SuperLU
LU=PrDrADcPc
here,
- Pr and Pc are row and col permutation matrices
- Dr and Dc are row and col scaling diagonal matrices
- L is unit lower triangular matrix (Lii=1)
- U is an upper triangular matrix
To solve
Ax=B
A is given by
A=Dr−1Pr−1LUPc−1Dc−1
A−1 is given by
A−1=DcPcU−1L−1PrDr
Therefore,
x=Dc(Pc(U