In an area called transformation optics, they transform Maxwell equations from one space coordinate system to another, and then somehow obtain the properties of background material $(\epsilon , \mu)$ in the first coordinate system, and this way find the required $(\epsilon , \mu)$ to direct EM waves in arbitrary directions.
From this paper, we have a transformation defined in cylindrical coordinates as : $$\rho'=R_1+\frac{R_2-R_1}{R_2}\rho$$ $$\phi'=\phi$$ $$z'=z$$
This transformation maps the circular region $0\leq\rho \leq R$ to the circular annular region $R_1\leq \rho' \leq R_2$. In the paper, the Jacobi matrix for this transformation is written as (equation 14): $$ A=\begin{pmatrix} (R_2-R_1)/R_2 & 0 & 0\\ 0 & \rho'/\rho & 0\\ 0& 0 & 1 \end{pmatrix}$$
We know that A is defined as $$ A=\begin{pmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} & \frac{\partial x'}{\partial z}\\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} & \frac{\partial y'}{\partial z}\\ \frac{\partial z'}{\partial x} & \frac{\partial z'}{\partial y} & \frac{\partial z'}{\partial z} \end{pmatrix}$$
Why the element $A_{22}=\rho'/\rho$ instead of $1$? This result is obtained here, equation (11), using more accurate mathematical notation.
In another transformation, defined as:
$$\rho'=\rho$$ $$\phi'=\frac{a}{2l}(z+l)+\phi$$ $$z'=z$$ The matrix $A$ is obtained as:
$$ A=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1& \frac{a}{2l}\rho'\\ 0& 0 & 1 \end{pmatrix}$$
In this example, why the $A_{23}$ element is not $\frac{a}{2l}$?
Note: I think it has something to do with metric tensor, etc. that I don't understand.
As far as I can see, the transformations are from cylindrical to cylindrical coordinates, and are purely linear in this representation. The coefficients in question must be $1$ and $\frac a{2l}$ as you say.
UPDATE:
As the coordinates are not Cartesian but curvilinear, differentiation isn't made the obvious way. There are scale factors to be taken into account. In the case of cylindrical coordinates, these are $1, \rho, 1$.
The corrected Jacobian is given by $$ \begin{pmatrix} 1 & 0 & 0\\ 0 & \rho'& 0\\ 0& 0 & 1 \end{pmatrix} [J] \begin{pmatrix} 1 & 0 & 0\\ 0 & \rho^{-1}& 0\\ 0& 0 & 1 \end{pmatrix}$$