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MathGroup Archive 2006

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Solve with dot products

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70723] Solve with dot products
  • From: Oliver Friedrich <xoliver.friedrich at tzm.dex>
  • Date: Tue, 24 Oct 2006 02:24:43 -0400 (EDT)

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


-- 
Regards

Oliver Friedrich

My email has no x!


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