Finding Inverses of Systems of Multi-Variable Equations

Let a function $f$:

$$f\left[\begin{array}{c}f\_1(x,y)\\ f\_2(x,y)\end{array}\right]=============\left[\begin{array}{c}xy\\ 3y-x\end{array}\right]=\left[\begin{array}{c}z\_1\\ z\_2\end{array}\right]$$

For this $f$ intend to find a function $g=f^{-1}$ such that

g\begin{bmatrix} f\_{1}(x,y) \\ f\_{2}(x,y) \end{bmatrix} = g\begin{bmatrix} z\_{1} \\ z\_{2} \end{bmatrix} ============= \left{\begin{matrix} \begin{bmatrix} \text{Something in direct relation to $(x,y)$} \\ \text{Something in direct relation to $(x,y)$} \end{bmatrix}\\ \text{or}\\ (x,y)\\ \end{matrix}\right.

Moreover we want to find an algorithm that can produce $g=f^{-1}$ for any invertible $f$.

Our attempt so far

Form the Jacobian

J\_{f} = \begin{bmatrix} \frac{\partial f\_{1}}{\partial x} & \frac{\partial f\_{1}}{\partial y} \\ \frac{\partial f\_{2}}{\partial x} & \frac{\partial f\_{2}}{\partial y} \end{bmatrix} \= \begin{bmatrix} y & x \\ -1 & 3 \end{bmatrix}

Form the inverse of the $2\times 2$ matrix.

$$J\_g=J\_f^{-1}=\frac{1}{\det J\_f}J\_f^T=\frac{1}{3y+x}\left[\begin{array}{cc}y& -1\\ x& 3\end{array}\right]$$

—

We now have the partial derivatives inside of $J\_g$ and assume that the rows of $J\_g$ are individually also inverses for $J\_f$ i.e. :

$$J\_g=\left[\begin{array}{cc}\frac{\partial g\_1}{\partial x}& \frac{\partial g\_1}{\partial y}\\ \frac{\partial g\_2}{\partial x}& \frac{\partial g\_2}{\partial y}\end{array}\right]⟹\text{The function g\_1 reconstructed from ∂x∂g\_1 and ∂x∂g\_2}$$ $$g\_1\text{satisfies}g\_1(z)=f\_1^{-1}(x,y)⟹g(f(x,y))\underbrace{\approx }\_\text{The output of g must be connected to (x, y) shw}(x,y)$$

—

Then we just have to reconstruct $g\_1$ and $g\_2$ from the partial derivatives of $J\_g$:

$$\int \frac{\partial g\_1}{\partial y}dx=f(x,y)+u(y)=-y\ln (3y+x)+u(y)$$

where

$$u(y)=\text{The terms of ∂x∂g\_2 independent of x (there are none)}$$

—

From our process $g\_1=-y\ln (3y+x)$ and $g\_2$ is found similarly. Then we have

g(z\_{1}, z\_{2}) = \begin{bmatrix} g\_{1}(z\_{1}) \\ g\_{2}(z\_{2}) \end{bmatrix} \= \left(f\begin{bmatrix} f\_{1}(x,y) \\ f\_{2}(x,y) \\ \end{bmatrix}\right)^{-1}

Critical Postulations

$$J\_gJ\_f=I$$

Doesn’t necessarily provide inverses to the separate rows of $J\_f$.

$$J\_g=\left[\begin{array}{cc}\frac{\partial g\_1}{\partial x}& \frac{\partial g\_1}{\partial y}\\ \frac{\partial g\_2}{\partial x}& \frac{\partial g\_2}{\partial y}\end{array}\right]$$

i.e.

$$\text{While}J\_gJ\_f=I\text{ }\text{it doesn’t mean that}g\_1f\_1=1\text{or}g\_1=f^{-1}\_1$$