LU Factorization

General Usage
1

I am just confused about one thing related to LU factorization. It is known that the solution of A.x=b is x = U\L\P*b. Why when I apply it in the below it gives another solution comparing with A\b, while LU\b gives the same result?

A = [1 3 3; 6 9 7;-3 3 4]; b = [1,2,3];

julia> A\b
3-element Vector{Float64}:
 -0.6000000000000001
  0.9333333333333331
 -0.39999999999999974

LU = lu(A);

julia> LU.U\LU.L\(LU.P)*b
3-element Vector{Float64}:
  46.00000000000013
  53.0000000000001
 -17.933333333333373

julia> LU\b
3-element Vector{Float64}:
 -0.6000000000000001
  0.9333333333333331
 -0.39999999999999974
2

Just need to add more parentheses:

julia> LU.U\(LU.L\(LU.P)*b)
3-element Vector{Float64}:
 -0.6
  0.9333333333333333
 -0.4000000000000001
2 Likes
3

\ is left-associative in Julia, so this is parsed as

(LU.U\LU.L)\(LU.P*b)

whereas you want:

LU.U\(LU.L\(LU.P*b))

or equivalently LU.U\(LU.L\b[LU.p]), which gives the same answer as A \ b.

2 Likes
4

Thank you very much! I got it