Lagrange, version C
The question says:
$ \Re_1 $ is obtained from $ \Re_0 $
using the following transforms:
1. a rotation $ Rot(i, 90) $ with respect to the fixed axes
We will call this HTM $ T_a $:
$ T_a = \begin{bmatrix} 1 & 0 & 0 & 0 \ 0&0&-1&0\ 0&1&0&0\ 0&0&0&1 \end{bmatrix} $
2. a translation $ \begin{bmatrix} 2&1&2\end{bmatrix}^T $ with respect to the fixed axes
We will call this HTM $ T_b $:
$ T_b = \begin{bmatrix} 1 & 0 & 0 & 2 \ 0&1&0&1\ 0&0&1&2\ 0&0&0&1 \end{bmatrix} $
3. a rotation $ Rot(j, 90) $ with respect to the mobile axes
We will call this HTM $ T_c $:
$ T_c = \begin{bmatrix} 0 & 0 & 1 & 0 \ 0&1&0&0\ -1&0&0&0\ 0&0&0&1 \end{bmatrix} $
We add a new frame of reference $ \Re_i $ that we consider as a zero frame. The given HTM matrix $ T_0 $ is the transform between $ \Re_0 $ and $ \Re_i$. In order to be clear we will rename $ T_0 $ as $ T_0^i $
Since $ \Re_1 $ is obtained from $ \Re_0 $, the $ T_0^i $ should not be involved. Indeed, even if the position or rotation of $ \Re_0 $ was different, the $ T_1^0 $ should be the same.
The given solution says that
$ T_1^0 = T_bT_a T_0^i * T_c$
This may be incorrect. The right HTM should be
$ T_1^0 = T_bT_a T_c$
In this last case
$ T_1^0 = \begin{bmatrix} 0 & 0 & 1 & 2 \ 1&0&0&1\ 0&1&0&2\ 0&0&0&1 \end{bmatrix} $
It should be noted that $ T_0^1 $ has a column $ t = \begin{bmatrix} 2 \1 \ 2\end{bmatrix}$ equal to the translation of step 2. This is ok since that was the only translation, and it was made w.r.t the fixed frame.