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. > >