wrong LUDecomposition - first row is the last one

Question

I am having a huge problem with my Mathematica notebook. I am trying to evaluate LUDecomposition and it keeps getting wrong - when I multiply l and u, I get my A matrix, but the first row is displayed as the last. I have no idea how to fix it...

My A matrix looks like this:

A = {{3, 2, 0, 0, 0, 0, 0, 0, 0},   {2, 8, 1, 0, 0, 0, 0, 0, 0},   {0, 1, 4, 1, 0, 0, 0, 0, 0},   {0, 0, 1, 3, 2, 0, 0, 0, 0},   {0, 0, 0, 2, 4, 1, 0, 0, 0},   {0, 0, 0, 0, 1, 5, 2, 0, 0},   {0, 0, 0, 0, 0, 2, 3, 1, 0},   {0, 0, 0, 0, 0, 0, 1, 8, 2},   {0, 0, 0, 0, 0, 0, 0, 2, 4}};

What I do is:

{lu, p, c} = LUDecomposition[A];
l = LowerTriangularize[lu, -1] + IdentityMatrix[Length[p]];
u = UpperTriangularize[lu];
P = Part[IdentityMatrix[Length[p]], p];
MatrixForm /@ {P.A, P, l, u, l.u};

And I get:
The last one should equal my A matrix, but obviously it does not. It seems like the first row is the last one. Do you have any idea why? I will be very grateful for any kind of help because it is getting me really crazy.

Answer 1

As Nasser has already mentioned, you also have to take into account the permuation p. It is however not entirely obvious how to do that. Here an example:

{LU, p, c} = LUDecomposition[A];
L = LowerTriangularize[LU, -1] + IdentityMatrix[Length[p]];
U = UpperTriangularize[LU];
Permute[L.U, p] == A

True

Admittedly, my first idea was to use L[[p]].U == A (which returns False). So in that respect, the documentation could be a bit more specific.