TITLE: How many $3 \times 3$ integer matrices are orthogonal?
QUESTION [9 upvotes]: 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.  
Source :- Tata Institute of Fundamental Research Graduate School Admissions 2016

REPLY [3 votes]: 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 $A$.