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 / c

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- Solution c

Solution

Swap the first two rows, then perform

R_{2} \leftarrow R_{2} + R_{1}

and

R_{3} \leftarrow R_{3} - 2 R_{1}

. These elementary row operations give

A \sim 100 - 2 00 2513102 - 1 - 10 - 2 .

Scale row 2 by

1/5

, subtract it from row 3, and finally subtract twice row 2 from row 1. The reduced row echelon form is

R = 100 - 2 00 010 - 1 20 3 - 2 0 .

Elementary row operations preserve the row space. The two nonzero rows of

R

are linearly independent, since their leading entries lie in different columns. Hence

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

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