Question about using Newton's Method to solve a system of equations

[Ax_xiom]

So the formula used to solve non-linear equations using the Newton-Raphson method is this $$ X_{k+1} = X_k - J^{-1}(X_k)F(X_k) $$ and instead of finding $J^{-1}(X_k)$, we solve for the change that will be applied to $X_k$ ($\Delta X_k$) using this relation $$J(X_k) \Delta X_k = -F(X_k)$$ and the next iteration will be this $$ X_{k+1} = X_k + \Delta X_k $$ My question is that wouldn't you need to solve for $J^{-1}(X_k)$ anyways when doing that? If so, why do we do that step in the first place?

 

[pasmith]

We don't solve linear systems numerically by finding an inverse matrix; that is incredibly inefficient. Instead we use other methods, like Gaussian elimination or LU decomposition.

 

[Ax_xiom]

We don't solve linear systems numerically by finding an inverse matrix; that is incredibly inefficient. Instead we use other methods, like Gaussian elimination or LU decomposition.

So what happens comptationally during the $J(X_k) \Delta X_k = -F(X_k)$ step? Do you express the problem as a series of linear equations and solve them?

Edit: I think that might be the case. I was under the impression that using Gaussian manipulation to solve a system of equations would give you the inverse matrix for free, but I don't think that is the case

 

[pasmith]

If $A$ is invertible, then solving $Ax = b$ by Gaussian elimination where $b$ consists of zeros except for a 1 in the $i$th row will give you the $i$th column of $A^{-1}$.

 

[Ax_xiom]

If $A$ is invertible, then solving $Ax = b$ by Gaussian elimination where $b$ consists of zeros except for a 1 in the $i$th row will give you the $i$th column of $A^{-1}$.

I'm assuming that this is computationally inefficient and it's much faster to just solve directly.

This is good to know but I attempted to implement this in Excel, and I was only doing it with a 3x3 matrix so it was much easier to just use Excel's MINVERSE() and MMULT() functions

 

[pasmith]

I'm assuming that this is computationally inefficient and it's much faster to just solve directly.

This is good to know but I attempted to implement this in Excel, and I was only doing it with a 3x3 matrix so it was much easier to just use Excel's MINVERSE() and MMULT() functions

That appears to be the only method Excel provides for solving $Ax = b$. The documentation does not explain what method Excel uses to calculate a matrix inverse, but assuming it is based on the standard LAPACK library it is using LU factorisation with partial pivoting.