I may well be reading you wrong, but this is my interpretation.
$${\bf X}_i \gets \begin{bmatrix} {\bf X}_i, & {\bf X}_i^2, & \log({\bf X}_i + {\bf I}), & ({\bf I} + \log {\bf X}_i)^{-1}, & {\bf 1} \end{bmatrix}$$
If this is the right interpretation, then many will understand and accept this. I am not sure how to understand $\log$ of a matrix which is not square, though.
EDIT: Since the operations are componentwise, I think it safer to do this
$${\bf X} \gets \begin{bmatrix} {\bf X}, & {\bf A}, & {\bf B}, & {\bf C}, & {\bf d} \end{bmatrix},$$
where
$$ a_{ij} = x_{ij}^2, \quad b_{ij} = \log(1 + x_{ij}), \quad c_{ij} = b_{ij}^{-1}, \quad d_i = 1.$$
If an index is absolutely necessary for the matrix ${\bf X}$, then let it be a superscipt, i.e. ${\bf X}^{(k)}$ and $x_{ij}^{(k)}$.
There is a group of people who will accept MATLAB notation in algorithms, but better safe than sorry.