Results 1 to 3 of 3

Math Help - System of nonlinear equations

  1. #1
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419

    System of nonlinear equations

    So I have that xy + 4xz - A = 0, \lambda y + 4\lambda z - yz = 0, \lambda x - xz = 0,  4\lambda x - xy = 0 and I want to solve for x,y,z,\lambda in terms of A, but I am absolutely stuck and do not know how to solve this system of equations, any help would be appreciated.

    Everytime I attempt to solve I get z = \lambda, which screws up my calculations (at least the way that I attempted)
    Last edited by Pinkk; October 6th 2010 at 07:25 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Well, the equation \lambda x-xz=0=x(\lambda-z) has only two possibilities: x=0 or \lambda=z. That's simply a requirement of your system.

    I was thinking along the lines of quadratic forms and diagonalization of matrices, but the four matrices associated with your quadratic system do not commute, so you can't simultaneously diagonalize them.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    Posts
    660
    Thanks
    133
    The third equation will factorize as x(\lambda-z)=0 implying that either x=0, or z=\lambda.

    Substituting z=\lambda into the first and second equations leads to
    xy+4\lambda x=A,\quad \lambda y + 4\lambda^{2}-\lambda y=0, from which (the second of these), \lambda (and therefore) z are both zero.
    Substitution of these values into the first and fourth equations produces xy=A and xy=0 implying that A=0. That leaves us with xy=0 from which one or the other is zero and the other is indeterminate.
    That gives us two solution sets for (x,y,z,\lambda), either (\mbox{indeterminate},0,0,0) or (0,\mbox{indeterminate},0,0).

    Moving to the other possibility x=0, substitution into the first equation implies that A=0 (again) while the second equation remains as is. (The fourth equation becomes 0=0).
    For that second equation you can let any two of y,z,\mbox{ or }\lambda take arbitrary (independent) values and solve for the third in terms of these.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. System of nonlinear equations
    Posted in the Math Software Forum
    Replies: 2
    Last Post: June 11th 2010, 06:29 AM
  2. Confused about nonlinear system of equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 28th 2009, 09:20 PM
  3. nonlinear system of equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 28th 2009, 09:18 PM
  4. solving a system of nonlinear equations
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: November 16th 2008, 11:39 AM
  5. Solver for a system of nonlinear equations
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: February 27th 2008, 07:38 PM

Search Tags


/mathhelpforum @mathhelpforum