{3} = s, x_{4} = t

. Then

x = s - 1 110 + t - 1 - 1 01 .

The coefficient vectors are linearly independent, because their third and fourth entries give independent coordinates. Consequently

Null (A) has basis ⎩ ⎨ ⎧ - 1 110, - 1 - 1 01 ⎭ ⎬ ⎫, dim Null (A) = 2.

Substitution in the original matrix verifies

A x = 0

for both basis vectors. The rank-nullity theorem supplies the independent dimension check

2 + 2 = 4

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