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MathGroup Archive 2005

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Re: Why does Inverse[M] hesitate?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54402] Re: [mg54370] Why does Inverse[M] hesitate?
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sun, 20 Feb 2005 00:08:02 -0500 (EST)
  • Reply-to: hanlonr at cox.net
  • Sender: owner-wri-mathgroup at wolfram.com

Works fine here

$Version

5.1 for Mac OS X (October 25, 2004)

A={{a11,a12},{a21,a22}};

B={{b11,b12},{b21,b22}};

Inverse[A].B

{{(a22*b11)/(a11*a22 - a12*a21) - 
    (a12*b21)/(a11*a22 - a12*a21), 
   (a22*b12)/(a11*a22 - a12*a21) - 
    (a12*b22)/(a11*a22 - a12*a21)}, 
  {(a11*b21)/(a11*a22 - a12*a21) - 
    (a21*b11)/(a11*a22 - a12*a21), 
   (a11*b22)/(a11*a22 - a12*a21) - 
    (a21*b12)/(a11*a22 - a12*a21)}}


Bob Hanlon

> 
> From: skirmantas.janusonis at yale.edu (Skirmantas)
To: mathgroup at smc.vnet.net
> Date: 2005/02/19 Sat AM 02:32:29 EST
> To: mathgroup at smc.vnet.net
> Subject: [mg54402] [mg54370] Why does Inverse[M] hesitate?
> 
> 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.
> 
> 


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