An arbitrary rotation in space can be constructed through the Euler angles. Imagine an $xyz$ axes, rotate around $z$ by $\phi$ so that the new axes is $\xi\eta\zeta$. Next rotate around $\xi$ by $\theta$ so that the new axes is $\xi'\eta'\zeta'$. Next rotate around $\zeta'$ by $\psi$ so that the new set of axes is $x'y'z'$.
The matrices of transformations are given by,
where $\rm{D}$ is the first $z$ rotation, $\rm{C}$ is the second $\xi$ rotation, and $B$ is the last $\zeta'$ rotation. In combination it gives,
where $\rm{A} = \rm{BCD}$.
One problem I encountered but I'm not sure how to answer is,
The body set of axes (all the axes I defined above) can be related to the space set (axes fixed in space, i.e. $xyz$) in terms of Euler’s angles also (instead of the above sequence) by the following set of rotations:
(a) Rotation about the x-axis by an angle $\theta$.
(b) Rotation about the z'-axis by an angle $\psi$.
(c) Rotation about the old z-axis by an angle $\phi$.
Show that this sequence leads to the same elements of the matrix of transformation as the sequence of rotations above. One key thing to note is that it is not necessary to carry out the explicit multiplication of the rotation matrices.
Taking this problem at face value, I tried calculating $\rm{BCD}-\rm{DBC}$ and found that it is not the zero matrix. How should I do this?
Edit:
The initial vector at the $xyz$-axes is $x$, acting $\rm{D}$ on it produces $\xi = \rm{D} x$, then acting $\rm{C}$ on it produces $\xi' = \rm{C} \xi$. Lastly, acting $\rm{B}$ on it produces $x' = \rm{B} \xi'$. So the complete matrix transformation is $x' = \rm{BCD} x = \rm{A} x$.
These images are taken from the book Classical Mechanics 3rd Edition by Goldstein.