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I found a solution for the case I am considering, which is when $A$ is a $3{\times}3$ cross product matrix of vector $a \in \mathbb{R}^3$: $$A = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix}, ~~~~~ a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} .$$ Using the explicit equation for the inverse of a $3{\times}3$ matrix, the inverse can be derived as: $$( I + A )^{-1} = \frac{1}{1+\|a\|^2} ((1 + \|a\|^2)I - A + A^2) .$$ I'm not sure if this can be extended to more general cases.
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