For vectors a, b in ℝ³, define a matrix M(a, b) = [a]ₓ [b]ₓ + [a × b]ₓ.
Here, [⋅]ₓ is the cross-product matrix map, so for c in ℝ³,
M(a, b) c = [a]ₓ [b]ₓ c + [a × b]ₓ c = a × (b × c) + (a × b) × c.
I can show that M(a, b) is non-singular unless a⊥b or a∥b.
Is there a simple expression for M(a, b)⁻¹, the inverse matrix?
UPDATE - What I have so far:
$\def\a{\mathbf{a}} \def\b{\mathbf{b}}$ Alternate formula for $M(\a, \b)$:
$$ M(\a, \b) \;=\; 2 \b \a^T - \a \b^T - (\a\cdot\b)I. $$
Effect on selected vectors: $$ M(\a, \b)\, \a \;=\; 2 (\a\times\b) \times \a \;=\; 2 ((\a\cdot\a)\,\b - (\a\cdot\b)\, \a), $$ $$ M(\a, \b)\, \b \;=\; \;(\a\times\b) \times \b \;=\; \;((\a\cdot\b)\,\b - (\b\cdot\b)\, \a), $$ $$ M(\a, \b)\, \a\times\b \;=\; -(\a\cdot\b)\,\a\times\b. $$
Using this to assemble a matrix expression: $$ M(\a, \b) = U \begin{bmatrix} -2\,\a\cdot\b & -\b\cdot\b & 0 \\ 2\,\a\cdot\a & \a\cdot\b & 0 \\ 0 & 0 & -\a\cdot\b \end{bmatrix} U^{-1}, $$ with $U = \begin{bmatrix}\a,&\b,&\a\times\b\end{bmatrix}$ and $$ U^{-1} = \begin{bmatrix} -((\a\times\b)\times\b)^T \\ ((\a\times\b)\times\a)^T \\ (\a\times\b)^T \end{bmatrix} \,/\, \|\a\times\b\|^2. $$
The inverse of the matrix expression is then $$ M(\a,\b)^{-1} = U \begin{bmatrix} \a\cdot\b\,/\,D & \b\cdot\b\,/\,D & 0 \\ -2\,\a\cdot\a\,/\,D & -2\,\a\cdot\b\,/\,D & 0 \\ 0 & 0 & -1/(\a\cdot\b) \end{bmatrix} U^{-1}, $$ where $D = 2\|\a\times\b\|^2$.
It's getting pretty close to something usable. It looks like it might simplify well, fingers crossed.
UPDATE -
The best I can come up with is $$ M(\a,\b)^{-1} \;=\; \frac{(\a\cdot\b)(\a\b^T - 2\b\a^T) - 2(\a\times\b)(\a\times\b)^T} {2\,(\a\cdot\b)\,\|\a\times\b\|^2} $$