A nifty trick I learned yesterday: suppose you have a big matrix with integer elements, like this one:

$M= \begin{bmatrix} 23 & 0 & 3 & 78 & 56\\ 12& 17 & 16& 20 & 100\\ 22& 14 & 111 & 1 & 15\\ 90& 32 & 54 & 29 & 12\\ 22 & 18 & 10 & 94 & 7\\ \end{bmatrix}$

and you want to prove that it’s invertible. What do you do? You consider the same matrix modulo 2:

$M \equiv \begin{bmatrix} 1 & 0 & 1 & 0 & 0\\ 0& 1 & 0& 0 & 0\\ 0& 0 & 1 & 1 & 1\\ 0& 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{bmatrix} (mod 2)$

now you can see that it’s triangular and therefore $det(M) \equiv 1 (mod 2)$, so the determinant is definitely non-zero, and $M$ really is invertible. If it turned out that $det(M) \equiv 0 (mod 2)$, it wouldn’t imply that $M$ is degenerate, it would mean that the results are inconclusive, and you could, for example, try doing the same thing modulo 3, and so on.