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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 ?

comlich
on 20 Oct 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
on 20 Oct 2014

SK
on 20 Oct 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.

John D'Errico
on 20 Oct 2014

SK
on 20 Oct 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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