we have the determinant:
$$
\begin{vmatrix}
6&1&1\\
4&-2&5\\
2&8&7\\
\end{vmatrix}
$$
A stumbled upon this question: Is the following matrix invertible?
If I apply the modulo 2 procedure to this matrix, I get a zero determinant, which is not true. Why doesn't that procedure work in this case?
Best Answer
The modulo $2$ idea does not work because this determinant, albeit nonzero, is divisible by $2$. (In fact, it is $-306$.) If something is nonzero modulo $2$, it is certainly nonzero, but if something is zero modulo $2$, this only means that it is even but doesn't mean it is zero.
You can still apply the same technique, but first divide the first column by $2$: you get the determinant:
$$\begin{vmatrix}3&1&1\\2&-2&5\\1&8&7\end{vmatrix}$$
which is, modulo $2$:
$$\begin{vmatrix}1&1&1\\0&0&1\\1&0&1\end{vmatrix}=1$$
so this new determinant is nonzero, and so the original determinant (which is twice the new determinant) must be nonzero as well.