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}


\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

  • 1
    $\begingroup$ 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. $\endgroup$
    – Dave
    Dec 1, 2016 at 12:51
  • 2
    $\begingroup$ 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. $\endgroup$ Dec 1, 2016 at 16:12
  • $\begingroup$ @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. $\endgroup$
    – Lelouch
    Dec 1, 2016 at 16:15

2 Answers 2


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.

But rotation matrices in right handed frame (if you would want only them in the future) we have only 24, because the third column is always calculated as the cross product of 1 and 2 column.


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 $A$.

  • $\begingroup$ 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? $\endgroup$
    – user464147
    Nov 11, 2017 at 10:20
  • $\begingroup$ I'll add that the question from the above comment was discussed here in chat. $\endgroup$ Nov 11, 2017 at 13:10

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