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.
>
>