I just answered this on gitter earlier today, but you can just take the
jacobian of the system to get its matrix form. For example:
In [1]: from sympy import *
In [2]: a, b, c, d = symbols('a, b, c, d')
In [3]: x1, x2, x3, x4 = symbols('x1:5')
In [4]: x = Matrix([x1, x2, x3, x4])
In [5]: system = Matrix([a*x1 + b*x2 + c,
...: c*x1 + d*x3 + 2,
...: c*x3 + b*x4 + a])
In [6]: A = system.jacobian(x)
In [7]: B = A*x - system
In [8]: A
Out[8]:
Matrix([
[a, b, 0, 0],
[c, 0, d, 0],
[0, 0, c, b]])
In [9]: B
Out[9]:
Matrix([
[-c],
[-2],
[-a]])
In [10]: assert A*x - B == system
The functionality I'm adding for my GSoC for linearizing a system of
equations will also be able to return these matrices in a convenient form.
But it's not terribly difficult to solve for these arrangements using the
current functionality.
On Thursday, June 5, 2014 4:22:52 PM UTC-5, Andrei Berceanu wrote:
>
> Was this implemented into sympy at any point? It could be the equivalent
> of Mathematica's CoefficientArrays function.
>
> On Thursday, November 14, 2013 5:56:22 AM UTC+1, Chris Smith wrote:
>>
>> I forgot that as_independent, without the as_Add=True flag will treat
>> Muls differently. The following will be more robust:
>>
>> def eqs2matrix(eqs, syms, augment=False):
>> """
>> >>> s
>> [x + 2*y == 4, 2*c + y/2 == 0]
>> >>> eqs2matrix(s, (x, c))
>> (Matrix([
>> [1, 0],
>> [0, 2]]), Matrix([
>> [-2*y + 4],
>> [ -y/2]]))
>> >>> eqs2matrix([2*c*(x+y)-4],(x, y))
>> (Matrix([[2*c, 2*c]]), Matrix([[4]]))
>> """
>> s = Matrix([si.lhs - si.rhs if isinstance(si, Equality) else si for
>> si in eqs])
>> sym = syms
>> j = s.jacobian(sym)
>> rhs = -(s - j*Matrix(sym))
>> rhs.simplify()
>> if augment:
>> j.col_insert(0, rhs)
>> else:
>> j = (j, rhs)
>> return j
>>
>>
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