Polynomial system of 6 variables

[Bruno Tolentino]

U = A a²
V = 2 A a b
W = A b²
u = 2 A a c + B a
v = 2 A b c + B b
w = A c² + B c + C

I'd like to solve this system for A, B, C, a, b, c. Is it possible!?

 

[Suraj M]

If you want those terms, in terms of all the other terms, then i don't think you can, if in terms of U,V,W,u,v,w included then ofcourse, it would be simple,
the basic problem is you have only one equation with U in it, or were you trying to say $U$instead of $u$?

 

[DDH]

Unless I'm mistaken, there are 12 variables and not 6 which means the system cannot be solved.

 

[Mark44]

Unless I'm mistaken, there are 12 variables and not 6 which means the system cannot be solved.

The system cannot be solved uniquely, which is different from saying that it can't be solved.

 

[DDH]

I stand corrected.

 

[mathman]

U = A a²
V = 2 A a b
W = A b²
u = 2 A a c + B a
v = 2 A b c + B b
w = A c² + B c + C

I'd like to solve this system for A, B, C, a, b, c. Is it possible!?

Assuming U,V,W,u,v,w are known, it might be possible. As a first step, the last three equations in A,B,C are linear, so you can get A,B,C, in terms of u,v,w,a,b,c. Substitute the expression for A into the first three equations. You now have polynomial expressions for a,b,c - good luck!

 

[Bruno Tolentino]

Yeah! (U, V, W, u, v, w) are known. I want to write (A, B, C, a, b, c) in terms of (U, V, W, u, v, w).
...\begin{bmatrix}<br /> u\\ <br /> v\\ <br /> w\\<br /> \end{bmatrix}<br /> <br /> =<br /> <br /> \begin{bmatrix}<br /> 2ac &amp; a &amp; 0\\ <br /> 2bc &amp; b &amp; 0\\ <br /> 1cc &amp; c &amp; 1\\<br /> \end{bmatrix}<br /> <br /> \begin{bmatrix}<br /> A\\ <br /> B\\ <br /> C\\<br /> \end{bmatrix}<br />
===> http://www.wolframalpha.com/input/?i=inverse+of+{{2+a+c,+a,+0},+{2+b+c,+b,+0},+{c^2,+c,+1}}

Oooooops...