System of equations for Lagrange multipliers

Question

I'm looking for all the critical points of $f(x,y,z) = 2xy + 2yz -2x^2 -2y^2 - 2z^2$ constrained to the surface of the unit sphere in $\mathbb{R}^3$. Thus far I've set up the Lagrange multiplier: $$\begin{bmatrix}2y-4x & 2x+2z-4y & 2y-4z \end{bmatrix} = \lambda\begin{bmatrix}2x&2y&2z\end{bmatrix}$$ So I have the system of equations: \begin{align*} y-2x&=\lambda x \\ x+z-2y&=\lambda y\\ y-2z&=\lambda z\\ x^2 + y^2 + z^2 &=1 \end{align*} So far I have thought of the case where $\lambda = 0$, which leads me to: \begin{align*} y&=2x \\ x+z&=2y\\ y&=2z\\ x^2 + y^2 + z^2 &=1 \end{align*} which appears to only have a solution $(x,y,z) = (0,0,0)$, but that violates the constraint. Is this reasoning sound? How would I go about solving this system in the case that $\lambda \neq 0$? My fallback thought process was to try to use Newton's method, but that seems very tedious.

Answer 1

The first three equations form a homogeneous system of three linear equations in three variables depending on a $\lambda$. If that system has exactly one solution, then that solution is $(0,0,0)$, which doesn't satisfy the constraint. So, take the matrix of the coefficients of the system, compute its determinant and work only with those $\lambda$'s for which that determinant is $0$.