 Inverse of a scalar plus a cross-product operator

For a unit vector $\mathbf{n}$, real $\alpha \ne 0$, and real $\beta$, the displayed linear equation has the unique vector solutionTo prove it, split vectors into components parallel and orthogonal to $\mathbf{n}$. On the parallel line, the linear map is multiplication by $\alpha$. On the perpendicular plane, the operator $N\mathbf{v}=\mathbf{n}\times \mathbf{v}$ obeys $N^2=-I$, by the vector triple product. Therefore $(\alpha I-\beta N)(\alpha I+\beta N)=({\alpha }^2+{\beta }^2)I$ on that plane. Combining its inverse with the parallel solution gives the formula. This avoids component calculations in equations with an axial cross product term.