`For the first step, since the maximal
element in the first column of is
we permute rows one and three, and store this
permutation as .
`

`As before, we multiply the first row by
() and subtract it from the
second (third) row in order to zero the elements and
respectively; the multipliers , are stored in in the
appropriate positions.
`

`Now we process the second column. The maximum between and
is ; no row permutation is necessary here,
hence . The second row is multiplied by and subtracted from the
third to get
`

`Now the product gives
`

`Note that, when pivoting is used, all the multipliers are less than or equal
to 1. Compactly we can represent the LU decomposition of as
`