solve system of equations (find the constants with respect to variable)

Hi,

I have four equations with four constants (a,b,c,d). how I can find values of the constant (a,b,c,d) with respect to x,y,z? I tried to used solver but I failed

These are my equations

 a*x+b=0
((a*y)+b)-((c*y)+d)=0
((a*z)+b)- 2*((c*z)+d))=0
 a*d-b*c=0

Regards

[John D'Errico]

Simple.

a=b=c=d=0

[Ali Kareem]

Hi,

Thank you for your reply. I do not think that will be the answer.

Regards

Answer 1

[John D'Errico]

Do it using pencil and paper? WTP?

syms a b c d x y z

First, solve for b, and substitute.

b = -a*x

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.

[Ali Kareem]

Thank you!