solving for unknown matrix element as function of an undetermined independent variable

Assume a matrix equation of the form A*x=b, where matrix A has elements that are functions of an independent variable "E" which is to be determined later (For example, for simplicity,let the four matrix elements be a11=@(E) 2*E, a12=@(E) E^2 and so on for a21 and a22 to have any form of functions)

vector x=[1,r] where r is unknown to be found as a function of E

vector b=[t,0] where t is unknown to be found as a function of E

the question goes: Is there a possible way to code the problem (as it is without further manual mathematical manipulations) to find "t" and "r" as functions of E

Answer 1

[John D'Errico]

Think about it. Write out what you are asking. You have a 2x2 matrix, where each element is a function of the unknown variable E. (Some of those elements may be a constant.) But assume that there is only one unknown in A, and that is the variable E. So write it out!!!!!!!!

A = [A11(E),A12(E);A21(E),A22(E)];

So what is A times some vector x? Lets use the example you gave. x = [1;r], b = [t;0]. Then A*x=b means this:

A11(E) + A12(E)*r = t

AND

A21(E) + A22(E)*r = 0

Here, we assume that r and t are fixed variables, although they are not given any known value. Is there a way to solve for E?

NO NO NO. I said it three times, so it must be true.

You are posing TWO equations in the ONE unknown variable E. In general, there will never be any solution to solve for E that satisfies BOTH equations, unless one of them is trivially true, or unless they are both essentially the same equation.

The point is a matrix equation of this form is equivalent to TWO equations. With only one variable to solve for, you have no way forward.

[mina asham]

Thanks for your reply, but actually I dont want to solve for E which would be determined later but I am solving for t and r as functions of E. To end the problem here consider two variables instead of one i.e. 'E' and 'x' where all the elements in the equations are functions of both of them. the question still holds.

Sir, I am looking for the code not the mathematics

[John D'Errico]

You can NEVER determine code to solve a problem, UNTIL you explain it sufficiently for someone to understand what you want, or until you understand the problem. In this case, what matters is that you needed to clearly explain what problem you wanted to solve.

If all you need to do is solve for t and r, then why is there a problem?

You have two equations, in two unknowns. That they are in matrix form is not important.

solve(A*x == b,r,t)

Solve will do all the work for you.

syms E t r
A = [E,E^2 + 2;1,E];
x = [1;r];
b = [t;0];
sol = solve(A*x==b,t,r)

sol = 
  struct with fields:

      t: [1×1 sym]
      r: [1×1 sym]

sol.t
ans =
-2/E

sol.r
ans =
-1/E

In thiscase, I was careful to choose A in a way that the solution alwaysexists, since we have

det(A)
ans =
-2

So A is never singular for any value of E. (One of the rare circumstances where I will used det.)

[mina asham]

One more question please, can the same code be done using function-handle @(E) instead of using syms and struct

[Matt J]

You would have to use fsolve for every fixed E of interest.

[Bruno Luong]

r = @(E) (-a21(E)./a22(E))
t = @(E) (a11(E) - a12(E).*a21(E)./a22(E))

Answer 2

[Matt J]

Yes, the solution would be,

       C=[a12(E) -1;a22(E) 0 ]; 
       d=[-a11(E);-a12(E)];

       xr= inv(C)*d  %xr=[x(E);t(E)]

[mina asham]

thanks for your reply but here you deduced another matrix yourself "C" and continued coding. this example is a simple one for a more complicated problem so I dont want to manipulate anything myself, just enter the original matrices and look for the results

[Matt J]

Couldn't you just use solve() ? However, as long as the problem has the general linear structure A*x=b (the structure you told us to assume) and as long as x,b are linear in the unknowns. I don't see why the solution wouldn't generalize easily.

Answer 3

[Bruno Luong]

If there is an solution it would be:

r = @(E) (-a21(E)./a22(E))
t = @(E) (a11(E) - a12(E).*a21(E)./a22(E))