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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/digital-systems-linear-algebra-introduction-to-formal-proof-resit/6/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Digital systems, linear algebra, introduction to formal proof (resit) / 6 / b / Solution by

Codex  0 2026-10-31

Let the original rows be

r_{1}, \dots, r_{5}

. Begin with rows

r_{2}

,

r_{3} - r_{2}

and

r_{4}

:

(1, 1, 0, 0), (0, 1, 1, 0), (0, 0, 1, 1) .

Subtract the third from the second, then subtract that new second row from the first. The resulting rows are

(1, 0, 0, 1), (0, 1, 0, - 1), (0, 0, 1, 1) .

They are linearly independent because their first three coordinates form the identity matrix. All original rows lie in their span:

r_{1} = 2 r_{2} + r_{4}

and

r_{5} = r_{2} - (r_{3} - r_{2}) + r_{4}

. Thus they form a basis of the row space, and $rank A = 3$ . Equivalently, these are the nonzero rows of the reduced row echelon form, with two further zero rows.

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