= Solution
{c}
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 is
$$
\left(\begin{array}{rrrr|r}
2&-1&4&2&3\\0&-1&3&2&1\\
0&0&-3/4&-3/2&0\\0&-2/5&21/5&-6/5&2/5
\end{array}\right).
$$
The second pivot is $-1$, already largest in its remaining column. Its multipliers are zero and $2/5$, giving row four $(0,0,3,-2\mid0)$. In column three, partial pivoting now requires exchanging rows three and four, since $3>3/4$. The final elimination multiplier is $-1/4$. Thus
$$
U=\begin{pmatrix}
2&-1&4&2\\0&-1&3&2\\0&0&3&-2\\0&0&0&-2
\end{pmatrix},\qquad
b_{\rm elim}=(3,1,0,0)^T.
$$
Back 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, are
$$
P=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix},\qquad
L=\begin{pmatrix}
1&0&0&0\\1/2&1&0&0\\1/5&2/5&1&0\\1/2&0&-1/4&1
\end{pmatrix}.
$$
Direct multiplication verifies $PA=LU$. Since $U$ is invertible and $A=P^TLU$,
$$
UP^TL=UAU^{-1}.
$$
The two matrices are similar, and similar matrices have the same characteristic polynomial and eigenvalues.
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