= Solution
Write $U=EE_5E_4E_3E_2E_1M$ and put $F_i=EE_iE$. Inserting $E^2=I$ between adjacent elimination <matrices> gives
$$
U=F_5F_4F_3F_2F_1(EM),\qquad EM=F_1^{-1}F_2^{-1}F_3^{-1}F_4^{-1}F_5^{-1}U.
$$
Each $E_i^{-1}$ has the restricted lower triangular pattern in part (b)(v). Also $F_i^{-1}=EE_i^{-1}E$, as direct multiplication verifies. Hence every $F_i^{-1}$ is lower triangular. Their product is lower triangular by the allowed closure property, proving the required <Pivoted LU decomposition> with $L=F_1^{-1}\cdots F_5^{-1}$.
Explicitly,
$$
\boxed{L=\begin{pmatrix}1&0&0&0\\1&1&0&0\\-1&-1&1&0\\2&1&0&1\end{pmatrix},\qquad U=\begin{pmatrix}1&1&1&1\\0&2&-1&1\\0&0&3&5\\0&0&0&4\end{pmatrix},\qquad EM=LU.}
$$
The multipliers in rows three and four of $L$ are permuted along with the rows of $M$; simply keeping the unpermuted elimination multipliers would give the wrong product. This is the mechanism in <conjugating elimination matrices across a row permutation>.
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