Re: Why does Inverse[M] hesitate?
- To: mathgroup at smc.vnet.net
- Subject: [mg54452] Re: [mg54370] Why does Inverse[M] hesitate?
- From: DrBob <drbob at bigfoot.com>
- Date: Sun, 20 Feb 2005 00:11:05 -0500 (EST)
- References: <200502190732.CAA06115@smc.vnet.net>
- Reply-to: drbob at bigfoot.com
- Sender: owner-wri-mathgroup at wolfram.com
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
- References:
- Why does Inverse[M] hesitate?
- From: skirmantas.janusonis@yale.edu (Skirmantas)
- Why does Inverse[M] hesitate?