Let $L$ denotes the set of all primes $p$ such that the following matrix is invertible I am stuck  on the following problem:  

Let $L$ denotes the set of all primes $p$ such that the following matrix is invertible when considered as a matrix with entries in $\Bbb Z/p \Bbb Z$ .   
$A=\begin{pmatrix}
1 &2  &0 \\ 
0 &3  &-1 \\ 
-2 &0  &2 
\end{pmatrix}$ 

Then how can I verify whether the following statements are true/false?   

  
*
  
*$L$ contains all the prime numbers greater than $10$    
  
*$L$ contains all the prime numbers other than $2$ and $5$   
  
*$L$ contains all the prime numbers  
  
*$L$ contains all the odd prime numbers. 

Can someone give explanation?  Thanks in advance for your time.
 A: Hint: a square matrix over any field is invertible iff its determinant is nonzero (as a member of the field).
A: We can compute the determinant of the matrix, finding that
\begin{align}
\det{A} &= 1 \det \left(\begin{array}{c} 3 & -1 \\ 0 & 2\end{array}\right) - 2 \det \left(
\begin{array}{c} 0 & -1 \\ -2 & 2\end{array}\right) \\
&= 1 \Big(3 \cdot 2 - (-1) \cdot 0\Big) - 2 \Big(0\cdot 2 - (-1)\cdot(-2)\Big) \\
&= 6 - 2 \cdot (-2) \\
&= 10
\end{align}
Note that it is an ever-so-slight abuse of notation to refer to things such as "$10$," since field elements are really equivalence classes modulo $p$ - but we can always reduce modulo $p$, so it's okay. Note that the symbol $10$ really refers to
$$10 = \underbrace{1 + 1 + ... + 1}_{\text{$10$ times}}$$
So as long as this is not $0$ in the field, the matrix is invertible.

In particular, we know that $10 = 0$ in $\Bbb{Z} / \Bbb{Z}_p$ if and only if $p | (10 - 0) = 10$; the only primes satisfying this are $2$ and $5$. So the final answer is
$$
\textbf{The matrix is invertible $\iff$ $p \notin \{2, 5\}$}
$$

Another way to see this is that if $p =3$, then $10 = 1 \ne 0$, and if $p = 7$, then $10 = 3 \ne 0$. On the other hand, if $p > 10$, it is clear that $10 \ne 0$.
