level $k$.

In this section, we aim to establish the following matrix identity on weighted Motzkin numbers.

Let $M=(m_{i,j})_{i,j\geq 1}$ be the lower triangular matrix such that the first column is the sequence of the number of $(k-t-1, t)$-Motzkin paths of length $n$ and $m_{i,j}$ satisfies the following recurrence relation for $j\geq 2$: $$\label{rc.5}
m_{i,j}=m_{i-1,j-1}+(k-t-1)m_{i-1,j}+tm_{i-1,j+1}.$$ Then we have $$\label{eq.5}
(m_{i,j})  \times
\begin{bmatrix} 1 \\ 1+t \\ 1+t+t^2  \\ \vdots\\
\end{bmatrix} = \begin{bmatrix} 1 \\ k \\ k^2  \\ \vdots
\end{bmatrix},$$

It is well known that the number of $2$-Motzkin paths of length $n$ is given by the Catalan number $C_{n+1}$. It follows that the matrix identity (\[eq2.2\]) is a special case of (\[eq.5\]) for $k=4,t=1$. Denote by $f(x)=\sum_{n\geq 0}f_{n}x^n$ the generating function for the number of $(3,2)$-Motzkin paths. Then it is easy to find the functional equation for $f(x)$: $$f(x)=1+3xf(x)+2x^2f^2(x).$$ It follows that $$f(x)={{1-3x-\sqrt{1-6x+x^2}}\over {4x^2}}.$$ From the above generating function, one see that the number of $(3,2)$-Motzkin paths of length $n$ equals the $n$-th little Schröder number. Therefore, the matrix identity (\[eq1.1\]) is a special case of (\[eq.5\]) for $k=6,t=2$.

Let us rewrite matrix identity (\[eq.5\]) in the following form $$\label{mi}
\sum_{j= 1}^i \, m_{i,j}(1+t+\cdots +t^{j-1}) =k^{i-1}.$$

The following combinatorial 