 Solution

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/2/b/solution

R_{2} \leftarrow R_{2} - R_{1}

and

R_{3} \leftarrow R_{3} + R_{1}

, then

R_{3} \leftarrow R_{3} - R_{2}

A \sim 100200 - 1 55 4 - 3 - 3 \sim 100200 - 1 50 4 - 3 0 .

These operations preserve the row space. The two nonzero rows of this row echelon form are linearly independent: only the first has a nonzero first entry, and the second is nonzero. Hence

a row-space basis is {(1, 2, - 1, 4), (0, 0, 5, - 3)}, rank A = 2.

100200010 17/5 - 3/5 0,

which will also identify the free variables in the null space calculation.

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