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## Re: LU decomposition: backsubstitution available?

**From**: |
Mario Storti |

**Subject**: |
Re: LU decomposition: backsubstitution available? |

**Date**: |
Wed, 25 Mar 1998 09:26:45 -0300 |

>*>>>> On Wed, 25 Mar 1998 10:27:40 +0100 (MET), *
>*>>>> Thomas Hoffmann <address@hidden> said:*
>* It is a known procedure for the solution of a sequence of*
>* systems of linear equations Ax=b with only varying rhs to*
>* make once a LU decompostion of A (function lu() of octave)*
>* and then backsubstitute for the several b's, e.g.*
>* x=backsubs(l,u,b)*
>* As e.g. RLaB has such an mechanism, I thought of similar*
>* functionality in Octave, but did not find any.*
>* Can anybody shed some light on this?*
>* Thomas.*
I faced this same problem before. It would be nice to have someone
writing a 'backsubs' routine. It seems to me that it should be written
in fortran in order to be efficient (do to the triangular structure of
l and u).
Meanwhile, I think that you make a significant save in computational
effort if compute the inverse of A, say
>* invA=inv(A);*
>* x1=invA*b1;*
>* x2=invA*b2;*
>* x3=invA*b3;*
>* .*
>* .*
since computing the inverse requires O(N^3) ops. and matrix-vector
multiplication requires only O(N^2). Of course, if you know all the
rhs's at once, then you can put them incolumns in a matrix B and then
make a:
X= A \ B;
and solve all the systems at once.
Mario
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Mario Alberto Storti | Fax: (54)(42) 55.09.44 |
Centro Internacional de Metodos Computacionales| Tel: (54)(42) 55.91.75 |
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