I wish to advertise the Method of Condensation in proving determinantal evaluation. The key is guess the answer, which is thanks to Ehud Meir. Then, generalize it a bit. Let $x_1, x_2,\dots$ be an infinite set of variables and modify the original matrix (by shifting variables) to $M^{a,b}(n)$ so that $$M_{i,j}^{a,b}(n):=\frac{x_{\max(i+a,j+b)}}{x_{\min(i+a,j+b)}}.$$ Convention: $M^{0,0}(n)=M(n)$.

Claim. If $a\neq b$ then $\det M^{a,b}(n)=0$, and if $a=b$ then $$\det M^{a,a}(n)=\prod_{r=2}^n\frac{x_{k-1+a}^2-x_{k+a}^2}{x_{k-1+a}^2}.\tag1$$ Proof. The case $a\neq b$ is easy - simply factor out a variable from $n^{th}$-column/row and another variable from the $(n-1)^{th}$-column/row. These new columns/rows are identical.

Inductive proofs neatly work with this Dodgson's recursive relation $$\det Z^{0,0}(n)=\frac{\det Z^{1,1}(n-1)\det Z^{1,1}(n-1)-\det Z^{0,1}(n-1)\det Z^{1,0}(n-1)}{\det Z^{1,1}(n-2)}$$ satisfied by any matrix (so long as denominators do not vanish). Thus, it holds for $\det M^{a,b}(n)$. So, it remains to prove that the (explicit) formula on the RHS of (1) does satisfy the same equation. However, this is quite a routine simplification (preferably with symbolic sofwares). The proof is complete. $\square$

I wish to advertise the Method of Condensation in proving determinantal evaluation. The key is guess the answer, which is thanks to Ehud Meir. Then, generalize it a bit. Let $x_1, x_2,\dots$ be an infinite set of variables and modify the original matrix (by shifting variables) to $M^{a,b}(n)$ so that $$M_{i,j}^{a,b}(n):=\frac{x_{\max(i+a,j+b)}}{x_{\min(i+a,j+b)}}.$$ Convention: $M^{0,0}(n)=M(n)$.

Claim. If $a\neq b$ then $\det M^{a,b}(n)=0$, and if $a=b$ then $$\det M^{a,a}(n)=\prod_{r=2}^n\frac{x_{k-1+a}^2-x_{k+a}^2}{x_{k-1+a}^2}.\tag1$$ Proof. The case $a\neq b$ is easy - simply factor out a variable from $n^{th}$-column/row and another variable from the $(n-1)^{th}$-column/row. These new columns/rows are identical.

Inductive proofs neatly work with this Dodgson's recursive relation $$\det Z^{0,0}(n)=\frac{\det Z^{1,1}(n-1)\det Z^{1,1}(n-1)-\det Z^{0,1}(n-1)\det Z^{1,0}(n-1)}{\det Z^{1,1}(n-2)}$$ satisfied by any matrix (so long as denominators do not vanish). Thus, it holds for $\det M^{a,b}(n)$. So, it remains to prove that the (explicit) formula on the RHS of (1) does satisfy the same equation. However, this is quite a routine simplification (preferably with symbolic sofwares). The proof follows. $\square$

I wish to advertise the Method of Condensation in proving determinantal evaluation. The key is guess the answer, which is thanks to Ehud Meir. Then, generalize it a bit. Let $x_1, x_2,\dots$ be an infinite set of variables and modify the original matrix (by shifting variables) to $M^{a,b}(n)$ so that $$M_{i,j}^{a,b}(n):=\frac{x_{\max(i+a,j+b)}}{x_{\min(i+a,j+b)}}.$$ Convention: $M^{0,0}(n)=M(n)$.

Claim. If $a\neq b$ then $\det M^{a,b}(n)=0$, and if $a=b$ then $$\det M^{a,a}(n)=\prod_{r=2}^n\frac{x_{k-1+a}^2-x_{k+a}^2}{x_{k-1+a}^2}.\tag1$$ Proof. The case $a\neq b$ is easy - simply factor out a variable from $n^{th}$-column/row and another variable from the $(n-1)^{th}$-column/row. These new columns/rows are identical.

Inductive proofs neatly work with this Dodgson's recursive relation $$\det Z^{0,0}(n)=\frac{\det Z^{1,1}(n-1)\det Z^{1,1}(n-1)-\det Z^{0,1}(n-1)\det Z^{1,0}(n-1)}{\det Z^{a+1,b+1}(n-2)}$$ satisfied by any matrix (so long as denominators do not vanish). Thus, it holds for $\det M^{a,b}(n)$. So, it remains to prove that the (explicit) formula on the RHS of (1) does satisfy the same equation. However, this is quite a routine simplification (preferably with symbolic sofwares). The proof is complete. $\square$

I wish to advertise the Method of Condensation in proving determinantal evaluation. The key is guess the answer, which is thanks to Ehud Meir. Then, generalize it a bit. Let $x_1, x_2,\dots$ be an infinite set of variables and modify the original matrix (by shifting variables) to $M^{a,b}(n)$ so that $$M_{i,j}^{a,b}(n):=\frac{x_{\max(i+a,j+b)}}{x_{\min(i+a,j+b)}}.$$ Convention: $M^{0,0}(n)=M(n)$.

Claim. If $a\neq b$ then $\det M^{a,b}(n)=0$, and if $a=b$ then $$\det M^{a,a}(n)=\prod_{r=2}^n\frac{x_{k-1+a}^2-x_{k+a}^2}{x_{k-1+a}^2}.\tag1$$ Proof. The case $a\neq b$ is easy - simply factor out a variable from $n^{th}$-column/row and another variable from the $(n-1)^{th}$-column/row. These new columns/rows are identical.

Inductive proofs neatly work with this Dodgson's recursive relation $$\det Z^{0,0}(n)=\frac{\det Z^{1,1}(n-1)\det Z^{1,1}(n-1)-\det Z^{0,1}(n-1)\det Z^{1,0}(n-1)}{\det Z^{1,1}(n-2)}$$ satisfied by any matrix (so long as denominators do not vanish). Thus, it holds for $\det M^{a,b}(n)$. So, it remains to prove that the (explicit) formula on the RHS of (1) does satisfy the same equation. However, this is quite a routine simplification (preferably with symbolic sofwares). The proof is complete. $\square$