Thanks to both of you!
Mladen
On Feb 7, 12:19 pm, Jason Grout <jason-s...@creativetrax.com> wrote:
> Craig Citro wrote:
> > Hi,
>
> > Yep, there are definitely easy ways of doing this. Here's one way:
>
> > sage: var('a b c d')
> > (a, b, c, d)
> > sage: A = matrix(2,[2,1,1,1])
> > sage: B = matrix(2,[a,b,c,d])
> > sage: C = A*B - B*A
>
> > sage: [ e == 0 for e in C.list() ]
> > [c - b == 0, d + b - a == 0, -d - c + a == 0, b - c == 0]
>
> > sage: solve([ e == 0 for e in C.list() ], N.list())
> > [[a == r2 + r1, b == r2, c == r2, d == r1]]
>
> Just to point out for those that need a guide through the above, Craig
> is basically creating a list of equations from each entry in the matrix.
> He doesn't show it, but I suppose that N is a matrix of the variables.
> You could also do:
>
> solve([ e == 0 for e in C.list() ], C.variables())
>
> That said, I think it would be a great thing if solve could recognize
> matrices and that two matrices are equal if each entry is equal. I
> believe MMA does this (but it's easier there; matrices are nothing more
> than nested lists). It'd certainly make certain things I do more
> natural if I could do:
>
> solve(matrixA==matrixB)
>
> and that was equivalent to:
>
> solve([i==j for i,j in zip(matrixA.list(), matrixB.list())])
>
> if the matrices were of the same dimensions.
>
> Okay, so now that I've written my piece, I suppose the next step is to
> open a trac ticket, write a patch to implement it, and post it for
> review :).
>
> The ticket ishttp://trac.sagemath.org/sage_trac/ticket/5201
>
> I won't cry if someone submits a patch before I get to it :).
>
> Jason
--~--~---------~--~----~------------~-------~--~----~
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support-unsubscr...@googlegroups.com
For more options, visit this group at
http://groups.google.com/group/sage-support
URLs: http://www.sagemath.org
-~----------~----~----~----~------~----~------~--~---
