Re: Solving linear equations with symbolic RHS
- To: mathgroup at smc.vnet.net
- Subject: [mg30154] Re: Solving linear equations with symbolic RHS
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Wed, 1 Aug 2001 02:19:09 -0400 (EDT)
- References: <9k5qn6$h9t$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Tony, How about some pre-expanding? Ainv=Inverse[N[A,16]]; soln=Ainv.Expand[N[b,16]]; Expand[soln]; -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Tony MacKenzie" <mackenzi at usq.edu.au> wrote in message news:9k5qn6$h9t$1 at smc.vnet.net... > Once again I would like to thank everyone who helped with my last post. This > question is more general and therefore probably much more difficult to > answer. > > I am trying to solve linear equations of the form A.x=b. A is a completely > numeric matrix but b is a symbolic vector. I have tried Solve and > LUDecomposition but neither of these seem to work very well when b is a > symbolic vector. What has worked the best for me is to invert the matrix A > (There is no problem with ill-conditioning) and to multiply by the symbolic > vector b. However, what slows this process down is when I expand the > result. > > Ainv=Inverse[N[A,16]]; > soln=Ainv.N[b,16]; > Expand[soln]; (This last steps slows down dramatically as the size of the > problem increases). > > Each element of the symbolic solution vector may have a large number of > terms. > > I realise this question I am posting is quite vague but if anyone has any > ideas it would be greatly appreciated. > > Tony MacKenzie > > >