TITLE: A q-rious identity
QUESTION [22 upvotes]: Let $[x]_q=\frac{1-q^x}{1-q}$,  $[n]_q!=[1]_q[2]_q\cdots[n]_q$ and ${\binom{x}{n}}_{q}=\frac{[x]_q[x-1]_q\cdots[x-n+1]_q }{[n]_q!}$.
Computer experiments suggest that
$$\det \left(q^\binom{i-j}{2}\left(\binom{i+r}{j}_{q}x+\binom{i+r-j}{j}_{q}\right)\right)_{i,j = 0}^{n - 1} = (1+x)^n$$
Any idea how to prove this?

REPLY [4 votes]: 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.