Find three real orthogonal matrices of order $3$ having all integer entries. 
Find three real orthogonal matrices of order $3$ having all integer entries.

I have no idea to solve the problem. I don't know how to start. If $A$ be such matrix then $AA^T=A^TA=I_3$. Please help me.
 A: Let $A$ be such a matrix, and $a_1$, $a_2$ and $a_3$ its columns. Given that $A$ is orthogonal means that 
$$\langle a_i,a_j\rangle =\left\{\begin{array}{ll}1&\text{ if } i=j\\0&\text{ if }i\neq j\end{array}\right.$$
So in particular $\langle a_i,a_i\rangle=1$. Because the $a_i$ have all integer coefficients, it follows that
$$a_1,a_2,a_3\in\{(1,0,0),(-1,0,0),(0,1,0),(0,-1,0),(0,0,1),(0,0,-1)\},$$
leaving six options for each of $a_1$, $a_2$ and $a_3$. For any given value of $a_1$ there are precisely $4$ values of $a_2$ such that $\langle a_1,a_2\rangle=0$. For any given values of $a_1$ and $a_2$ there are precisely $2$ values of $a_3$ such that $\langle a_1,a_3\rangle=\langle a_2,a_3\rangle=0$. Hence there are $6\times4\times2=48$ such matrices in total.
A: Any three distinct $3\times 3$ permutation matrices would work.
A: If $a,b,c$ are the entries of the first row of a matrix, then $a^2+b^2+c^2$ should be equal to one. this equation has the following integer solutions : (1,0,0),(0,1,0),(0,0,1),(-1,0,0),(0,-1,0),(0,0,-1). Thus, each row of the desired matrix should be filled with exactly one of these solutions. So, you have 120 possible matrices.
