Transforming a matrix from cartesian to spherical coordinates Consider a variable matrix
$$\left[\begin{array}{ccc}a_{11}(x,y,z) \quad a_{12}(x,y,z) \quad a_{13}(x,y,z)\\
                          a_{21}(x,y,z) \quad a_{22}(x,y,z) \quad a_{23}(x,y,z)\\
    a_{31}(x,y,z) \quad a_{32}(x,y,z) \quad a_{33}(x,y,z)\\\end{array}\right] ,$$
    elements are real functions. How can we transform this matrix to spherical coordinates based on point $S(x_0,y_0,z_0)$, so that we get another matrix
$$\left[\begin{array}{ccc}a_{11}(r,\varphi,\psi) \quad a_{12}(r,\varphi,\psi) \quad a_{13}(r,\varphi,\psi)\\
                          a_{21}(r,\varphi,\psi) \quad a_{22}(r,\varphi,\psi) \quad a_{23}(r,\varphi,\psi)\\
    a_{31}(r,\varphi,\psi) \quad a_{32}(r,\varphi,\psi) \quad a_{33}(r,\varphi,\psi)\\\end{array}\right] ?$$
    I know that transform only the elements of matrix $A$ is not correct. Could you advise me what to do in this case?
 A: Based on the problem as you have stated we have a matrix: A that can be described as
$$A(x,y,z)$$
where we input x, y, z and get out a matrix A. From here it becomes clear that changing variables is independent of the matrix definition: in other words
$$A(x,y,z) = A(r \sin(\phi_1) \cos(\phi_2 ) , r \sin(\phi_1)\sin(\phi_2), r \cos(\phi_1)) $$
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I'm going to venture to guess you are doing something involving differentiation/integration on this matrix: since you suggested in your comments "I should have multiplied matrix A [to other matrices]"
Which we can explore here: 
Suppose we wish to evaluate
$$\int \int \int  A(x,y,z) \partial x \partial y \partial z $$
If we wish to convert to another coordinate system not only does the argument of $A$ change, but... the expression $\partial x \partial y \partial z  $ is going to have change as well:
The article: http://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables
goes into more detail but we will sketch it out with your example:
The reason we can't simply throw down $\partial r$ $\partial \phi_1$ $ \partial \phi_2$ is that the infinitesmal change in volume by integrating with respect to this system is different than standard cartesian coordinates. Recall from calculus I:
that to grab an area under a curve $f$ in cartesian coordinates we simply evaluate:
$$ \int_{a}^{b} y(x) dx$$
but in polar the formula became
$$ \frac{1}{2} \int_{a}^{b} r(\phi)^2 d\phi$$
notice that the answer really should've been just r but instead become $r^2$ where the extra r is the scaling factor and we need to calculate that for all integrals.
To compute the scaling factor simply take function:
$$[x(r, \phi_1 ,\phi_2),y(r, \phi_1 ,\phi_2),z(r, \phi_1 ,\phi_2) $$
and compute its jacobian which is the matrix of partial derivatives with respect to $r$, $\phi_1$, and $\phi_2$. And then take the determinant of said matrix (followed by absolute value of it) and multiply it with your answer.
if you go through the work you get a scale factor of: $r^2 \sin(\phi_2)$ that needs to be multipl
