problem of positive semi-definite matrix Let $a$ be a real number and consider the square matrix of order 3, 
$$ N=\left[
\begin{array}{ccc}
  a&1&1\\
  1&a&1\\
  1&1&a
\end{array}
\right] $$
Determine all values of $a$ for which:
$(a)$ The matrix $N$ is invertible.
$(b)$ For every $v \in \mathbb R^3$, $v^TNv\ge0.$
please check if the solution is ok, if there is an error please correct, thanks
Solution :
$a)$ We show for what values of $a$ the determinant of N is zero,
$$\begin{vmatrix}
  a&1&1\\
  1&a&1\\
  1&1&a
\end{vmatrix}=0
 $$
$\Rightarrow \quad$  $(a-1)^2(a+2)=0$  $\quad \Rightarrow \quad$  $a=1\quad $ or $\quad a=-2$.
therefore  $N$ is invertible for all $a\in \mathbb R -\{1,-2\} $ 
$b)$ we show for what values of $a$, the matrix N is positive semi-definite.
$i)$ $a\ge0$ $\quad ($in particular for $\quad v = e_i, \quad i=1,2,3 \quad \Rightarrow $, ${e_i}^TNe_i\ge0.$)
$ii) $all the eigenvalues of N are greater than or equal to zero.
$\quad $the characteristic polynomial of N is given by,
$\quad p(\lambda)=det(N-\lambda I)$
$$\begin{vmatrix}
  a-\lambda&1&1\\
  1&a-\lambda&1\\
  1&1&a-\lambda
\end{vmatrix}=0
 $$
$\quad \Rightarrow \quad $  $(a-\lambda-1)^2(a-\lambda+2)=0$  $\quad \Rightarrow \quad$  $a-\lambda-1=0\quad $ or $\quad a-\lambda+2=0$.
$\quad \Rightarrow \quad $ $\lambda =a-1\quad $ or $\quad \lambda = a+2$
as the eigenvalues must be non-negative so that the matrix N is positive semi-definite.
$\quad \Rightarrow \quad $  $a\ge 1 \quad$ or $\quad a\ge -2  $
then of $(i)$ and $(ii)$we have
$a\ge 0$, $a\ge 1$ and $a\ge -2$.
therefore
$a\ge 1$.
please check if the solution is ok, if there is an error please correct, thanks

Your solution seems fine. As mentioned in one comment though, the first part of your writings at $(b)$ isn't needed, as all you need to do is find the cases for which the eigenvalues are non-negative.
