RE: Re: Solving an equation
- To: mathgroup at smc.vnet.net
- Subject: [mg34516] RE: [mg34476] Re: Solving an equation
- From: "DrBob" <majort at cox-internet.com>
- Date: Fri, 24 May 2002 02:42:30 -0400 (EDT)
- Reply-to: <drbob at bigfoot.com>
- Sender: owner-wri-mathgroup at wolfram.com
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: [mg34516] [mg34476] 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?
>
>
>
>
>
>
>