1:02 PM
Why did he select only ±e type columns? will it eliminate all the possibilities? Can you please help me, How did you do the permutation calculations?
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Let $S$ be the set of all $3 \times 3$ matrices $A$ with integer entries such that the product $AA^t$ is the identity matrix. Then $|S|$ is : 23. 24 48 60 Answer is $48$ but I don't know how to solve it.

@ManeeshNarayanan You noticed that there is a requirement that the entries have to be integers right?

> 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).
Well, so if you have a vector $(a,b,c)$ where all coordinates are integers, then you get $$a^2+b^2+c^2=1.$$
If you add a condition that a,b,c have to be integers, what possibilities do you have for them?

yes. it is clear now. what about the permutation part?

You mean this?
> 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.
You agree that you see that columns can only be $\pm e_1$, $\pm e_2$, $\pm e_3$.

1:07 PM
yes. this much clear. How to calculate the number?'
yes
we need to consider independence also. right?

So it is clear that if we ignore signs then the columns are some permutation of $e_1$, $e_2$, $e_3$. (Simply because we cannot have any of them twice.)
Well, orthogonal vectors are automatically linearly independent.

> This results in a total of $3! \cdot 2^3 = 6 \cdot 8 = 48$ options for $A$.
You have $3!$ permutations of three vectors.
In each of the three spots you can choose one of two possible signs.

In each of the three spots you can choose one of two possible signs. that gives $2^3$. right?

Yes.
BTW good catch that there is a duplicate. I have added a comment below yours linking to this chat. (So that people seeing your comment know that you already have an answer.)

1:12 PM
Now it is clear. Thank you.