= Solution
Expanding the formula from part (a) in components gives $y_i=x_i+(\alpha-1)n_i n_jx_j$, with repeated indices summed. Hence the <matrix of a linear map> and its inverse, when $\alpha\ne0$, have entries
$$
\boxed{A_{ij}=\delta_{ij}+(\alpha-1)n_i n_j,\qquad
B_{ij}=\delta_{ij}+(\alpha^{-1}-1)n_i n_j.}
$$
Here $\delta_{ij}$ is the <Kronecker delta>. To check the <matrix inverse>, write $P=nn^T$; $P^2=P$ because $|n|=1$. Thus $A=(I-P)+\alpha P$ and $B=(I-P)+\alpha^{-1}P$, whose product is the <identity matrix>.
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