Solving a system with 4 variables and 4 equations

Question

I am trying to solve a linear system of equations with Mathematica. My variables are x1, x2, x3, x4. I use the code shown below, but it gives error saying that:

Equations may not give solutions for all "solve" variables.

Even when I used LinearSolve, I got the same error.

Solve[{(x1 - ((-m3* n2 + m2 * z3 - 0.5*m2*z3)* x2 + (0.5 *n2* z3)*
       x3 + (0.5 *m2 * z3)*x4)/((-1.5) n2* z3)) == (2/
     1.5)   ,  (x2 -  ((m3* n2 - 0.5*m3 *n2 - m2 * z3)*
       x1 + (0.5* m3* n2)* x3 + (0.5* m2* m3)* 
       x4)/((-1.5) (m2* m3)) ) == (-m3* n2 + m2* z3 )/((-1.5) m2* 
      m3)  ,    (x3 - ((-n2*  z1 + .5*n2 * z1)*
       x1 + (.5* m2 * z1 - m2 * z1)*x2 + (m1 *n2  - 0.5* m2* z1)*
       x4 )/((1.5) n2 z1) ) ==  (2/1.5 )  ,   ( 
    x4 - ((-m1 *n2 + 0.5*m1 *n2 )*x1 + (-0.5 *m1* m2) *
       x2 + (-.5* m1* n2 + m2 *z1)*x3)/((1.5) m1* m2) ) == ( 
     m1* n2 - m2* z1)/((1.5)* m1* m2)} , {x1, x2, x3, x4}]

Can anyone help me please. I really appreciate your comments.

Answer 1

Extended comment.

eq = Rationalize@{(x1 - ((-m3*n2 + m2*z3 - 0.5*m2*z3)*x2 + (0.5*n2*z3)*
          x3 + (0.5*m2*z3)*x4)/((-1.5) n2*z3)) == (2/
      1.5), (x2 - ((m3*n2 - 0.5*m3*n2 - m2*z3)*x1 + (0.5*m3*n2)*
          x3 + (0.5*m2*m3)*x4)/((-1.5) (m2*m3))) == (-m3*n2 + 
       m2*z3)/((-1.5) m2*
       m3), (x3 - ((-n2*z1 + .5*n2*z1)*x1 + (.5*m2*z1 - m2*z1)*
          x2 + (m1*n2 - 0.5*m2*z1)*x4)/((1.5) n2 z1)) == (2/
      1.5), (x4 - ((-m1*n2 + 0.5*m1*n2)*x1 + (-0.5*m1*m2)*
          x2 + (-.5*m1*n2 + m2*z1)*x3)/((1.5) m1*m2)) == (m1*n2 - 
       m2*z1)/((1.5)*m1*m2)};
sol = Solve[eq, {x1, x2, x3, x4}][[1]];

The generic solution looks like this:

{x1, x2, x3, x4} /. sol

{x1, -(z3/m3) + (x1 z3)/ m3, -((m1 x1 (m3 n2 + m2 z3))/(m3 (m1 n2 + m2 z1))) - (-2 m1 m3 n2 - m2 m3 z1 - m1 m2 z3)/(m3 (m1 n2 + m2 z1)), (z1 (m3 n2 + m2 z3))/( m3 (m1 n2 + m2 z1)) - (x1 z1 (m3 n2 + m2 z3))/(m3 (m1 n2 + m2 z1))}

Let's check that these really are solutions by substituting the solutions rules sol into the equation eq:

eq /. sol // Simplify

{True, True, True, True}

So the equations do not specify a single solution. Because there are 4 equations and four variables, that means that the equations are somewhat redundant. Usually, one would expect to get only finitely many (or countably many) solution. This is not the case here: We have a continuum of solutions. Because that is somewhat unexpected (and because this often causes problems), Solve warns us. That's all. The message is a warning, not an error.