= Solution
Eliminate the first column using multipliers $3$ and $2$. The two residual rows have second entries zero and four, so interchange their order for the next <pivot>. One resulting <Pivoted LU decomposition> is
$$
\boxed{P=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix},\quad
L=\begin{pmatrix}1&0&0\\2&1&0\\3&0&1\end{pmatrix},\quad
U=\begin{pmatrix}3&1&4\\0&4&-2\\0&0&-11\end{pmatrix}.}
$$
The factors below the first <pivot> must follow the permuted row order, which is why $L$ contains $2$ before $3$. Direct multiplication gives rows $(3,1,4)$, $(6,6,6)$ and $(9,3,1)$ in both $PA$ and $LU$. This checks the permutation convention rather than confusing $PA=LU$ with a different placement of $P$.
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