Number of matrices Let $A$ be a $3*3$ matrix consisting of twelve distinct elements $1,2,3,4,5,6,7,8,9, \iota, 2\iota, -\iota $. Therefore the total number of distinct matrices that can be formed are $ 12^{9} $. Then, the number of matrices that are


*

*Singular 

*Non-Singular 

*Symmetric 

*Skew-symmetric 

*Hermitian   

*Skew-Hermitian 

*Orthogonal
is?

 A: Items 3-6 are easy. For item 7, if you mean real orthogonal matrix, the answer is clearly zero, as the dot product of two entrywise positive vectors must be positive. If you mean complex orthogonal matrix, the answer happens to be zero, too, because a brute force search reveals that, for $z^Tz=1$, $z^T$ (up to permutation) must be $(\pm i, 1, 1), (2i,1,2)$ or $(2i, 2i, 3)$. If
$$
A=\begin{pmatrix}\pm i&1&1\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix},
$$
since $z^T=(\pm i,a_{21},a_{31})$ must also satisfy $z^Tz=1$, we must have $a_{21}=a_{31}=1$. By considering the second row, this further implies that $(a_{22},a_{23})=(\pm i,1)$ or $(2i,2)$ (up to permutation) and hence by considering the second and third columns, we conclude that $(a_{32},a_{33})=(1,\pm i)$ or $(2,2i)$. But then you may verify that $A$ is not complex orthogonal. For other two choices of the first row of $A$, you can refute the possibility in a similar manner.
Items 1 and 2 are equivalent problems. For a problem size of $12^9$, I think the best way to solve it is by brute force search.
A: According to the OEIS, the number of singular $3\times3$ matrices with entries taken from $\{{1,2,\dots,9\}}$ is 5902335. How this is affected by the availability of $\pm i$ and $2i$, I do not know, but in any event we now have a nontrivial lower bound. 
