Find the value of $s$ for which the matrix is positive definite I am given the following Matrix 
$$
A=        \begin{pmatrix}
        s & -4 & -4 \\
        -4 & s & -4 \\
        -4 & -4 & s \\
        \end{pmatrix}
$$
and asked to find $s$ such that $A$ is positive definite. In finding it, I simply set the determinant to $0$ and solved for $s$, and in doing so, I got $0 = s^2-4s-32$ which solves to $s=-4,8$. I'm not sure whether this is the correct process though, and if I use either value for $s$ I have at least one $0$ for each eigenvalue of $A$. Am I on the right track? Thanks in advance. 

You could also approach it like this: A symmetric matrix is postive definite iff all eigenvalues are positive.
Take $s=-4$ and we have a rank $1$-matrix $A = \begin{pmatrix}-4&-4&-4\\-4&-4&-4\\-4&-4&-4\end{pmatrix}$. Such a matrix has two different eigenvalues: $0$ and the trace, which is $-12$.
For arbitrary $s$ our matrix is of the form $(s+4)\operatorname{Id}+A$, hence the eigenvalues are $s+4$ and $s+4-12=s-8$. This shows that the matrix is positive definite iff $s > 8$.
