= Solution
Use <Gaussian elimination> without row exchanges. First apply $R_2\gets R_2-3R_1$ and $R_3\gets R_3-2R_1$; then apply $R_3\gets R_3-R_2$:
$$
\begin{pmatrix}1&2&4\\3&8&14\\2&6&13\end{pmatrix}\longrightarrow\begin{pmatrix}1&2&4\\0&2&2\\0&2&5\end{pmatrix}\longrightarrow U=\begin{pmatrix}1&2&4\\0&2&2\\0&0&3\end{pmatrix}.
$$
The elimination multipliers are $m_{21}=3$, $m_{31}=2$ and $m_{32}=1$. Put them below the diagonal of a <unit lower triangular matrix>:
$$
\boxed{L=\begin{pmatrix}1&0&0\\3&1&0\\2&1&1\end{pmatrix},\qquad U=\begin{pmatrix}1&2&4\\0&2&2\\0&0&3\end{pmatrix},\qquad A=LU.}
$$
To see why these multipliers reconstruct the rows, row 2 of $LU$ is three times row 1 of $U$ plus row 2 of $U$, giving $(3,8,14)$. Row 3 of $LU$ is twice row 1 plus row 2 plus row 3 of $U$, giving $(2,6,13)$. This directly verifies the <LU factorization>.
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