Define the domain of symbolic variables for a matrix when solving a system of linear equations?

Questions
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1

Hi! I have a problem about solving a system of linear equations with 8 equations and 8 unknowns in the form AX = 0. I solved Det[A] =2(b^2 - 1)c^3(x - 2)^2Sin[Pix]^3*(bfSin[Pix] + Cos[Pix]), but this matrix has parameters c > 0, -0.5 < b < 0.5, f ≠ 0, and 0 < x < 1. So there is only one solution for x: x=ArcTan[1/b/f]/Pi. But I brought this solution into A using the NullSpace to solve gives me the output { }. I guess it could be that the matrix does not define the domains of the parameters a,b,c,f. Is that something that I should specify in the code? I want to get the solution to the system of equations when x = ArcTan[1/b/f]/Pi.

Here is my code:

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2

Your value x=ArcTan[1/b/f]/Pi seems wrong to me. I get this:

{{x -> ConditionalExpression[(
    2 ArcTan[b f + Sqrt[1 + b^2 f^2]])/\[Pi], -(1/2) < b < 1/2 && 
     c > 0]}}
3

Thank you! Your method gives the correct result. However, when I change some parameters to get a new matrix, the solution for x becomes a complex transcendental equation, and it is not possible to get a solution for x. How do I find the solution X to the system of linear equations in this case? The parameters also satisfy -1 < a < 1, -0.5 < b < 0.5, c > 0, f ≠ 0, and 0 < x < 1.
Here is the new matrix:

mat={{(-2 + x) Cos[1/2 \[Pi] (-2 + x)] - 
   f (-2 + x) Sin[1/2 \[Pi] (-2 + x)], -f (-2 + x) Cos[
     1/2 \[Pi] (-2 + x)] + (2 - x) Sin[1/2 \[Pi] (-2 + x)], 
  x Cos[(\[Pi] x)/2] - 
   f (-2 + x) Sin[(\[Pi] x)/2], -f (-2 + x) Cos[(\[Pi] x)/2] - 
   x Sin[(\[Pi] x)/2], 0, 0, 0, 
  0}, {(-2 + x) Cos[1/2 \[Pi] (-2 + x)], (2 - x) Sin[
    1/2 \[Pi] (-2 + x)], 
  x Cos[(\[Pi] x)/2], -x Sin[(\[Pi] x)/2], (2 - x) Cos[
    3/2 \[Pi] (-2 + x)], (2 - x) Sin[3/2 \[Pi] (-2 + x)], -x Cos[(
    3 \[Pi] x)/2], -x Sin[(3 \[Pi] x)/2]}, {1/
   2 (-2 + a c - b c) (-2 + x) Cos[1/2 \[Pi] (-2 + x)], 
  1/2 (2 - a c + b c) (-2 + x) Sin[1/2 \[Pi] (-2 + x)], 
  1/2 (-2 + a c - b c) (-1 - (-2 + c - b c)/(2 - a c + b c) + 
     x) Cos[(\[Pi] x)/2], 
  1/2 (2 - a c + b c) (-1 - (-2 + c - b c)/(2 - a c + b c) + 
     x) Sin[(\[Pi] x)/2], (-2 + x) Cos[
    3/2 \[Pi] (-2 + x)], (-2 + x) Sin[
    3/2 \[Pi] (-2 + x)], (-1 + 1/2 (2 - c - a c) + x) Cos[(3 \[Pi] x)/
    2], (-1 + 1/2 (2 - c - a c) + x) Sin[(3 \[Pi] x)/2]}, {Sin[
   1/2 \[Pi] (-2 + x)], Cos[1/2 \[Pi] (-2 + x)], Sin[(\[Pi] x)/2], 
  Cos[(\[Pi] x)/2], Sin[3/2 \[Pi] (-2 + x)], -Cos[3/2 \[Pi] (-2 + x)],
   Sin[(3 \[Pi] x)/2], -Cos[(3 \[Pi] x)/2]}, {0, 1, 0, 1, 0, -1, 
  0, -1}, {-2 + x, 0, x, 0, 2 - x, 0, -x, 0}, {0, 
  1/2 (2 - a c + b c) (-2 + x), 0, 
  1/2 (2 - a c + b c) (-1 + (-2 + c - b c)/(2 - a c + b c) + x), 0, 
  2 - x, 0, 1 + 1/2 (2 - c - a c) - x}, {1/2 (2 - a c + b c) (-2 + x),
   0, 1/2 (2 - a c + b c) (-1 - (-2 + c - b c)/(2 - a c + b c) + x), 
  0, 2 - x, 0, 1 + 1/2 (-2 + c + a c) - x, 0}}

1-1.nb (25.1 KB)

4

Do you really need the full symbolic solution with all parameters? It seems hard.

5

Yes, I would like to have the solution with all parameters.The solution can be complicated. Anyway, thank you very much for your help!!

6

The determinant is of the first degree with respect to f. If you insist on solving for x it may be unfeasible. Here is a numeric case:

det = Det[mat] // Simplify // Factor
det1 = Simplify[det /. {a -> -1/2, b -> 1/4, c -> 1, f -> 1}]
Plot[det1, {x, 0, 1}]
Solve[det1 == 0 && 0 < x < 1]
7

Thanks! If the value of f is changed from f ≠ 0 to -1/10 < f < 1/10, or even f = 0, while keeping the range of the other parameters constant and x as a real number, will this provide a solution to the system of equations?

8

You can try. Good luck!