Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Digital systems, linear algebra, introduction to formal proof / 5 / a / ii / Solution

The second pivot is two. Row three has second entry two, so subtract row two; row four has second entry minus two, so add row two. Therefore $d=-1$, $e=1$. Their trailing row pairs become $(0,4)$ and $(3,5)$, respectively; the zero in the third pivot position is why a later row interchange is needed.