I'm trying to write a c++ programme to find the solutions to a nonlinear system using the simple iteration method described on page 120 here. It says: Given a system of nonlinear equations
$$\left\{\begin{array}{l} f_{1}\left(x_{1}, \ldots, x_{m}\right)=0 \\ f_{2}\left(x_{1}, \ldots, x_{m}\right)=0 \\ \vdots \\ f_{m}\left(x_{1}, \ldots, x_{m}\right)=0 \end{array}\right.$$
If we let $$\mathbf{F}=\left(\begin{array}{c} f_{1}(\mathbf{x}) \\ f_{2}(\mathbf{x}) \\ \vdots \\ f_{m}(\mathbf{x}) \end{array}\right): \mathbb{R}^{m} \rightarrow \mathbb{R}^{m}$$
Then we can rewrite the first expression as $\mathbf{F}(\mathbf{x}) = 0, \qquad \mathbf{x} = \mathbf{G}(\mathbf{x}) \qquad \mathbf{G}: \mathbb{R}^m \to \mathbb R^m$.
Solution $\boldsymbol{\alpha}: \boldsymbol{\alpha}=\mathbf{G}(\boldsymbol{\alpha})$ is called a fixed point of G. Example: $\mathbf{F}(\mathbf{x})=0 \space ,$ $\mathbf{x}=\mathbf{x}-A \mathbf{F}(\mathbf{x})=\mathbf{G}(\mathbf{x}) \quad$ for some non singular matrix $A \in \mathbb{R}^{m \times m}$.
Iteration: initial guess $x_{0}$ $$ \mathbf{x}_{n+1}=\mathbf{G}\left(\mathbf{x}_{n}\right), \quad n=0,1,2, \ldots $$
An example is given on page 135.
Solve $\left\{\begin{array}{l}f_{1} \equiv 3 x_{1}^{2}+4 x_{2}^{2}-1=0 \\ f_{2} \equiv x_{2}^{3}-8 x_{1}^{3}-1=0\end{array}, \text { for } \boldsymbol{\alpha} \text { near }\left(x_{1}, x_{2}\right)=(-.5, .25)\right.$
The iterative solution given is $$ \left[\begin{array}{c} x_{1, n+1} \\ x_{2, n+1} \end{array}\right]=\left[\begin{array}{c} x_{1, n} \\ x_{2, n} \end{array}\right]-\left[\begin{array}{cc} .016 & -.17 \\ .52 & -.26 \end{array}\right]\left[\begin{array}{c} 3 x_{1, n}^{2}+4 x_{2, n}^{2}-1 \\ x_{2, n}^{3}-8 x_{1, n}^{3}-1 \end{array}\right] $$
The notes don't explain how to find the matrix A. How can I find the matrix A?