TITLE: Is Wolfram Alpha Wrong? (LU Decomposition)
QUESTION [5 upvotes]: Ok, I've been working on a matrix solver for a while (no, its not going well, thanks for asking), and just discovered something that worrys me alot;
The problem at hand is LU Decomposition of a square matrix, for example
$\begin{matrix}
1&2&3\\
2&-1&1\\
3&4&-1\\
\end{matrix}$
WolframAlpha's answer is completely different from the example source
Wolfram
$\begin{matrix}
1&2&3\\
3&-2&-10\\
2&2.5&20\\
\end{matrix}$
Source
$\begin{matrix}
1&2&3\\
2&-5&-5\\
3&0.4&-8\\
\end{matrix}$
What's going on?

REPLY [9 votes]: It looks like Wolfram Alpha is doing an LU decomposition with partial pivoting. So, it's actually solving a system like
$$\newcommand{\bm}[1]{\mathbf{#1}}
\bm{P} \bm{A} = \bm{L} \bm{U}
$$
where $\bm{P}$ is a permutation matrix.
Curiously, it appears that Alpha simply isn't listing the permutation used in solving the system. In this case it is
$$
\bm{P} = \left(
\begin{matrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{matrix}
\right) .
$$
This is a common technique to enhance numerical stability.
You can recover $\bm{A}$ via
$$
\bm{A} = \bm{P}^{-1} \bm{L} \bm{U} = \bm{P L U}
$$
since in this case $\bm{P} = \bm{P}^{-1}$.