Jacobian Matrix set-up:

$J=\begin{bmatrix} \dfrac{\partial F}{\partial x} & \dfrac{\partial F}{\partial y} \\ \dfrac{\partial G}{\partial x} & \dfrac{\partial G}{\partial y}\end{bmatrix}$ , where $J$ is the coefficient of linearized equations. For this example let’s take the two equations $F(x,y)=x+y-2y^2=4$ and $G(x,y)=x^2+y^2=8$. (Use $x_0=1$ and $y_0=1$ for initial approximations.)

Let’s do the partial derivative calculate to plug into our $J$ matrix.

• $\partial F/\partial x=1$
• $\partial F/\partial y= 1-4y_0$
• $\partial G/\partial x= 2x_0$
• $\partial G/\partial y= 2y_0$

---

Define our $X$ matrix as $X= \begin{bmatrix} dx \\ dy \end{bmatrix}$, and our $K$ matrix as $K= \begin{bmatrix} -4-(x_0+y_0-2y_0^2) \\ 8-(x_0^2+y_0^2) \end{bmatrix}$. Remember that to solve for $X$ matrix, where $X=J^{\\-1}K$. Plugging in $x_0$ and $y_0$ gives us the following $J$ matrix: $J= \begin{bmatrix} 1 & -3 \\ 2 & 2 \end{bmatrix}$, where ${J}^{\\-1} = \frac{1}{1\cdot2-(-3)\cdot2} \times\begin{bmatrix} 2 & 3 \\ -2 & 1 \end{bmatrix} = ⅛ \times \begin{bmatrix} 2 & 3 \\ -2 & 1 \end{bmatrix}$. This would make $J^{\\-1}=\begin{bmatrix} ¼ & ⅜ \\ -¼ & ⅛ \end{bmatrix}$ and $K= \begin{bmatrix} -4\\ 6 \end{bmatrix}$.

So this means that $\begin{bmatrix} ¼ & ⅜ \\ -¼ & ⅛ \end{bmatrix}\times\begin{bmatrix} -4\\ 6 \end{bmatrix}=X= \begin{bmatrix} dx \\ dy \end{bmatrix}= \begin{bmatrix} 1¼ \\ 1¾ \end{bmatrix}$