How Julia use Lufact to solve Ax=b

Question

I want to recreate a solve function(solve Ax = b for x) for sparse matrix. In the Julia documentation, it says that when we applied a sparse matrix to lufact(), it returns the following:

L, U, p, q, Rs = F[:(:)]

With the given formula in Julia doc: LU = Rs.*A[p,q], I did some algebra and obtained the following formula:

x = U \ ( L \ (Rs.*b[p]) )
ipermute!(x,q)

This formula matched with the default F\b solver in Julia when the matrix is dense but the result is off when the matrix is sparse. Does anyone know why?

It turns out the routine above is correct except for missing a bracket in the scaling factors. x = U \ ( L \ ((Rs.*b)[p]) ) ipermute!(x,q) — Junwei su, Jul 15 '17 at 00:06

score 2 · Answer 1 · edited Dec 31 '21 at 18:05

2

using LinearAlgebra, then B = lu(A); B\b. Julia returns a type and its dispatch on \ handles the rest.

edited Dec 31 '21 at 18:05

user14717

4,757
2
44
68

answered Jul 14 '17 at 20:07

Chris Rackauckas

18,645
3
50
81

Yes, but I want to develop my own solving routine using the factor (L,U,p,q,Rs) – Junwei su Jul 14 '17 at 22:11
1

Then check `@which B\b` and you'll get a link to the code. – Chris Rackauckas Jul 14 '17 at 22:15

How Julia use Lufact to solve Ax=b

1 Answers1