The jacobian and the change of coordinates Here is a problem with solution from Gadea's book. I have a little question about the solution.

My problem: I do not know why $A\left(\begin{array}{c}
f_{1}\\
f_{2}\\
f_{3}
\end{array}\right)=\left(\begin{array}{c}
2\\
-1\\
3
\end{array}\right)
 $
It is probably something quite trivial or fundamental, but I do not get it.
 A: The matrix $A$ is the jacobian of the map, 
$$\Phi(\rho,\theta,z) = (\rho \cos \theta, \rho \sin \theta,z) = (x,y,z)$$
$$ A: T_{(\rho, \theta,z)}\mathbb{R}^3 \to T_{(x,y,z)}\mathbb{R}^3,\ \begin{pmatrix} \rho \\ \theta \\ z \end{pmatrix} \mapsto \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$
In the above definition, we choose to not distinguish between a vector in the tangent space (which is expressed in terms of $\partial^j$) and the coefficients i.e to the vector $X$ we have the associated vector,
$$\begin{pmatrix} \ \ 2 \\ -1 \\ \ \ 3 \end{pmatrix}$$
In terms of the tangent space basis generated by $\rho, \theta,z$ we have that,
$$X = f_1 \frac{\partial}{\partial \rho} +f_2 \frac{\partial}{\partial \theta} +f_3 \frac{\partial}{\partial z} $$
Hence, if we apply $A$ to the vector $\begin{pmatrix} f_1 & f_2 & f_3 \end{pmatrix}$ then its image should be the vector $X$ i.e its coordinate representation. This is true since $\Phi$ is a diffeomorphism and so $A$ is an isomorphism i.e $(f_1, f_2, _3)^t$ is the unique preimage. 
$$ A\begin{pmatrix} f_1 \\ f_2 \\ f_3 \end{pmatrix} = \begin{pmatrix} \ \ 2 \\ -1 \\ \ \ 3 \end{pmatrix}$$
$\textbf{Big Picture}$: All you need to think about is that the represenation of $X$ in cylindrical is the velocity vector of a curve $\gamma$ in $\mathbb{R}^3$ where this $\mathbb{R}^3$ is the spherical coordinate one. Next, we take $(\Phi \circ \gamma)'(0)$ which is now the representation of $X$ in cartesian coordinates i.e the flat or standard $\mathbb{R}^3$. Hence we can say $(\Phi \circ \gamma)'(0) = (2,-1,3)$ i.e,
$$ (f_1,f_2,f_3) = A^{-1}(2,-1,3)$$
