Some tridiagonal determinants (Q6158147)

Let \(C=C_{n+1}(y,a,b,h)=(c_{ij})\) be the \((n+1)\times(n+1)\) tridiagonal determinant with \begin{align*} &c_{jj}=y+(n-j)a+jb+2j(n-j)h,&0\le j\le n, \\ &c_{j,j+1}=(n-j)(a+jh),&0\le j\le n-1, \\ &c_{j+1,j}=(j+1)\big(b+(n-j-1)h\big),&0\le j\le n-1. \end{align*} The author proves the following factorization: \[ C=\prod_{j=0}^n\Big(y+j\big(a+b+(j-1)h\big)\Big). \] Further, let \(D=D_n(x,a,z,h,k,q)=(d_{ij})\) be the \(n\times n\) tridiagonal determinant with \begin{align*} &d_{jj}=x-[k]_qazq^{j-1}-[j-1]_qh,&1\le j\le n, \\ &d_{j-1,j}=[j-1]_qa,&2\le j\le n, \\ &d_{j+1,j}=[n-j]_qzq^{k+j-1}(az-h),&1\le j\le n-1, \end{align*} where \[ [m]_q=\frac{1-q^m}{1-q},\,q\ne 1,\quad [m]_q=m,\,q=1. \] In this case, the factorization reads as \[ D=\prod_{j=0}^{n-1}\big(x-[k+n-2j-1]_qazq^j-[j]_qh\big). \] As special cases of these results, certain well-known determinants are obtained.