How to find steady-state probabilities in a queuing system? Queuing system set
$$\begin{matrix}\text{intake intensity} & \lambda=26 \\
\text{service channels} & m=4\\
\text{service intensity} & \mu=8\\
\text{maximum queue size} &n=18\end{matrix}$$
It is required to

Draw up a graph of a Markov process, write down the Kolmogorov system
of equations and find the steady-state probabilities.

I understand what the queuing system graph looks like

But how can we find the probabilities?
Because this is a homogeneous ergodic process, we can solve it using the Kolmogorov equation by equating the derivative to 0.
$$\dot{p}_i(t)=\sum\limits_j p_j(t)\lambda_{ji}(t)-p_i(t)\sum\limits_j\lambda_{ij}(t),\qquad (i=1,2,3...)\\
\frac{\partial p_i(t)}{\partial t}=\dot{p}_i(t)=0$$
Then we get
$$\sum\limits_j p_j\lambda_{ji}-p_i\sum\limits_j\lambda_{ij}=0\\$$
In the tutorial I found the so called system of equations for n = 3
$$\begin{equation*}
 \begin{cases}
   p_2\lambda_{21}+p_3\lambda_{31}-p_1(\lambda_{12}+\lambda_{13})=0\\
   p_1\lambda_{12}+p_3\lambda_{32}-p_2(\lambda_{21}+\lambda_{23})=0\\
   p_1\lambda_{13}+p_2\lambda_{23}-p_3(\lambda_{31}+\lambda_{32})=0
 \end{cases}
\end{equation*}$$
$$M=\begin{pmatrix}
-(\lambda_{12}+\lambda_{13}) & \lambda_{21} & \lambda_{31} \\
\lambda_{12} & -(\lambda_{21}+\lambda_{23}) & \lambda_{32} \\
\lambda_{13} & \lambda_{23} & -(\lambda_{31}+\lambda_{32})
\end{pmatrix}, X=\begin{pmatrix}p_1\\p_2\\p_3\end{pmatrix}, B=\begin{pmatrix}0\\0\\0\end{pmatrix}$$
$$M = \Lambda^T-D\text{, where } D = diag(\sum\nolimits_j\lambda_{1j},\sum\nolimits_j\lambda_{2j},\sum\nolimits_j\lambda_{3j})\\ \text{- 
Diagonal matrix of the sums of the rows of the $\Lambda$ transition intensities matrix}$$
The solution will only give the proportions between $p_i$ we need to replace any string with a string of ones
$$\begin{equation*}
 \begin{cases}
   p_2\lambda_{21}+p_3\lambda_{31}-p_1(\lambda_{12}+\lambda_{13})=0\\
   p_1\lambda_{12}+p_3\lambda_{32}-p_2(\lambda_{21}+\lambda_{23})=0\\
   \textbf{p}_1+\textbf{p}_2+\textbf{p}_3=\textbf{1}
 \end{cases}
\end{equation*}$$
$$M_*=\begin{pmatrix}
-(\lambda_{12}+\lambda_{13}) & \lambda_{21} & \lambda_{31} \\
\lambda_{12} & -(\lambda_{21}+\lambda_{23}) & \lambda_{32} \\
\textbf{1} & \textbf{1} & \textbf{1}
\end{pmatrix}, B=\begin{pmatrix}0\\0\\\textbf{1}\end{pmatrix}$$
$$M_*\times X = B$$
Whence multiplying both sides on the left by the inverse matrix we get
$$X = M^{-1}_*\times B$$
But how to compose this system in our case? How many rows and columns do we have? And what are different lambdas equal to? We essentially have the same value and everywhere will be the same number or not?

As I understand it, the values of the matrix are derived from the queuing system graph.
Namely we have $m+n+1$ rows and $m+n+1$ columns since we also take cell with the zero state
$$ \Lambda = \begin{pmatrix} 0 & \lambda & 0 & 0 & \dots & 0 & 0\\
\mu & 0 & \lambda & 0 & \dots & 0 & 0 \\
0 & 2\mu & 0 & \lambda & \dots & 0 & 0\\
0 & 0 & 3\mu & 0 & \dots & 0 & 0\\
& & & & \dots & & \\
0 & 0 &0 & m\mu & 0 & \lambda & 0\\
0 & 0 &0 &0 & m\mu & 0 & \lambda \\
0 & 0 &0 &0 & 0 & m\mu & 0
\end{pmatrix} $$
In our case
$$ \Lambda_{22\times 22} = \begin{pmatrix} 0 & 26 & 0 & 0 & \dots & 0 & 0\\
1\cdot 8 & 0 & 26 & 0 & \dots & 0 & 0 \\
0 & 2\cdot 8 & 0 & 26 & \dots & 0 & 0\\
0 & 0 & 3\cdot 8 & 0 & \dots & 0 & 0\\
& & & & \dots & & \\
0 & 0 &0 & 4\cdot 8 & 0 & 26 & 0\\
0 & 0 &0 &0 & 4\cdot 8 & 0 & 26 \\
0 & 0 &0 &0 & 0 & 4\cdot 8 & 0
\end{pmatrix} $$
