Re: Solve with dot products
- To: mathgroup at smc.vnet.net
- Subject: [mg70733] Re: Solve with dot products
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Wed, 25 Oct 2006 01:39:59 -0400 (EDT)
- References: <ehkch0$isb$1@smc.vnet.net>
Hi,
you must write out your equations explicit because
Mathematica will not know that a, b,c and d
are matrices and not scalars, vectors or higher order
tensors ?
That's why the output of Mathematica
is nonsense because the Dot[] product
has no inverse and Mathematica thinks
that a,b,c,d are scalars.
Regards
Jens
Oliver Friedrich wrote:
> Hallo,
>
> I have some difficulties with interpretation of a solution that Solve
> returns.
>
> In[2]:=
> Solve[a.b.c==d,a]
>
> From In[2]:=
> Solve::ifun: Inverse functions are being used by Solve, so some solutions
> may \
> not be found.
>
> Out[2]=
> {{a\[Rule]InverseFunction[Dot,1,3][d,b,c]}}
>
> When I do that equation by hand I get something that I can't put into
> correspondence with the solution above.
>
> * means Inverse of a
> (a.b)*=b*.a*
>
> a.b.c=d |(_)*
>
> c*.b*.a*=d* |c.
>
> b*.a*=c.d* |b.
>
> a*=b.c.d* |(_)*
>
> a=d.c*.b*
>
> b and c are swapped compared to the Mathematica solution.
>
> 1) Where's my error?
> 2) What's the interpretation of such an InverseFunction expression, I
> don't get along the docu.
> 3) What's the z in InverseFunction[x,y,z][args] good for when I could
> count the number of arguments in args?
>
> Thanks a lot for your help
>
>