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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

  1. Singular
  2. Non-Singular
  3. Symmetric
  4. Skew-symmetric
  5. Hermitian
  6. Skew-Hermitian
  7. Orthogonal is?
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@hardmath, you're right ($12^9$ is correct!) post was more an expanded comment, anyway, hence deleted. – amWhy Jan 28 '13 at 15:02
What is $\iota$? – Harald Hanche-Olsen Jan 28 '13 at 15:13
@Harald: As was clarified in a comment on the deleted answer, it is supposed to be imaginary unit $i$. – hardmath Jan 28 '13 at 15:19
Okay. Seems to me that #1 is very hard, #2 is trivially answered by #1, and all the others are simple counting exercises. The answer to #7 is obviously zero. – Harald Hanche-Olsen Jan 28 '13 at 15:49
The answer to #7 is zero because no two rows (or columns) with all positive entries can be orthogonal. (Also, an orthogonal matrix has only real entries by definition.) – Harald Hanche-Olsen Jan 28 '13 at 16:52

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.

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1 and 2 are definitely equivalent problems, but can they be proved independently of each other ? – wamiq reyaz Jan 29 '13 at 3:03
As I don't have a proof for any one of them, I can't say whether they can be proved independently, not to mention that what is meant by "independent" is ambiguous here. – user1551 Jan 29 '13 at 3:22

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.

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