Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / P / 1 / Solution

Use Gaussian elimination with partial pivoting, choosing the largest absolute available entry in the current pivot column. The first pivot is 2, with multipliers $1/2,1/2,1/5$. After eliminating that column, the augmented system isThe second pivot is $-1$, already largest in its remaining column. Its multipliers are zero and $2/5$, giving row four $(0,0,3,-2\mid 0)$. In column three, partial pivoting now requires exchanging rows three and four, since $3>3/4$. The final elimination multiplier is $-1/4$. ThusBack substitution gives $\boxed{(w,x,y,z)=(1,-1,0,0)}$.

The permutation and recorded multipliers, with the earlier multiplier rows swapped at the third pivot, areDirect multiplication verifies $PA=LU$. Since $U$ is invertible and $A=P^{T}LU$,The two matrices are similar, and similar matrices have the same characteristic polynomial and eigenvalues.