# How to apply Newton's method to this system of two equations?

I'm a bit stuck. We're given the following system of equations: $$3x-2y+1=0$$ $$12x+3y-18=0$$

And they ask us whether Newton's Method converges after only one iteration. It's similar to this question I asked a few days ago Newton's Method for linearly dependent system of equations but here the two equations are not linearly dependent, so the Jacobian is invertible.

The Jacobian is $$\begin{bmatrix} 3 & -2 \\ 12 & 3 \end{bmatrix}$$ and its determinant is 33.

I'm a bit stuck about what to do next. I googled heavily but could not find an answer to my question.

• @MattiP. In this case, $B= (-1, 18)^T, A^{-1}=1/33 (3 \ \ 3, -12 \ \ 3)^T$. The problem is that I'm not given any initial guess $X_n$. If we have no initial guess, how could we possibly use the formula ? Mar 31 '20 at 9:51
• (Revised because I made a typo) Newton's method works exactly the same way for multiple variables as it does for a single variable. You have an equation of the form $f(X)=0$, where $X$ is an array of variables; $X=(x,y)^T$. In this case $$f(X) = AX-B$$ where $A$ and $B$ are matrices. The only trick is calculating the "inverse of the derivative". In reality you will use the Jacobian $Jf(X_n)$, and the iteration formula becomes $$X_{n+1} = X_n - (Jf(X_n))^{-1} f(X_n)$$ Notice that this formula is applicable to all multivariable equations of the form $f(X)=0$. Mar 31 '20 at 9:53
• I tried my numerical Newton's and it converged after one iteration to $$(x, y) = (1, 2)$$
• @Moo : Of course $x_1=x_0-A^{-1}(Ax_0-b)=A^{-1}b$ is independent of the input $x_0$ and gives the solution in one step. Mar 31 '20 at 14:36