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Re: non linear system, 6 unknowns, 6 equations (quadratic)
Posted:
Jun 17, 2012 12:43 PM
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> Hi,
>
> x1, x2, x3 are fixed vectors of R^3
> a=(a1,a2,a3) and t=(t1,t2,t3) are two unknown vectors
> of R^3. I is the 3,3 identity matrix
>
> Solve this system (the matrices in parenthesis are
> symmetric: only 6 eqns.):
> a1(x1*x1'-x1*t'-t*x1')+a2(x2*x2'-x2*t'-t*x2')+a3(x3*x3
> '-x3*t'-t*x3')=I
>
> Before trying to solve that numerically,
> any analytical result would be great
> (e.g., is there at least one solution ?).
>
> Thank you!
>
> Michel.
OK, reading some of the other posts I think I know what you mean. Evidently x_i are column vectors and you are using ' to mean transpose (looks weird to me).
Writing x1 as [ x11, x12, x13 ]', I got the 6 equations to be those printed below.
This system is solvable symbolically with resultants. For example, the resultant for a1 factors into polynomials of 6, 1, 12, and 5566 terms. If you are interested, let me know.
Robert H. Lewis
Fordham University
equations, set each to 0:
-2*x31*a3*t1 - 2*x21*a2*t1 - 2*x11*a1*t1 + x31^2*a3 + x21^2*a2 + x11^2*a1 - 1 ,
- x31*a3*t2 - x21*a2*t2 - x11*a1*t2 - x32*a3*t1 - x22*a2*t1 - x12*a1*t1 + x31*x32*a3 + x21*x22*a2 + x11*x12*a1 ,
- x31*a3*t3 - x21*a2*t3 - x11*a1*t3 - x33*a3*t1 - x23*a2*t1 - x13*a1*t1 + x31*x33*a3 + x21*x23*a2 + x11*x13*a1 ,
-2*x32*a3*t2 - 2*x22*a2*t2 - 2*x12*a1*t2 + x32^2*a3 + x22^2*a2 + x12^2*a1 - 1 ,
- x32*a3*t3 - x22*a2*t3 - x12*a1*t3 - x33*a3*t2 - x23*a2*t2 - x13*a1*t2 + x32*x33*a3 + x22*x23*a2 + x12*x13*a1 ,
-2*x33*a3*t3 - 2*x23*a2*t3 - 2*x13*a1*t3 + x33^2*a3 + x23^2*a2 + x13^2*a1 - 1
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