# System of nonlinear equations

• Oct 6th 2010, 07:13 PM
Pinkk
System of nonlinear equations
So I have that $\displaystyle xy + 4xz - A = 0, \lambda y + 4\lambda z - yz = 0, \lambda x - xz = 0, 4\lambda x - xy = 0$ and I want to solve for $\displaystyle x,y,z,\lambda$ in terms of $\displaystyle A$, 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 $\displaystyle z = \lambda$, which screws up my calculations (at least the way that I attempted)
• Oct 7th 2010, 01:43 AM
Ackbeet
Well, the equation $\displaystyle \lambda x-xz=0=x(\lambda-z)$ has only two possibilities: $\displaystyle x=0$ or $\displaystyle \lambda=z.$ 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.
• Oct 7th 2010, 05:19 AM
BobP
The third equation will factorize as $\displaystyle x(\lambda-z)=0$ implying that either $\displaystyle x=0$, or $\displaystyle z=\lambda.$

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

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