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
>
> 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
>
>
>

```

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