singular matrix in numerical I am trying to find all values for $\alpha$ and $\beta$ for which 
$$ A(\alpha, \beta)= 
\left[
\begin{matrix}
3&0&-2\\\alpha&3&2\\-2&2&\beta 
\end{matrix}
\right]
$$
is singular.
My understanding is that I need to find values $\beta$ and $\alpha$ for which $det(A)=0$
Therefore so,
$$\det(A) = 3
\begin{bmatrix}
3&2\\2&\beta
\end{bmatrix} - 0 + (-2)\left[
\begin{matrix}
\alpha&3\\-2&2
\end{matrix}
\right]\\= 3(3\beta - 4) - 2(2\alpha +6) \\ = 9\beta -12 -4\alpha -12$$
If my workings are correct, then what is $\beta,\alpha?$
 A: Yes, you are correct. But there is no unique answer. 
For each $\alpha $ and $\beta$ such that $9\beta -4\alpha =24$ the matrix is singular.
So for each $\alpha $ you take, you can calculate $\beta $ (it is a linear function of $\alpha$): $$\beta = {4\over 9}\alpha +{8\over 3}$$

If those are integer then we have some more restriction, say $4∣β$ and $3∣α$:
Proof: $$9\beta =4\alpha +24 = 4(\alpha+6)\implies 4\mid 9\beta \implies 4\mid \beta $$ so $\beta =4y$. Now we have $$9y-\alpha =6 \implies 3(3y-2)=\alpha \implies 3\mid \alpha$$ So $\alpha = 3x$. So, $x$ and $y$ are connected with formula $3y-x=2$ and now however you choose integer $y$ you can calculate integer $x$ (by again linear formula) $$ x= 3y-2$$ 
A: You want $\det(A)=0$, so you have to solve the linear diophantine equation
$$9b-4a=24\tag{1}$$
We should first find a solution to the linear diophantine equation
$$9b-4a=1$$
which we can easily see is solved by $(a,b)=(2,1)$. Thus, we have that $(a,b)=(48,24)$ solves $(1)$, and by Bézout's Identity, we may parametrize all solutions as
$$(a,b)=(48+9t,24+4t)$$
where $t\in\mathbb Z$.
