Star operator in matrices and new math to find inverse? Example Calculation
I realized I could make some progress on an open problem if I introduced a new operation: $f(*) x = f(x)$, where $f$ is a function and $x$ is a variable, then I could make progress. It seems initially just like some new notation but its not as trivial. Consider the following transformation:
$$
\left( \begin{array}{cc}
x \\
y
\end{array} \right)=
%
\left( \begin{array}{cc}
e^{*} & 0 \\
0 & e^{*}
\end{array} \right) \cdot(
\left( \begin{array}{cc}
\ln(*) &\ln(\cos(*)) \\
\ln(*) & \ln(\sin(*))
\end{array} \right)
\left( \begin{array}{cc}
r \\
\theta
\end{array} \right))
$$
Note:
$$
\left( \begin{array}{cc}
e^{*} & 0 \\
0 & e^{*}
\end{array} \right) \cdot
\left( \begin{array}{cc}
\ln(*) &\ln(\cos(*)) \\
\ln(*) & \ln(\sin(*))
\end{array} \right)
\neq
\left( \begin{array}{cc}
* &\cos(*) \\
* & \sin(*)
\end{array} \right)$$
As one gets the wrong answer then as it seems associative property of matrices breaks down. Why? Because if one assumes it doesn't then one gets the answer:
$ x = r + \cos \theta$ and $ y = r + \sin \theta$ whereas if one solves the bracketed matrices first one gets:
$ x = r  \cos \theta$ and $ y = r  \sin \theta$. But it seems we can multiply by inverses that takes precedence over other operations (inverse precedence conjecture):
Multiplying both sides by an inverse of the leftmost matrix on the R.H.S:
$$
\left( \begin{array}{cc}
\ln(*) & 0 \\
0 & \ln(*)
\end{array} \right) \cdot
\left( \begin{array}{cc}
x \\
y
\end{array} \right)=
%
\left( \begin{array}{cc}
\ln(*) &\ln(\cos(*)) \\
\ln(*) & \ln(\sin(*))
\end{array} \right)
\left( \begin{array}{cc}
r \\
\theta
\end{array} \right)
$$
Once again, multiplying with an inverse (the inverse was using a clever guess in this case):
$$
(\left( \begin{array}{cc}
1 & 0 \\
0 & \tan^{-1} e^*
\end{array} \right) \cdot(
\left( \begin{array}{cc}
\exp(*)^2 & \exp(*)^2 \\
-1 & 1
\end{array} \right) \cdot
(\left( \begin{array}{cc}
\ln(*) & 0 \\
0 & \ln(*)
\end{array} \right) \cdot
\left( \begin{array}{cc}
x \\
y
\end{array} \right))))=
\left( \begin{array}{cc}
r \\
\theta
\end{array} \right)
$$
Questions
Is there any case where the inverse precedence conjecture fails? Is there a general procedure find an inverse of a matrix with $*$ operators in it? 
 A: Lets take a 'general case' of an mxn matrix 'multiplying' a n element column vector, that is
$$
\left(
\begin{array}{cccc} 
f_{11}(\star) & f_{12}(\star) & \ldots & f_{1n}(\star) \\
\vdots \\
f_{m1}(\star) & f_{m2}(\star) & \ldots & f_{mn}(\star)
\end{array} 
\right)
\left(
\begin{array}{c} 
x_1\\
\vdots \\
x_n
\end{array} 
\right) 
=
\left(
\begin{array}{cccc} 
f_{11}(x_1) + f_{12}(x_2) +\ldots + f_{1n}(x_n) \\
\vdots \\
f_{m1}(x_1) + f_{m2}(x_2) + \ldots + f_{mn}(x_n)
\end{array} 
\right)
$$
Thus we see that this notation is equivalent to a function
$$
f
\left(\left(
\begin{array}{c} 
x_1\\
\vdots \\
x_n
\end{array} 
\right) \right) 
=
\left(
\begin{array}{cccc} 
f_{11}(x_1) + f_{12}(x_2) +\ldots + f_{1n}(x_n) \\
\vdots \\
f_{m1}(x_1) + f_{m2}(x_2) + \ldots + f_{mn}(x_n)
\end{array} 
\right)
$$
and we are considering functions of column vector expressable as a sum of functions of each element, the matrix notation being useful as book-keeping.
To obtain an inverse function of $f$ is not possible in general. However, if $m=n$ and the function-matrix has one non-zero element per row and collumn, so only $f_{1\sigma(1)},\ldots,f_{n\sigma(n)}$ are non zero (and invertable), where $\sigma$ is a permutation function on $1,\ldots,n$, then
$$
f
\left(\left(
\begin{array}{c} 
x_1\\
\vdots \\
x_n
\end{array} 
\right) \right) 
=
\left(
\begin{array}{cccc} 
f_{1\sigma(1)}(x_{\sigma(1)})\\
\vdots \\
f_{n\sigma(n)}(x_{\sigma(n)} )
\end{array} 
\right)
$$
$$
f^{-1} 
\left(\left(
\begin{array}{c} 
y_1\\
\vdots \\
y_n
\end{array} 
\right) \right) 
=
\left(
\begin{array}{c} 
f_{\sigma^{-1} (1) 1 }^{-1}(y_{\sigma^{-1} (1)})\\
\vdots \\
f_{\sigma^{-1} (n) n }^{-1}(y_{\sigma^{-1} (n)})
\end{array}
\right) 
$$
One of the cases considered in the opening question corrosponds to the case $n=2$, $\sigma(i)=i$.
Explicit example:
Consider the case $n=m=2$ where we have a 'matrix'
$$
\left(
\begin{array}{cc} 
f_{11}(\star) & f_{12}(\star) \\
f_{21}(\star) & f_{22}(\star)
\end{array}
\right)
=
\left(
\begin{array}{cc} 
0 & \sin(\star) \\
\cos(\star) & 0
\end{array}
\right)
$$
so that there is exactly one non-zero element in each row and column. The action of this matrix can be summed up in a function
$$
f
\left(\left(
\begin{array}{c} 
x_1\\
x_2
\end{array} 
\right) \right)
=
\left(
\begin{array}{cc} 
f_{11}(\star) & f_{12}(\star) \\
f_{21}(\star) & f_{22}(\star)
\end{array}
\right)
\left(
\begin{array}{c} 
x_1\\
x_2
\end{array} 
\right)
=
\left(
\begin{array}{cc} 
0 & \sin(\star) \\
\cos(\star) & 0
\end{array}
\right)
\left(
\begin{array}{c} 
x_1\\
x_2
\end{array} 
\right)
=
\left(
\begin{array}{c} 
\sin(x_2)\\
\cos(x_1)
\end{array} 
\right)
$$
To find an inverse function we write
$$
\left(
\begin{array}{c} 
y_1\\
y_2
\end{array} 
\right)
=
\left(
\begin{array}{c} 
\sin(x_2)\\
\cos(x_1)
\end{array} 
\right)
$$
$$
\Rightarrow
\left(
\begin{array}{c} 
x_1\\
x_2
\end{array} 
\right)
=
\left(
\begin{array}{c} 
\arccos(y_2)\\
\arcsin(y_1)
\end{array} 
\right)
$$
$$
\Rightarrow
f^{-1}
\left( \left(
\begin{array}{c} 
y_1\\
y_2
\end{array} 
\right) \right)
=
\left(
\begin{array}{c} 
\arccos(y_2)\\
\arcsin(y_1)
\end{array} 
\right)
=
\left(
\begin{array}{cc} 
0 & \arccos(\star)\\
\arcsin(\star) & 0
\end{array} 
\right)
\left(
\begin{array}{c} 
y_1\\
y_2
\end{array}
\right)
$$
$$
\Rightarrow
\left(
\begin{array}{cc} 
0 & \sin(\star) \\
\cos(\star) & 0
\end{array}
\right)^{-1}
=
\left(
\begin{array}{cc} 
0 & \arccos(\star)\\
\arcsin(\star) & 0
\end{array} 
\right)
$$
The important things to note:


*

*the inverse of the 'matrix' contains the inverse functions

*the functions are permuted (I expect this corresponds to the matrix being 'transposed') to preserve function action


Tackling manipulation
Let us suppose now that we have some 'matrices' $F(\star)$ and $G(\star)$ with corresponding functions $f$ and $g$. Moreover, we have an expression
$$
y = F(\star) G(\star) x
$$
where $x$ and $y$ are column vectors. If we can find inverse functions for $f$ and $g$ then we can perform manipulations
$$
y = F(\star) G(\star) x = f(g(x))
$$
$$
\Rightarrow
f^{-1}(y) = g(x)
$$
$$
\Rightarrow
g^{-1}(f^{-1}(y)) = x
$$
If these inverse functions can be written in matrix form then 
$$
F^{-1}(\star) y = G(\star) x
$$
and
$$
G^{-1}(\star)F^{-1}(\star)y = x
$$
Hopefully this sheds some light on your 'inverse precedence conjecture'
