I have the following matrix:
A = \begin{bmatrix} 1 & 2 & -3 \\ -5 & -4 & 13 \\ 2 & 1 & -5 \\ \end{bmatrix}
If you attempt to find the LU using what my book calls "storage of multipliers" or I assume to really be pivoting you run into a problem, when using this technique. That is U cannot have a value that is 0 in the diagonal entries.
In which case the following should be the wrong answer?
L = \begin{bmatrix} 1 & 0 & 0 \\ -5 & 1 & 0 \\ 2 & -1/2 & 1 \\ \end{bmatrix}
U = \begin{bmatrix} 1 & 2 & -3 \\ 0 & 6 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}
In which case one would have to use a permutation matrix P? (This wasn't covered for us but I assume it's the only way to solve the problem correctly, since U is not of the correct form?)
I'm basically confused as to if my answer is wrong or it's wrong and there is a more appropriate way to solve the problem. Thanks.