Identify geometrically and find the inverse of the linear map defined by $\mathbf x \mapsto\mathbf x'=a\mathbf x +b(\mathbf{n\times x})$ 
Let $\mathcal M:\mathbb R^3 \rightarrow \mathbb R^3$ be the linear map
  defined by $$\mathbf x \mapsto\mathbf x'=a\mathbf x +b(\mathbf{n\times
 x})$$ where $a$ and $b$ are positive scalar constants and $\mathbf n$
  is a unit vector.
$(i)$ By considering the effect of  $\mathcal M$ on $\mathbf n$ and on
  a vector orthogonal to $\mathbf n,$ describe geometrically the action
  of $\mathcal M$.
$(ii)$
   Find, in the general case, the inverse map.

For
$(i)$
I feel as though I should know immediately but I am actually struggling to see anything, I've done as the question says but it doesn't resemble anything to me and I am struggling to find anything on it online.
For
$(ii)$
I think I have, for
$M$,
the matrix of
$\mathcal M, M_{ij}=a\delta_{ij}+b\epsilon_{ipj}n_p$
and I'm looking to find the inverse of this, but I'm not sure how to.
Thank you
 A: Hint: Note that 
$$
\newcommand{\mb}[1]{\mathbf{#1}}
\mathbf n \mapsto a \mathbf n
$$
On the other hand: we can take $\mb e_1$ to be a unit vector orthogonal to $\mathbf n$ and define $\mb e_2 = \mathbf n \times \mb e_1$.  We then have
$$
\mb e_1 \mapsto a \mb e_1 + b \mb e_2\\
\mb e_2 \mapsto a \mb e_2 - b \mb e_1
$$
The matrix of the transformation with respect to $\{\mb{n,e_1,e_2}\}$ is
$$
\pmatrix{a&0&0\\0&a&b\\0&-b&a}
$$
You may notice that this looks a lot like a rotation (although that's not quite what it is).  We can write this matrix as the composition of two transformations that are easy to understand if we write
$$
\pmatrix{a&0&0\\0&a&b\\0&-b&a} = 
\pmatrix{a&0&0\\0&\sqrt{a^2 + b^2}&0\\0&0&\sqrt{a^2 + b^2}}
\pmatrix{1&0&0\\0&\cos \theta &\sin \theta \\0&-\sin \theta&\cos \theta}
$$
where $\tan \theta = b/a$.
A: Hint. Choosing $\mathbf{m}$ orthogonal to $\mathbf{n}$ and unitary then, $B=\{\mathbf{n},\mathbf{m},\mathbf{m}\times \mathbf{n}\}$ is an orthonormal basis of $\mathbb{R}^3.$ Then,
$$\begin{cases}\mathcal{M}(\mathbf{n)}=a\;\mathbf{n}\\\mathcal{M}(\mathbf{m})=a\;\mathbf{m}+b\;(\mathbf{m}\times \mathbf{n}) \\
\mathcal{M}(\mathbf{m}\times \mathbf{n})=-b\;\mathbf{m}+a\;(\mathbf{m}\times \mathbf{n})\end{cases}\Rightarrow [\mathcal{M}]_{B}=\begin{bmatrix}{a}&{0}&{\;\;0}\\{0}&{a}&{-b}\\{0}&{b}&{\;\;a}\end{bmatrix}.$$
