 Solution

Use Gaussian elimination with partial pivoting, choosing a largest absolute entry in the uneliminated part of each pivot column. The first pivot is two in row one, with multipliers $1/2,1/2,1/5$ for rows two, three and four. The second-column candidates are then $-1,0,-2/5$, so row two stays as pivot and the remaining nonzero multiplier is $2/5$ for row four. The augmented matrix at this stage isThe third pivot is three, so swap rows three and four. The final multiplier is $-1/4$, leaving last pivot $-2$. Carrying that swap through the previously stored multipliers givesThe transformed right side is $(3,1,0,0{)}^T$. Back substitution gives $z=0$, $y=0$, $x=-1$ and $w=1$, hence $\boxed{(w,x,y,z)=(1,-1,0,0)}$.

For the eigenvalue claim, put $S=P^{T}L$. It is invertible because a permutation matrix and a unit lower triangular matrix are invertible. The factorization gives $A=SU$, soSimilar matrices have the same characteristic polynomial and therefore the same eigenvalues with algebraic multiplicities. This proves the claim without changing the prescribed elimination method.