Do it using pencil and paper? WTP?
First, solve for b, and substitute.
Then solve for d, using your second equation.
d = a*y - a*x - c*y
(a*z-a*x)- 2*((c*z)+ a*y - a*x - c*y)=0
Again, substitute into the other equations. This now reduces to two equations
a*x - 2*a*y + a*z + 2*c*y - 2*c*z = 0
a*(a*y - a*x - c*y) - a*x*c = 0
In the latter equation, we can see that EITHER a == 0, in which case we have b = 0, or a is non-zero. in the latter case, as long as a is non-zero...
a*x - 2*a*y + a*z + 2*c*y - 2*c*z = 0
a*y - a*x - c*y - x*c = 0
These form a homogeneous system. And as we decided above, we have a non-zero. In fact though, we can choose any value for a since the system is linear and homogeneous in the unknowns {a,c}.
Therefore, we may arbitrarily choose a == 1.
x - 2*y + z + 2*c*y - 2*c*z = 0
y - x - c*y - x*c = 0
As you see, that results in an inconsistency, since that implies two independent values for c.
c = (y-x)/(y+x)
c = (x - 2*y + z)/(2*(z-y))
So as long as a is non-zero, AND (x-y) ~= 0, AND (z-y) ~= 0, then we have a problem, UNLESS it is also true that
(y-x)/(y+x) = (x - 2*y + z)/(2*(z-y))
So unless this last relation holds for x,y,z, there is no non-trivial solution.
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