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matrixjuliadecompositionmatrix-factorization
Junwei su
Junwei su

Reputation: 138

How Julia use Lufact to solve Ax=b

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?

Upvotes: 0

Views: 1336

Answers (1)

Chris Rackauckas
Chris Rackauckas

Reputation: 19162

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

Upvotes: 2

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