Inverse of a scalar plus a cross-product operator ($\alpha \mathbf{x}+\beta \mathbf{n}\times \mathbf{x}=\mathbf{b}$)

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.

For any real three-dimensional vectors $a,b$, the displayed linear equation has the unique solutionTaking a dot product with $b$ first yields $(1+|b{|}^2)(b\cdot x)=a\cdot b$. Taking a cross product with $b$ and using the vector triple product identity givesSubstitution into the original equation cancels the longitudinal terms and proves the formula. Conversely direct substitution verifies it for all $b$, including $b=0$, since the denominator is positive. If $C_{b}x=b\times x$, the underlying identity for linear maps is $(I-C_b)(I+C_b+b{b}^T)=(1+|b{|}^2)I$, using $C_b^2=b{b}^T-|b{|}^{2}I$.