Re: How to solve a simple quadratic system of 3 equations?

john_ramsden_at_sagitta-ps.com
Date: 02/19/05 

Date: 19 Feb 2005 14:01:15 -0800

arturo wrote:
>
> Thanks for your trouble.

For some strange reason I don't get the tree view in
Google on my PC at home. (Any suggestions welcome.)
So I've no idea who you are replying to. Everyone
most likely, but in case it was me I should add that
in fact you don't need polynomials for _any_ case
of your set of quadratics: Expressing these as:

  a_1.x_2^2 + 2.b_1.x_2.x_3 + c_1.x_3^2 = d_1 [1]

  a_2.x_3^2 + 2.b_2.x_3.x_1 + c_2.x_1^2 = d_2 [2]

  a_3.x_1^2 + 2.b_3.x_1.x_2 + c_3.x_2^2 = d_3 [3]

The ternary quadratic p.[1] + q.[2] has determinant

  p.q.| c_2 0 b_2 |
      | |
      | 0 a_1 b_1 |
      | |
      | b_2.q b_1.p c_1.p + a_2.q |

which equals - p.q times:

  c_2.(b_1^2 - a_1.c_1).p + a_1.(b_2^2 - a_2.c_2).q

(For three general quadratics in the x_i I think it
would be a homogenous cubic in p, q. But, to answer
your question, the special "variable separated" form
of these quadratics does simplify things with this
approach.)

If b_r^2 - a_r.b_r = 0 for one or more indexes r
then the respective equation[s] are each the
algebraic square of a linear form.

Also, if c_2 = a_1 = 0 then d_2.[1] - d_1.[2] gives:

  x_3.(2.b_2.d_1.x_1 - 2.b_1.d_2.x_2

      - (a_2.d_1 - c_1.d_2).x_3) = 0

Otherwise (the general case of my previous reply),
there's a unique ratio p:q such that the determinant
is zero, and the combination form is an algebraic
square.

So in _all_ cases you can derive three pairs of linear equations, with
simple coeficients, from which to choose
one from each in 8 ways and obtain the requisite number
of solutions (simplified by the fact that solutions of
opposite sign occur in pairs, i.e. x,y,z and -x,-y,-z,
except when x, y, z are all zero).

These days computer algebra aficionados tend to scoff
at "hand crafted" solutions, on the grounds that your
system could just as well have had ten variables
instead of three and then where would you be? But
I think there are two answers (at least) to that.

Firstly, these kinds of solutions are usually more
informative. Also, if you need to calculate solutions
many times for the same kind of system, for example
to control a robot arm in real time, they tend to be
a damned site more efficient!

ObProblem: Indicate how to find three planes, through
the origin, on at least one of which all 8 solutions
must lie.

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