how i can solve a system of 5 equation where there is numbers

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bassam shallak
bassam shallak el 20 de Oct. de 2014
Editada: SK el 20 de Oct. de 2014
hi there i have a system of equation of 5 varaibales as follows :
-(l1+l2)*p1 +p2*(u2)+p3*0+p4*u2=0
l1*p1-p2*(u2+u1)+p3*0+p5*u2==0)
(p2*l2-p3*u1==0)
(l2*p1+p3*u1-p4*(l1+u2)==0)
(l1*p4-u2*p5==0)
(p1+p2+p3+p4+p5==1)
p1 p2 p3 p4 p5 are the unkown while l1 l2 l3 l4 l5 and u1 u2 u3 u4 u5 are the values but I the unkown in terms of l1 l2 l3 l4 l5 and u1 u2 u3u u4 u5 is this possible or not ?

Respuestas (4)

SK
SK el 20 de Oct. de 2014
Check the symbolic toolbox and "Mupad" that comes with it.

comlich
comlich el 20 de Oct. de 2014
Hello,
Unfortunately, there is no a way to find an analytical solution in such a complex situation or system of equations. The main reason is that, the solution depends on whether the matrix of the system (when rewritten in the matrix form: Ap=b) is invertible or not.
I have had such a problem with differentials equation which have unknown parameters, see the attached part of a report's draft.

comlich
comlich el 20 de Oct. de 2014
By 'analytical solution', of course I mean analytical solution which depends on the parameters l_i and u_i. You can sill give it a try with Matlab, but I fear that it will return an useless solution (even by using symbolics).

SK
SK el 20 de Oct. de 2014
You should be able to get a symbolic solution - makes no difference what the values of the parameters are. They don't come into play at all. After all it is just Cramer's rule. Having said that, note:
1. You have 6 equations in 5 unknowns so you wont get a symbolic solution.
2. The Symbolic toolbox is S***. Unmentionable word here. It will take you some time to get used to its quirks. Plus the confusion between the Mupad documentation and the matlab - the syntax is totally inconsistent. Best of luck trying to use it. I have written a few m-files to construct and solve some specific large linear systems. Works, but the frustration is not worth the effort.
  2 comentarios
John D'Errico
John D'Errico el 20 de Oct. de 2014
Problems with 6 equations in 5 unknowns are NOT directly solvable by Cramer's rule, and even if you could do so, Cramer's rule would be a poor way to solve the problem.
SK
SK el 20 de Oct. de 2014
Editada: SK el 20 de Oct. de 2014
That is exactly what I said in point 1. No need to repeat it.
As for Cramer's rule I mentioned it just to indicate that there is a symbolic solution which can be found using solve or perhaps linsolve or whatever. I didn't recommend actually using it to solve the equations.
Moreover, I suspect, but I'm not sure, that the question poster intended to have 5 equations but made some sort of a typo in his post.

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