Re: Solve with dot products

• To: mathgroup at smc.vnet.net
• Subject: [mg70741] Re: Solve with dot products
• From: dh <dh at metrohm.ch>
• Date: Wed, 25 Oct 2006 01:40:08 -0400 (EDT)
• References: <ehkch0\$isb\$1@smc.vnet.net>

```Hi Oliver,
the manual and you are both right. The manual says about InverseFunction:
"InverseFunction[f][y] gives the value of x for which f[x] is equal to y"
therefore, InverseFunction[Dot,1,3][d,b,c] means, that value of x, that
makes Dot[x,b,c] equal to d and this is exactly your a. This solution is
implicite, wheras your solution is explicite and because of the
inversion, b and c are reversed.

Daniel

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
>
>

```

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