Re: Re: Solving an equation
- To: mathgroup at smc.vnet.net
- Subject: [mg34528] Re: Re: Solving an equation
- From: "PSi" <no at eee.gr>
- Date: Mon, 27 May 2002 01:15:58 -0400 (EDT)
- References: <ackoi2$8es$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Thank you. I gave only an example. The other cases are similar. Note that, under the specified conditions on b, c (b.c==c.b and b not a scalar multiple of the unit matrix), the given equation always has a solution. "DrBob" <majort at cox-internet.com> wrote in message news:ackoi2$8es$1 at smc.vnet.net... > PSi, > > You've changed the problem by making b diagonal. The same solution > technique works even if you hadn't done that... but gives a solution > that involves division by b3 (which you've now made 0). Hence the > solution for non-diagonal b doesn't include the other solution as a > special case. > > Also, with diagonal b, notice that unless c is diagonal as well, the > condition c.b==b.c implies b1==b4. (Evaluate c.b-b.c to see what I > mean.) Yet the solution involves division by b1-b4. Hence this > solution is valid only if c is also diagonal. You can verify this as > follows: > > a = {{1, 0}, {0, 1}}; > b = {{b1, 0}, {0, b4}}; > c = {{c1, c2}, {c3, c4}}; > Solve[{a x + b y == c, c.b == b.c}, {x, y, c2, c3}] > > {{c2 -> 0, c3 -> 0, > x -> -((b4*c1 - b1*c4)/ > (b1 - b4)), > y -> -((-c1 + c4)/ > (b1 - b4))}} > > By the way, > > b.c - c.b > > {{0, b1 c2 - b4 c2}, {-b1 c3 + b4 c3, 0}} > > Rather than making b or c diagonal, you might solve for b2 and b3 this > way: > > b = {{b1, b2}, {b3, b4}}; > c = {{c1, c2}, {c3, c4}}; > Solve[{c.b == b.c}, {b2, b3}] > > {{b2 -> ((b1 - b4)*c2)/ > (c1 - c4), b3 -> > ((b1 - b4)*c3)/(c1 - c4)}} > > In general... I think there's more going on here than you've realized. > > Bobby Treat > > -----Original Message----- > From: PSi [mailto:psino at tee.gr] To: mathgroup at smc.vnet.net > Subject: [mg34528] Re: Solving an equation > > Nevermind, it's simple! > a = {{1, 0}, {0, 1}} > b = {{b1, 0}, {0, b4}} > c = {{c1, c2}, {c3, c4}} > Solve[{a x + b y == c, c.b == b.c}, {x, y}] > "PSi" <psino at tee.gr> wrote in message news:... > > > > I want to solve the following equation with Mathematica 4.1: > > a*x+b*y=c > > where x, y are the unknown scalars, > > a={{1,0},{0,1}}, > > b={{b1,b2},{b3,b4}}, > > c={{c1,c2},{c3,c4}}, > > the matrices b, c commute, and the matrix b is not a scalar multiple > of the > unit > > matrix a. > > Could anybody help? > > > > > > > > > > > > > > > > > > > > > >