Past exam of the options I course of the University of Oxford / 2010 / P / P1

The first elimination multipliers are $l_{21}=2$ and $l_{31}=-1$. Subtracting those multiples of row one gives lower rows $(0,2,4)$ and $(0,-1,3)$. The next multiplier is $l_{32}=-1/2$, and the final elimination produces $(0,0,5)$. ThusFor an upper triangular system $Ux=b$ with nonzero diagonal, back substitution computesFor the supplied triangular system, the last row gives $x_3=10/5=2$; the second gives $x_2=(6-4\cdot 2)/2=-1$; the first gives $x_1=3-2(-1)-2(2)=1$. Hence