Given two matrices $$A = \begin{pmatrix} 3 & -\frac{1}{2} & 0 \\ -\frac{1}{2} & 1 & -\frac{1}{2} \\ 0 & -\frac{1}{2} & 1 \\ \end{pmatrix}, \qquad D = \begin{pmatrix} \alpha & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}, \quad\alpha \neq 0$$
I'm first asked to bound the eigenvalues of $\mathrm{A}$. And then using the fact that $\mathrm{D^{-1}AD}$ has the same eigenvalues as A finding a better bound for the eigenvalues of $\mathrm{A}$. I did this also with applying Gershgorin's theorem on $\mathrm{D^{-1}AD}$. Finally I'm asked to conclude that $\mathrm{A}$ is nonsingular.
I don't know why I can conclude that $\mathrm{A}$ is nonsingular after applying Gershgorin's theorem on $\mathrm{D^{-1}AD}$. I know that a strictly diagonally dominant matrix is nonsingular and positive definite, but this is not the case for $\mathrm{A}$.
Hint: Define $$ A(\alpha) = [D(\alpha)]^{-1}AD(\alpha) = \pmatrix{3 & -1/2\alpha & 0\\-\alpha/2 & 1 & -1/2\\0 & -1/2 & 1} $$ Find a value of $\alpha$ for which $A(\alpha)$ is diagonally dominant.