Past exam of the options I course of the University of Oxford / 2004 / P / P1

The first-column largest pivot element is four, so eliminate with multipliers $1/2,0,1/4$ below it. The augmented matrix becomesPartial pivoting in the second column swaps rows two and three because two is largest in modulus. The next elimination multipliers are $1/4$ and $1/2$. This givesThe third pivot element is already largest and the final multiplier below it is zero. Back substitution gives $z=1$, $y=3$, $x=-6$, $w=-1/2$. ThereforeThe row swap must also reorder the previously stored first-column multipliers in $L$. The factors areMultiplying these factors verifies $PA=LU$. This is the required elimination with partial pivoting, not an alternative solution method.