Find the sum $ \sum\limits_{a,b=0}^ \infty \begin{vmatrix} x^{a+b} & y^{a+b} & z^{a+b}\\ x^b & y^b & z^b \\ 1 & 1 & 1 \end{vmatrix} u^a v^b. $

Find a close determinant expression for the sum $$ \sum_{a,b=0}^ \infty \begin{vmatrix} x^{a+b} & y^{a+b} & z^{a+b}\\ x^b & y^b & z^b \\ 1 & 1 & 1 \end{vmatrix} u^a v^b. $$

Is is not too hard expand the determinant and get that

\begin{gather*} \sum_{a,b=0}^ \infty \begin{vmatrix} x^{a+b} & y^{a+b} & z^{a+b}\\ x^b & y^b & z^b \\ 1 & 1 & 1 \end{vmatrix} u^a v^b=\\=\sum_{a,b=0}^ \infty ({x}^{a+b}{y}^{b}+{y}^{a+b}{z}^{b}-{x}^{a+b}{z}^{b}-{ x}^{b}{y}^{a+b}+{x}^{b}{z}^{a+b}-{z}^{a+b}{y}^{b}) u^a v^b= \\= {\frac {1}{ \left( xu-1 \right) \left( vxy-1 \right) }}+{\frac {1}{ \left( yu-1 \right) \left( vyz-1 \right) }}-{\frac {1}{ \left( xu-1 \right) \left( vxz-1 \right) }}-\\-{\frac {1}{ \left( yu-1 \right) \left( vxy-1 \right) }}+{\frac {1}{ \left( zu-1 \right) \left( vxz-1 \right) }}-{\frac {1}{ \left( zu-1 \right) \left( vyz-1 \right) }}=\\= {\frac {uv \left( y-z \right) \left( x-z \right) \left( x-y \right) \left( uvxyz-1\right) }{ \left( zu-1 \right) \left( vyz-1 \right) \left( xu-1 \right) \left( vxy-1 \right) \left( yu-1 \right) \left( vxz-1 \right) }} \end{gather*}

Question. Is it possible to go back to determinant and express the result

$$ {\frac {uv \left( y-z \right) \left( x-z \right) \left( x-y \right) \left( uvxyz-1\right) }{ \left( zu-1 \right) \left( vyz-1 \right) \left( xu-1 \right) \left( vxy-1 \right) \left( yu-1 \right) \left( vxz-1 \right) }} $$

again as determinant ( or sum ( product) of determinants)?

Ant help?

Solution 1:

$$ \sum_{a,b=0}^ \infty \begin{vmatrix} x^{a+b} & y^{a+b} & z^{a+b}\\ x^b & y^b & z^b \\ 1 & 1 & 1 \end{vmatrix} u^a v^b = \\ \sum_{a,b=0}^ \infty \begin{vmatrix} u^ax^{a+b} & u^ay^{a+b} & u^az^{a+b}\\ x^b & y^b & z^b \\ 1 & 1 & 1 \end{vmatrix} v^b =\\ \sum_{b=0}^ \infty \begin{vmatrix} \frac{1}{1-xu}x^{b} & \frac{1}{1-yu}y^{b} & \frac{1}{1-zu}z^{b}\\ x^b & y^b & z^b \\ 1 & 1 & 1 \end{vmatrix} v^b =\\ \sum_{b=0}^ \infty \begin{vmatrix} \frac{1}{1-xu} & \frac{1}{1-yu} & \frac{1}{1-zu}\\ 1 & 1 & 1 \\ x^{-b} & y^{-b} & z^{-b} \end{vmatrix} (vxyz)^b =\\ \sum_{b=0}^ \infty \begin{vmatrix} \frac{1}{1-xu} & \frac{1}{1-yu} & \frac{1}{1-zu}\\ 1 & 1 & 1 \\ (yzv)^{b} & (xzv)^{b} & (xyv)^{b} \end{vmatrix} =\\ \begin{vmatrix} \frac{1}{1-xu} & \frac{1}{1-yu} & \frac{1}{1-zu}\\ 1 & 1 & 1 \\ \frac{1}{1-yzv}& \frac{1}{1-xzv} & \frac{1}{1-xyv} \end{vmatrix} $$