Question on reducing 3 equations with 3 unknowns

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This discussion focuses on reducing a system of three quadratic equations with three unknowns (A, B, C) to fewer variables. The equations are structured as a_1 A^2 + b_1 B^2 + c_1 C^2 + d_1 = f_1, a_2 A^2 + b_2 B^2 + c_2 C^2 + d_2 = f_2, and a_3 A^2 + b_3 B^2 + c_3 C^2 + d_3 = f_3. Participants suggest transforming the variables to X = A^2, Y = B^2, Z = C^2, allowing the equations to be expressed linearly. The discussion concludes that it is feasible to reduce the system to two equations in two unknowns or one quartic equation in one unknown through systematic elimination.

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Say I have a system of equations of the following form:

a_1 A^2 + b_1 B^2 + c_1 C^2 + d_1 = f_1
a_2 A^2 + b_2 B^2 + c_2 C^2 + d_2 = f_2
a_3 A^2 + b_3 B^2 + c_3 C^2 + d_3 = f_3

Where a,b,c,d,f are coefficients, and A,B and C are unknown variables.

1. Can we write them in terms of two variables only?
2. Can we write them in terms of one variable only?
 
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If you replace the A^2 with AA and similarly with B and C then you have 3 linear eqns in AA, BB, and CC that can be easily solved using matrix methods.
 
I was asking because I was reading a document where the author deals with 3 equations and 3 unknown parameters, each equation has quadratic dependence on each of the 3 parameters. Then the author says: with these three equations and three parameters we can reduce the equations to two second order equations with respect to two unknowns (call this case 1), or to one quartic equation with respect to one unknown (call this case 2). So I wanted to know how to do this. I'll do the exercise by eliminating one of the variable and see what I get.
 
Ad jedishrfu said, in the given equation, since only the squares of the unknown values occur, we can treat them as, say X=A^2, Y= B^2, and Z= C^2. Then the equations become a_1X+ b_1Y+ c_1Z= f_1, a_2X+ b_2Y+ c_2Z= f_2, and a_3X+ b_3Y+ c_3Z= f_3. (There is no need for the separate 'd_1' and 'f_1'- we can always subtract d_1 from both sides of the equation.)

There are many ways to solve a "system of equations" but, yes, they all basically involve reducing from three equations in three unknowns to two equations in two unknowns and then to one equation in one unknown. For example, if a_2 is non-zero, I can solve the first equation for X: a_1X= f_1- b_1Y- c_1Z so X= (f_1-b_1X- c_1Z)/a_1. Now replace "X" in the other two equations by that to get two equations in Y and Z only.

Or:
1) multiply the first equation by a_2 to get
a_1a_2X+ b_1a_2Y+ c_1a_2Z= f_1a_2 and
2) multiply the second equation by a_1 to get
a)a_1a_2X+ a_1b_2Y+ a_2c_3Z= a_2f_3. Now
3) subtract those two equation. Since the X term in each equation has the same coefficient, a_1a_2, they cancel leaving a single equation in Y and Z.

Do the same with, say, the first and third equation, to get a second equation in Y and Z.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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