This question has been asked before, but in this case, we should do a little bit different thing.
Assume that we have this lower Hessenberg matrix.
$$A = \begin{bmatrix} CB & 0 &0 & 0 & 0\\ CAB & CB & 0& 0 &0 \\ CA^2B &CAB & CB & 0 & 0\\ CA^3B & CA^2B &CAB & CB & 0\\ CA^4B & CA^3B & CA^2B & CAB & CB \end{bmatrix}$$
Where $CB \in \Re^{n x n}$ and $CA^iB \in \Re^{n x n}$ and $n = 1$
We want to solve $Ax = b$, classic. But what if we have $n > 1$ ? E.g matrix $A$ looks like this:
0.07741 0.23210 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
0.23865 123.33000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
0.12238 29.62902 0.07741 0.23210 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
-0.00611 15.82693 0.23865 123.33000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000
0.10198 28.81903 0.12238 29.62902 0.07741 0.23210 0.00000 0.00000 0.00000 0.00000
-0.05920 -12.09672 -0.00611 15.82693 0.23865 123.33000 0.00000 0.00000 0.00000 0.00000
0.07206 21.47042 0.10198 28.81903 0.12238 29.62902 0.07741 0.23210 0.00000 0.00000
-0.05632 -15.31863 -0.05920 -12.09672 -0.00611 15.82693 0.23865 123.33000 0.00000 0.00000
0.04746 14.49060 0.07206 21.47042 0.10198 28.81903 0.12238 29.62902 0.07741 0.23210
-0.04167 -12.22749 -0.05632 -15.31863 -0.05920 -12.09672 -0.00611 15.82693 0.23865 123.33000
Here we can see that the dimension and column dimension is $n=2$. Solving if $n=1$ is easy. Then I use Gaussian Elimination.
But how should I do in this case when $n > 1$ and $A$ is still lower traingular hessenbeg matrix? What is the easiest way in this case when $n > 1$?
If you have a system like this you can use forward substitution, $Ax=b$ then
$$x_{1} = \frac{b_{1}}{a_{1,1}} $$
and
$$ x_{i} = \frac{b_{i} - \sum_{j=1}^{i-1} a_{i,j} x_{j} }{ a_{i,i} }$$