# How many $3 \times 3$ integer matrices are orthogonal?

Let $S$ be the set of $3 \times 3$ matrices $\rm A$ with integer entries such that $$\rm AA^{\top} = I_3$$ What is $|S|$ (cardinality of $S$)?

The answer is supposed to be 48. Here is my proof and I wish to know if it is correct.

So, I am going to exploit the fact that the matrix A in a set will be orthognal, so if the matrix is of the form \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix}

Then each column and row will have exactly one non-zero element which will be +1 or -1. Thus, I have split possibilities for the first column into three cases and counted the possibilities in each case as follows :- $$a_{11} \neq 0$$ or $$a_{21} \neq 0$$ or $$a_{31} \neq 0$$

In case 1), we obviously have two possibilities(+1 or -1) so we consider the one where the entry is +1. Now, notice that the moment we choose the next non-zero entry, all the places for non-zero entries will be decided because of the rule 'each column and row will have exactly one non-zero element'. Meaning, if b and c are remaining two non-zero entries, we only have two possibilities left \begin{bmatrix} 1 & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \\ \end{bmatrix}

or

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & c \\ 0 & b & 0 \\ \end{bmatrix}

Using the fact that b and c are simply $$\pm1$$

In each of the above matrices, we get 4 possibilities for each of the matricies. Thus, 8 possibilities in totality. Basically, we are getting 8 possibilities on the assumption that $$a_{11} = 1$$

Thus, we get 16 possibilities on the case that $$a_{11} \neq 0$$

Following, the second and third cases analogously, we get a total of 16 possibilities in each of them and 48 possibilities in total.

• That looks good to me. The only other way I would think to do it is with binomial coefficients, but it is essentially the same as what you've done here. – Dave Dec 1 '16 at 12:51
• The simplest argument, in my opinion: You have $3!=6$ ways of permuting the columns of the identity matrix, and you multiply this by the $2^3=8$ possible ways of choosing the sign (plus or minus) for the $1$ in each column. – Hans Lundmark Dec 1 '16 at 16:12
• @HansLundmark :- I believe that's exactly what I did but I thank you for making it look so much simpler rather than what it does now! Edit : - I wasn't being sarcastic, just in case it appeared so. – Lelouch Dec 1 '16 at 16:15

Dot product of columns can be used (it must be $0$) The first column of matrix $A$ has 6=2*3 possibilities for location 1 or -1, after location 1 or -1 the rest of entries must be zeros and they give possibilities for the second column - we have here only 2*2 = 4 possible choices for location 1 or -1, and the third column stays with only 2.
Note that each column of $A$ must be an integer vector of unit length which means that each column is of the form $\pm e_i$ for some $1 \leq i \leq 3$ (where $e_i$ are the standard basis vectors). Thus, we need to pick a permutation of the $e_i$'s to put as columns and then, for each column independently, decide whether it gets a plus or a minus sign. This results in a total of $3! \cdot 2^3 = 6 \cdot 8 = 48$ options for $A$.
• Why did you select only $\pm e$ type columns? will it eliminate all the possibilities? Can you please help me, How did you do the permutation calculations? – Unknown x Nov 11 '17 at 10:20