# linear algebra – Rotation Matrix Equivalences? Given the following rotation matrix:

$$R_{BA} = begin{bmatrix} 0.2362 & 0.9412 & 0.2414\ 0.7558 & -0.3341 & 0.5631\ 0.6107 & 0.0494 & -0.7903 end{bmatrix}$$

Is it possible to reconstruct the above transformation with only two rotations?

Thoughts:

My intuition tells me no simply because there is no entry that contains a $$0$$ value so the rotations had to have been a symmetric/asymmetric 3 rotation sequence (ie. 3-1-3, 3-2-1, etc) but how can I show this mathematically? Is it enough to simply construct a 2 rotation DCM and show there’s a $$0$$ value? For example:

3-1 rotation rotation would lead to the DCM
$$R_{BA} = begin{bmatrix} 1 & 0 & 0\ 0 & cos(i) & sin(i)\ 0 & -sin(i) & cos(i) end{bmatrix} begin{bmatrix} cos(omega) & sin(omega) & 0\ -sin(omega) & cos(omega) & 0\ 0 & 0 & 1 end{bmatrix} = begin{bmatrix} cos(omega) & sin(omega) & 0\ -cos(i)sin(omega) & cos(i)cos(omega) & sin(i)\ sin(i)sin(omega) & -sin(i)cos(omega) & cos(i) end{bmatrix}$$ 