I have seen on this site and in lectures the following condition:
$A$ is a $n\times n$ matrix. Then $\det A \not =0$ iff all of its diagonal elements are nonzero.
But if we consider the following example:
$A=\begin{pmatrix}2&3&9\\4&5&6\\1&0&0\end{pmatrix}$. Then certainly $\det A \not =0$, but nonetheless $A_{33}$ is zero. So has this condition been quoted improperly? Should it include reference to the fact that you should check whether it is possible to move the rows around (inside the determinant calculation) so that the criterion is satisfied?
I think I am missing something quite obvious here, and am probably being rather daft!
Yes, you are right and your exemple is correct. I suppose that what you have in mind is that an upper triangular matrix has determinant equal to $0$ if and only if there is some $0$ in the main diagonal.
And a much simpler example would be $\left[\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right]$.