Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / P / 1 / Solution

Eliminate the first column with multipliers ${\ell }_{21}=2$ and ${\ell }_{31}=-1$. The new second and third rows are $(0,2,4)$ and $(0,-1,3)$. The next multiplier is ${\ell }_{32}=-1/2$, giving final row $(0,0,5)$. Recording these multipliers gives the LU decompositionFor an upper-triangular matrix with nonzero diagonal, the back substitution algorithm works from bottom to top:In the separate triangular system printed in the question, $5x_3=10$, $2x_2+4x_3=6$, and $x_1+2x_2+2x_3=3$. Hence