Let $A_n$ be the matrix involved in the problem and let $L_n=\left((-1)^{i-j}\binom{i}{j}_q\right)_{i,j=0}^{n-1}$.
Observe that $L_n$ is lower-triangular with 1's in the diagonal. Multiplying, we have:
$$L_nA_n=\left(x\,u_{i,j}(r)+u_{i,j}(r-j)\right)_{i,j=0}^{n-1}$$
where
$$u_{i,j}(r)=\sum_{k=0}^i(-1)^{i-k}q^{\binom{k-j}{2}}\binom{i}{k}_q\,\binom{k+r}{j}_q\,.$$

Now
\begin{eqnarray*}
u_{i,j}(r)&=&\sum_{k=0}^i(-1)^{i-k}q^{\binom{k-j}{2}}\binom{i}{i-k}_q\,\binom{k+r}{k+r-j}_q \\
&=&(-1)^{r+i-j}\,q^{\binom{r+1}{2}}\,\sum_{k=0}^i q^{k(k+r-j)}\binom{i}{i-k}_q\,\binom{-j-1}{k+r-j}_q \\
&=&(-1)^{r+i-j}\,q^{\binom{r+1}{2}}\,\sum_{k=0}^i q^{k(k+r-j)}\binom{i}{i-k}_q\,\binom{-j-1}{k+r-j}_q \\
&=&(-1)^{r+i-j}\,q^{\binom{r+1}{2}}\,\binom{i-j-1}{i-j+r}_q
=q^{\binom{i-j}{2}}\,\binom{r}{j-i}_q\,.
\end{eqnarray*}
Hence $u_{i,j}(r)=0$ when $i>j$ and $u_{i,i}(r)=1$.
Thus $L_nA_n$ is upper triangular with $x+1$ as elements of the diagonal.

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