$$Ax=b;\qquad b_k=\sum^n_{m=1}A_{k.m};\qquad A-\mathrm{HilbertMatrix}(funkcja\quad \mathbf{hilb}())$$
Also, $k = k + 1$ until it reaches $n$ value just like with $m$ value.
I've understood this equation as if $n = 3$, then we sum matrix $1\times1$ with matrix $2\times2$ and we sum it again with matrix $3\times3$.
I know that I can't sum matrix of different size.
Orignal image of the problem:
Let's start with an example: Suppose $$A = \begin{bmatrix}A_{11}' & A_{12}' & \ldots & A_{1m}' \\ A_{21}' & A_{22}' & \ldots & A_{2m}' \\ \vdots & \vdots& \ddots & \vdots \\ A_{n1}' & A_{m2}' & \ldots & A_{nm}' \end{bmatrix}; x =\begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_m\end{bmatrix} $$
Then we have: $$Ax = \begin{bmatrix}A_{11}'x_1 + A_{12}'x_2 + \ldots + A_{1m}'x_m \\ A_{21}'x_1 + A_{22}'x_2 + \ldots + A_{2m}'x_m \\ \vdots \\ A_{n1}'x_1 + A_{n2}'x_2 + \ldots + A_{nm}'x_m\end{bmatrix} = \begin{bmatrix}\displaystyle\sum_{i=1}^m A_{1i}'x_i \\ \displaystyle\sum_{i=1}^m A_{2i}'x_i \\ \vdots \\ \displaystyle\sum_{i=1}^m A_{ni}'x_i\end{bmatrix}$$ If $Ax = b$, then we have: $$b = \begin{bmatrix}\displaystyle\sum_{i=1}^m A_{1i}'x_i \\ \displaystyle\sum_{i=1}^m A_{2i}'x_i \\ \vdots \\ \displaystyle\sum_{i=1}^m A_{ni}'x_i\end{bmatrix} = \begin{bmatrix}b_1 \\ b_2 \\ \vdots \\ b_n\end{bmatrix}$$ So for row $k$ we can say: $$b_k = \sum_{i=1}^m A_{ki}'x_i$$ Defining $A_{ki} = A_{ki}' x_i$ gives us: $$b_k = \sum_{i=1}^m A_{ki}$$ The difference between this and what you have must be that you're dealing with a square matrix $A$ in which case $m = n$. Then you get the desired result (upto changing indices)