Is the matrix $A$ positive (negative) (semi-) definite? Given, $$A = \begin{bmatrix}
2 &-1  & -1\\ 
 -1&2  & -1\\ 
 -1&  -1& 2
\end{bmatrix}.$$
I want to see if the matrix $A$ positive (negative)  (semi-) definite.
Define the quadratic form as $Q(x)=x'Ax$. 
Let $x \in \mathbb{R}^{3}$, with $x \neq 0$.
So, $Q(x)=x'Ax = \begin{bmatrix}
x_{1} &x_{2}  &x_{3} 
\end{bmatrix} \begin{bmatrix}
2 &-1  & -1\\ 
 -1&2  & -1\\ 
 -1&  -1& 2
\end{bmatrix} \begin{bmatrix}
x_{1}\\x_{2} 
\\x_{3} 
\end{bmatrix}$.
After multiplying out the matrices I am left with $$Q(x) = 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{2} - x_{1}x_{3}-x_{2}x_{3}).$$
Not sure what I can do with this result. Any suggestions on how to proceed would be appreciated.

If you want to proceed with this solution, you should complete the square. It is important that you "complete one variable completely every time". We write
$$\begin{aligned}
x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3&=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}x_2^2+\frac{3}{4}x_3^2-\frac{3}{2}x_2x_3\\
&=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}\bigl(x_2-x_3\bigr)^2.
\end{aligned}
$$
Can you conclude from here?
