= Solution
Use <columnwise partial pivoting>: select the largest absolute entry in the current column and interchange rows. In the first column the pivot is $4$, so interchange rows 1 and 2. The multipliers are $1/2$ and $-1/2$, and elimination gives
$$
\begin{pmatrix}4&1&0\\0&1/2&1\\0&5/2&1\end{pmatrix}.
$$
The second pivot is $5/2$, so interchange the two remaining rows. Their previously stored first-column multipliers must also be interchanged. The final multiplier is $(1/2)/(5/2)=1/5$, giving the <LU decomposition>
$$
\boxed{PA=LU,\qquad
P=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix},\quad
L=\begin{pmatrix}1&0&0\\-1/2&1&0\\1/2&1/5&1\end{pmatrix},\quad
U=\begin{pmatrix}4&1&0\\0&5/2&1\\0&0&4/5\end{pmatrix}.}
$$
The <permutation matrix> orders the original rows as $2,3,1$. Multiplication verifies the displayed identity.
If “column pivoting” instead denotes interchanging columns rather than searching within a column, that convention gives $AQ=\widetilde L\widetilde U$ with
$$
Q=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix},\quad
\widetilde L=\begin{pmatrix}1&0&0\\2&1&0\\-1&-1&1\end{pmatrix},\quad
\widetilde U=\begin{pmatrix}2&1&1\\0&-2&-1\\0&0&2\end{pmatrix}.
$$
This second identity uses a different pivot convention; it does not change the row-pivoted answer above.
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