The number of $3\times 3$ non-singular matrices, with four entries as $1$ and all other entries as $0$, is:

A $3\times 3$ matrix with four 1s and five 0s.For non-singularity, no row or column can be all zeros.Possible configurations:One row has two 1s, two rows have one 1.Example:$\left(\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)$ has determinant $-1$.By permuting rows and columns, we can find many such matrices.There are $3$ ways to choose the row with two 1s, and $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$ ways to place them. The remaining two 1s must be in different rows and different columns from each other and the existing 1s to avoid singularity.Enumeration shows the number is quite high ($>7$).