Let $E=\mathbb{R}^3$ an euclidean space and let the following matrix :
$$\dfrac{1}{9}\cdot\begin{pmatrix}8&1&4\\-4&4&7\\1&8&-4\end{pmatrix}$$
Since $AA^T=I_3$ I conclude that $A\in O(E)$, $\det A$ gives $-1$, so $A\in O(E)\backslash SO(E)$
It seems to be a combination of one rotation and one symmetry, but I don't know how to find the rotation axe.
In general if $r$ is a rotation and $s$ is a symmetry, I don't know how to study $r\circ s$
Your matrix has one real eigenvalue $-1$ and two non-real complex conjugate eigenvalues. An eigenvector $v$ for eigenvalue $-1$, i.e. a vector in the null space of $A + I$, is transformed to $-v$ by $A$, while the plane orthogonal to that vector is rotated by $A$. Thus $A$ consists of reflection across this plane together with a rotation around the axis $v$.