Question: Why doesn't solve give any indication that there can be a non-trivial solution to this system of two linear equations?

I created the following worksheet to illustrate my question.

restart

We have two equations. w and w__0 are parameters and we wish to solve for A__1 and A__2.

eq1 := A__1*(w^2-3*w__0^2)+A__2*w__0^2 = 0 = A__1*(w^2-3*w__0^2)+A__2*w__0^2 = 0NULL

eq2 := A__1*w__0^2+A__2*(w^2-w__0^2) = 0 = A__1*w__0^2+A__2*(w^2-w__0^2) = 0NULL

solve({eq1, eq2}, {A__1, A__2}) = {A__1 = 0, A__2 = 0}NULL

 

Maple says the only solution is the trivial solution.

If we check the determinant of the matrix of the system we see it can be zero for certain values of w given w__0.

M := Matrix([[w^2-3*w__0^2, w__0^2], [w__0^2, w^2-w__0^2]]) = Matrix(%id = 36893488152149895276)NULL

d := LinearAlgebra:-Determinant(M) = w^4-4*w^2*w__0^2+2*w__0^4NULL

solutions := solve(d = 0, w)

(2+2^(1/2))^(1/2)*w__0, -(2+2^(1/2))^(1/2)*w__0, (2-2^(1/2))^(1/2)*w__0, -(2-2^(1/2))^(1/2)*w__0

(1)

If w is one of these values then the system of equations is singular and has non-zero solutions.``

 

For example, subbing the first value above into the equations and solving Maple gives us non-trivial solutions.

 

solve({subs(w = solutions[1], eq1), subs(w = solutions[1], eq2)}, {A__1, A__2})

{A__1 = -(1+2^(1/2))*A__2, A__2 = A__2}

(2)

``

Why didn't Maple give us any indication that there could be non-zero solutions in the first call to solve?


For context, the system of equations comes from a calculation involving coupled oscillators.

I had a system of differential equations, guessed at a solution, plugged it in and got the equations shown in the worksheet. The guess isn't a correct solution in general, but it is a solution if w is one of the values computed in the worksheet (the values of the variable "solutions").

Download solve-nonzero-solutions.mw

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