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Re: Newbie question: big matrix calculations

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  • Subject: [mg9268] Re: Newbie question: big matrix calculations
  • From: hans.steffani at (Hans Steffani)
  • Date: Mon, 27 Oct 1997 02:46:41 -0500
  • Organization: University of Technology Chemnitz, FRG
  • Sender: owner-wri-mathgroup at

dscott at (Dennis Wayne Scott) writes:
>I have three 100x100 sparse (diagonal) matrices (matrxD, matrxU, and 
>matrxL) and three 100x1 matrices (b, x2, and x1), and I'm performing 
>several operation on them:

>x2 = -Inverse[matrxD].(matrxL+matrxU).x1+Inverse[matrxD].b;

To inverte a matrix is often a bad idea and not necessary.

x2 = Inverse[matrxD].( (matrxL+matrxU).x1 + b);

needs only one inversion instead of two and will be faster than your

That is the same as
v = ( (matrxL+matrxU).x1 + b);
x2 = Inverse[matrxD].v

which does not help much but shows us the way. It is obvious that x2 is
the solution of the matrix equation

matrxD.x2 == v

To solve such kind of equations LinearSolve is implemented in
Mathematica. We may know that it is not necessary to compute the
inverse to solve a matrix equation. Therefore we can hope that
LinearSolve[] handles this faster that your solution.

This all is not tested!

Hans Friedrich Steffani

>I have to do it MANY, MANY times...  with a 5x5 matrix it takes a few
>minutes, but with a 100x100 it takes longer than eight hours (and still
>running).  Does anyone (offhandedly) know of a way to reduce the time
>this takes for Mathematica to solve?

>I'm using a P133 and Mathematica 2.2.

>The entire hunk of code is viewable at:
>  solved for a 5x5 matrix

>PS-I've already tried decreasing the "accrcy" and changing my initial


>Dennis W. Scott, Jr. 	
>University of Illinois at Urbana          dscott at
>     ----------------------------------------------------------------
>Aspiring Electrical Engineer              "I want to know God's

Hans Friedrich Steffani
Institut fuer Elektrische Maschinen und Antriebe, TU Chemnitz
mailto:hans.steffani at

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