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.

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