On the contrary, that is precisely the "L" in the "LU" form.

Here's how I would get L: 'zero' out the area above the diagonal using row operations and **keeping track** of the row operations.

We can change \(\displaystyle \begin{pmatrix}1 & 3 & 1 \\ 3 & 8 & 4 \\ 1 & 4 & 5\end{pmatrix}\) to

\(\displaystyle \begin{pmatrix}1 & 3 & 1 \\ 0 & -1 & 1 \\ 0 & 1 & 4\end{pmatrix}\)

by "subtract 3 times the first row from the second" and "subtract 1 times the first row from the third"

We can change that matrix to

\(\displaystyle \begin{pmatrix}1 & 3 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 5\end{pmatrix}\)

in "upper triangular" form by "add the second row to the third row".

Now, do the **opposite** operations, in the **opposite** order, to the identity matrix.

That is, first "**subtract** the second row from the third row" to get

\(\displaystyle \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1\end{pmatrix}\)

and then "**add** 1 times the first row to the third" and "**add** 3 times the first row to the second" to get

\(\displaystyle \begin{pmatrix}1 & 0 & 0 \\3 & 1 & 0 \\1 & -1 & 1\end{pmatrix}\)

Now you can check that

\(\displaystyle \begin{pmatrix}1 & 0 & 0 \\3 & 1 & 0 \\1 & -1 & 1\end{pmatrix}\begin{pmatrix}1 & 3 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 5\end{pmatrix}= \begin{pmatrix}1 & 3 & 1 \\ 3 & 8 & 4 \\ 1 & 4 & 5\end{pmatrix}\)

or LU= A.