# The inverse of a symbolic matrix (with reciprocal binomials) has Laurent entries

Recalling the $$q$$-binomials (Gaussian polynomials). Let $$[n]_q!=\prod_{j=1}^n\frac{1-q^j}{1-q}$$ and $$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}$$.

Now, consider the $$n\times n$$ matrix $$\mathbf{M}_n(q)$$, with entries $$\frac1{\binom{i+j}j_q}$$, for $$0\leq i,j.

QUESTION. Is this true? The inverse of the matrix $$M_n(q)$$ has (almost) polynomial entries. To help out with this, one may try to prove the claim $$\det\mathbf{M}_n(q)=(-1)^{\binom{n}2}q^{(n-1)\binom{n}2}\prod_{j=1}^{n-1}\frac{1+q^j}{\binom{2j}j_q^2}.$$

Caveat. We said "almost" to mean that all entries are, indeed, polynomials except the lingering denominators that are simply some powers $$q$$; i.e. of the form $$q^m$$. Or, they are Laurent polynomials. For example, if $$n=2$$ then $$\mathbf{M}_2^{-1}(q)=\begin{pmatrix}-\frac1q&\frac{1+q}q \\ \frac{1+q}q&-\frac{1+q}q \end{pmatrix}.$$

We have $$\frac1{{i+j\choose j}_q}=\frac{[i]_q![j]_q!}{[i+j]_q!}= \frac{[i]_q![j]_q!}{[n-1+j]_q!}\cdot \frac{[n-1+j]_q!}{[i+j]_q!}\\ =\frac{[i]_q![j]_q!}{[n-1+j]_q!(q-1)^{n-1-i}}\cdot\prod_{s=1}^{n-1-i}(q^{j+i+s}-1)=A(i)B(j)P_{n-1-i}(q^j),$$ where $$P_{n-1-i}$$ is a polynomial of degree $$n-1-i$$. So, your matrix is a product of two diagonal matrices with diagonals $$A$$, $$B$$ and the Vandermonde type matrix with entries $$P_{n-1-i}(q^j)$$. This gives a determinant formula immediately, for the inverse matrix formula certain extra work is needed (the inverse of Vandermonde matrix is known, and our Vandermonde type matrix is a product of the genuine Vandermonde and a unipotent matrix).