This question is related to the previous problem How to find a solution to the linear equations? . Now, I just want to discuss the ''boundedness'' of the determinant, since it's difficult to find a solution to the linear equation.
Let $$\begin{aligned} &D_n(R^l,R^{-l}):=\\&\left|\begin{array}{ccccccccccc} R^l&R^{-l}&0&0&0&0&\cdots&0&0&0&0\\ {r_n}^l&{r_n}^{-l}&-{r_n}^{l}&-{r_n}^{-l}&0&0&\cdots&0&0&0&0\\{r_n}^l&-{r_n}^{-l}&-k_n{r_n}^l&k_n{r_n}^{-l}&0&0&\cdots&0&0&0&0\\ 0&0&{r_{n-1}}^l&{r_{n-1}}^{-l}&-{r_{n-1}}^l&-{r_{n-1}}^{-l}&\cdots&0&0&0&0\\ 0&0&{r_{n-1}}^l&-{r_{n-1}}^{-l}&-k_{n-1}{r_{n-1}}^l&k_{n-1}{r_{n-1}}^{-l}&\cdots&0&0&0&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\ 0&0&0&0&0&0&\cdots& {r_2}^l&{r_2}^{-l}&-{r_2}^{l}&-{r_2}^{-l}\\ 0&0&0&0&0&0& \cdots&{r_2}^l&-{r_2}^{-l}&-k_2{r_2}^l&k_2{r_2}^{-l}\\ 0&0&0&0&0&0&\cdots&0&0&(k_1-1){r_1}^l&(k_1+1){r_1}^{-l}\end{array}\right| \end{aligned}$$ Then we know $$\begin{aligned} D_1(R^l,R^{-l})&=(k_1+1)R^l{r_1}^{-l}-(k_1-1)R^{-l}{r_1}^l\\ \newline\\ D_2(R^l,R^{-l})&= R^l{r_2}^{-l}\left(-(k_2+1){r_2}^l(k_1+1){r_1}^{-l}-(k_2-1){r_2}^{-l}(k_1-1){r_1}^l\right)\\ &- R^{-l}{r_2}^{l}\left(-(k_2-1){r_2}^l(k_1+1){r_1}^{-l}-(k_2+1){r_2}^{-l}(k_1-1){r_1}^l\right)\\ &=R^l{r_2}^{-l} D_1(\color{red}{-(k_2+1){r_2}^l,(k_2-1){r_2}^{-l}})-R^{-l}{r_2}^{l}D_1(\color{red}{-(k_2-1){r_2}^l,(k_2+1){r_2}^{-l}})\\ \newline\\ D_3(R^l,R^{-l})&=R^l{r_3}^{-l}\left(-(k_3+1){r_3}^l{r_2}^{-l}\left(-(k_2+1)(k_1+1){r_2}^l{r_1}^{-l}-(k_2-1)(k_1-1){r_2}^{-l}{r_1}^l\right)\right.\\ &\quad\quad\quad\quad \left.-(k_3-1){r_3}^{-l}{r_2}^l\left(-(k_2-1)(k_1+1){r_2}^l{r_1}^{-l}-(k_2+1)(k_1-1){r_2}^{-l}{r_1}^l\right)\right)\\ &-R^{-l}{r_3}^l\left(-(k_3-1){r_3}^l{r_2}^{-l}\left(-(k_2+1)(k_1+1){r_2}^l{r_1}^{-l}-(k_2-1)(k_1-1){r_2}^{-l}{r_1}^l\right)\right.\\ &\quad\quad\quad\quad \left.-(k_3+1){r_3}^{-l}{r_2}^l\left(-(k_2-1)(k_1+1){r_2}^l{r_1}^{-l}-(k_2+1)(k_1-1){r_2}^{-l}{r_1}^l\right)\right)\\ &=R^l{r_3}^{-l} D_2(\color{red}{-(k_3+1){r_3}^l,(k_3-1){r_3}^{-l}})-R^{-l}{r_3}^{l}D_2(\color{red}{-(k_3-1){r_3}^l,(k_3+1){r_3}^{-l}}) \end{aligned}$$ Noticed that the red parts in $D_1\,,\,D_2$ are similar, hence by induction we can obtain $$D_n(R^l,R^{-l})=R^l{r_n}^{-l} D_{n-1}(\color{red}{-(k_{n}+1){r_{n}}^l,(k_{n}-1){r_{n}}^{-l}})-R^{-l}{r_{n}}^{l}D_{n-1}(\color{red}{-(k_n-1){r_n}^l,(k_n+1){r_n}^{-l}})$$
For $D_1$, we know $$|D_1|=(k_1+1)X^l\left[1-\frac{k_1-1}{k_1+1}\left(\frac{1}{X^{2l}}\right)\right]\leq 2(k_1+1) X^l$$ where $X:=\frac{R}{r_1}$.
The question is that does $D_n$ have the similar estimating? We still let $X=\frac{R}{r_1}$, and we want to find $C$, depend on $k_i$, such that $$|D_n|\leq C X^l$$