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Re: Why does Inverse[M] hesitate?


I see no such "hesitation". For example:

A = Array[a, {2, 2}];
B = Array[b, {2, 2}];
Inverse[A].B

{{(a[2, 2]*b[1, 1])/
      ((-a[1, 2])*a[2, 1] +
       a[1, 1]*a[2, 2]) -
     (a[1, 2]*b[2, 1])/
      ((-a[1, 2])*a[2, 1] +
       a[1, 1]*a[2, 2]),
    (a[2, 2]*b[1, 2])/
      ((-a[1, 2])*a[2, 1] +
       a[1, 1]*a[2, 2]) -
     (a[1, 2]*b[2, 2])/
      ((-a[1, 2])*a[2, 1] +
       a[1, 1]*a[2, 2])},
   {-((a[2, 1]*b[1, 1])/
       ((-a[1, 2])*a[2, 1] +
        a[1, 1]*a[2, 2])) +
     (a[1, 1]*b[2, 1])/
      ((-a[1, 2])*a[2, 1] +
       a[1, 1]*a[2, 2]),
    -((a[2, 1]*b[1, 2])/
       ((-a[1, 2])*a[2, 1] +
        a[1, 1]*a[2, 2])) +
     (a[1, 1]*b[2, 2])/
      ((-a[1, 2])*a[2, 1] +
       a[1, 1]*a[2, 2])}}

Bobby

On Sat, 19 Feb 2005 02:32:29 -0500 (EST), Skirmantas <skirmantas.janusonis at yale.edu> wrote:

> I'm puzzled by the following in Mathematica 5.0 and 5.1:
> If I define a symbolic matrix A and a symbolic matrix B and ask to
> calculate Inverse[A].B, the output is the same input operation with
> the A and B expanded. I have re-input this output, I finally get the
> result.
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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