# Thread: System of nonlinear equations

1. ## 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)

2. 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.

3. 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.