Finding the exterma of $x^2+y^2+z^2-yz-zx-xy$ s.t. $x^2+y^2+z^2-2x+2y+6z+9=0$ using Lagrange's multiplier, 
Using Lagrange's multiplier method, obtain the maxima and minima of $$x^2+y^2+z^2-yz-zx-xy$$ subject to the condition $$x^2+y^2+z^2-2x+2y+6z+9=0$$

My attempt:
I formed the expression
$$F=x^2+y^2+z^2-yz-zx-xy+\lambda(x^2+y^2+z^2-2x+2y+6z+9)=0$$
Differentiated partially wrt $x$, $y$ and $z$, and equated to $0$.
I get the following equations
$$(2\lambda+2)x-y-z=2\lambda$$
$$-x+(2\lambda+2)y-z=-2\lambda$$
$$-x-y+(2\lambda+2)z=6\lambda$$
In matrix form:
$$\begin{bmatrix} 2\lambda+2 & -1 & -1 \\-1 & 2\lambda+2 & -1 \\ -1 & -1 & 2\lambda+2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\lambda \begin{bmatrix}2 \\ -2 \\ -6 \end{bmatrix} $$
When I try to find $x$, $y$ and $z$ from these equations, it gets complex. Is there an easier way to ascertain their values?
 A: You solve the system and get
$\left[x= -\dfrac{3-2 \lambda}{2 \lambda+3},\;y= -1,\;z= -\dfrac{3 (2 \lambda+1)}{2 \lambda+3}\right]$
then you use the constraint
$x^2+y^2+z^2-2x+2y+6z+9=0$
and substitute. Simplify
$(2 \lambda-3) (2 \lambda+9)=0$
$\lambda=\dfrac{3}{2};\;\lambda=-\dfrac{9}{2}$
Substitute again in the solutions and get
$x= 2,\;y= -1,\;z= -4$ we have a maximum $27$
and
$x = 0,\; y = -1,\; z = -2$ we have a minimum $3$
hope this helps
A: I get the following equations
$$2x-y-z=\lambda(2x-2) \\-x+2y-z=\lambda(2y+2) \\-x-y+2z=\lambda(2z+6) $$
or equivalently
$$\begin{bmatrix} 2-2\lambda & -1 & -1 \\-1 & 2-2\lambda & -1 \\ -1 & -1 & 2-2\lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\lambda \begin{bmatrix}-2 \\ 2 \\ 6 \end{bmatrix} $$
If the matrix is invertible, there is only one solution $(x,y,z)$. Otherwise, its determinant is zero, and that gives you the values of $\lambda$.
A: You already got the system of equations. So the solution simply follows from there.
You should get $(x,y,z)=(2,-1,-4)$ and $(x,y,z)=(0,-1,-2)$ which gives you the maximum (27) and the minimum (3), respectively. 
