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
$$R_{BA} =
begin{bmatrix}
1 & 0 & 0\
0 & cos(i) & sin(i)\
0 & -sin(i) & cos(i)
end{bmatrix}
begin{bmatrix}
cos(omega) & sin(omega) & 0\
-sin(omega) & cos(omega) & 0\
0 & 0 & 1
end{bmatrix}
=
begin{bmatrix}
cos(omega) & sin(omega) & 0\
-cos(i)sin(omega) & cos(i)cos(omega) & sin(i)\
sin(i)sin(omega) & -sin(i)cos(omega) & cos(i)
end{bmatrix}
$$

Thanks for your help!