 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/digital-systems-linear-algebra-introduction-to-formal-proof/5/f/solution

x_{1} - 2 x_{2} - x_{4} + 3 x_{5} = 0, x_{3} + 2 x_{4} - 2 x_{5} = 0.

s = x_{2}

t = x_{4}

u = x_{5}

. Then

x = s 21000 + t 10 - 2 10 + u - 3 0201 .

Thus

null-space basis {(2, 1, 0, 0, 0)^{T}, (1, 0, - 2, 1, 0)^{T}, (- 3, 0, 2, 0, 1)^{T}}, dim Null (A) = 3.

Each listed vector lies in the null space, and every solution has the displayed form. Their coordinates in positions 2, 4 and 5 prove linear independence. The dimension also follows from the rank-nullity theorem:

5 - 2 = 3

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