Matrix $A_n$ can be generalized to $\big[(j+f(i))^{i-1}\big]_{i,j=1}^n$ for any function $f(i)$ not just $f(i)=i-1$.

Since $$(j+f(i))^{i-1} = \sum_{k=0}^{i-1} c_{i,k}\cdot j^k,$$ where $c_{i,k} := \binom{i-1}k f(i)^{i-1-k}$, we have $$\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = \begin{bmatrix} c_{1,0} & 0 & 0 & \dots & 0\\ c_{2,0} & c_{2,1} & 0 & \dots & 0\\ c_{3,0} & c_{3,1} & c_{3,2} & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ c_{n,0} & c_{n,1} & c_{n,2} & \dots & c_{n,n-1} \end{bmatrix}\cdot \big[j^{i-1}\big]_{i,j=1}^n.$$ Taking determinants, we get $$\det\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = c_{1,0}c_{2,1}\cdots c_{n,n-1}\cdot \det\big[j^{i-1}\big]_{i,j=1}^n=\det\big[j^{i-1}\big]_{i,j=1}^n.$$

For $B_n$ things work even for a simpler reason: $$q^{(i+j-1)(i-1)} = q^{(i-1)^2} q^{j(i-1)}$$ and we can take out a factor of $q^{(i-1)^2}$ from the $i$-th row of matrix $B_n$.

However for matrix $C_n$ neither binomial expansion work, nor we can take out any common factors from its rows. So, despite of visual similarity $C_n$ does not possess the properties of $A_n$ or $B_n$.

Matrix $A_n$ can be generalized to $\big[(j+f(i))^{i-1}\big]_{i,j=1}^n$ for any function $f(i)$ not just $f(i)=i-1$.

Since $$(j+f(i))^{i-1} = \sum_{k=0}^{i-1} c_{i,k}\cdot j^k,$$ where $c_{i,k} := \binom{i-1}k f(i)^{i-1-k}$, we have $$\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = \begin{bmatrix} c_{1,0} & 0 & 0 & \dots & 0\\ c_{2,0} & c_{2,1} & 0 & \dots & 0\\ c_{3,0} & c_{3,1} & c_{3,2} & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ c_{n,0} & c_{n,1} & c_{n,2} & \dots & c_{n,n-1} \end{bmatrix}\cdot \big[j^{i-1}\big]_{i,j=1}^n.$$ Taking determinants, we get $$\det\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = c_{1,0}c_{2,1}\cdots c_{n,n-1}\cdot \det\big[j^{i-1}\big]_{i,j=1}^n=\det\big[j^{i-1}\big]_{i,j=1}^n.$$

For $B_n$ things work even for a simpler reason: $$q^{(i+j-1)(i-1)} = q^{(i-1)^2} q^{j(i-1)}$$ and we can take out a factor of $q^{(i-1)^2}$ from the $i$-th row of matrix $B_n$.

However for matrix $C_n$ neither binomial expansion work, nor we can take out any common factors from its rows. So, despite visual similarity $C_n$ does not possess the properties of $A_n$ or $B_n$.

In Question 1, matrix $A_n$ can be generalized to $\big[(j+f(i))^{i-1}\big]_{i,j=1}^n$ for any function $f(i)$ not just $f(i)=i-1$.

Since $$(j+f(i))^{i-1} = \sum_{k=0}^{i-1} c_{i,k}\cdot j^k,$$ where $c_{i,k} := \binom{i-1}k f(i)^{i-1-k}$, we have $$\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = \begin{bmatrix} c_{1,0} & 0 & 0 & \dots & 0\\ c_{2,0} & c_{2,1} & 0 & \dots & 0\\ c_{3,0} & c_{3,1} & c_{3,2} & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ c_{n,0} & c_{n,1} & c_{n,2} & \dots & c_{n,n-1} \end{bmatrix}\cdot \big[j^{i-1}\big]_{i,j=1}^n.$$ Taking determinants, we get $$\det\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = c_{1,0}c_{2,1}\cdots c_{n,n-1}\cdot \det\big[j^{i-1}\big]_{i,j=1}^n=\det\big[j^{i-1}\big]_{i,j=1}^n.$$

Matrix $A_n$ can be generalized to $\big[(j+f(i))^{i-1}\big]_{i,j=1}^n$ for any function $f(i)$ not just $f(i)=i-1$.

Since $$(j+f(i))^{i-1} = \sum_{k=0}^{i-1} c_{i,k}\cdot j^k,$$ where $c_{i,k} := \binom{i-1}k f(i)^{i-1-k}$, we have $$\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = \begin{bmatrix} c_{1,0} & 0 & 0 & \dots & 0\\ c_{2,0} & c_{2,1} & 0 & \dots & 0\\ c_{3,0} & c_{3,1} & c_{3,2} & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ c_{n,0} & c_{n,1} & c_{n,2} & \dots & c_{n,n-1} \end{bmatrix}\cdot \big[j^{i-1}\big]_{i,j=1}^n.$$ Taking determinants, we get $$\det\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = c_{1,0}c_{2,1}\cdots c_{n,n-1}\cdot \det\big[j^{i-1}\big]_{i,j=1}^n=\det\big[j^{i-1}\big]_{i,j=1}^n.$$

For $B_n$ things work even for a simpler reason: $$q^{(i+j-1)(i-1)} = q^{(i-1)^2} q^{j(i-1)}$$ and we can take out a factor of $q^{(i-1)^2}$ from the $i$-th row of matrix $B_n$.

However for matrix $C_n$ neither binomial expansion work, nor we can take out any common factors from its rows. So, despite of visual similarity $C_n$ does not possess the properties of $A_n$ or $B_n$.