Ytukyg

I have a system of $N$ linear equations, $Ax=b$, in $N$ unknowns (where $N$ is large).

If I am interested in the solution for only one of the unknowns, what are the best approaches?

For example, assume $N=50,000$. We want the solution for $x_1$ through $x_{100}$ only. Is there any trick that does not require $O(n^{3})$ (or $O$(matrix inversion))?

There is a way to reduce the complexity and make the system solvable in parallel.
It is called Diakoptics (a method invented by Gabriel Kron). The methods primary use is for large electrical networks that have few interconnections like power grids. But you should be able to adapt it.

The complexity (for the case below) is reduced from $O(n^3)$ to $O(2(frac{n}{2})^3)$ or $O(frac{1}{4}n^3)$, the impact can be much greater if the system is divided it into more subsystems. For that case the complexity is ($s$-subsysems, $c$-interconnection points) $O(c^3)+O((frac{n^3}{s²}))$, if the systems is divided into equaly sized subsystems. I'm not sure about the notation for multiple variables, but you should get the point.

In short:

Lets assume you have a $N times N$ system, lets say you can divide the system into two systems with 1 connection point(plus reference when you look at electrical systems). The connection points are $m$ and $n$. Lets assume these systems are of the size $N_1=N/2$ and $N_2=N/2$ (for simplicitys sake). You should now solve them separately.

$mathbf A_1^{-1}=mathbf B_1$

$mathbf A_2^{-1}=mathbf B_2$

The next step is to put them back together, that is done with the help of the so called "Thevenin Matrix"(in our case it is 1$times$1). You can look up the exact principle for higher orders(more connection points), but for this example it looks like:
begin{align}
mathbf{B_{TH}}=B_{1mm}+B_{2nn}-2B_{mn}
end{align}
For our case we have $B_{mn}=0$. Now we need the solutions $x_1$ and $x_2$ to form the coefficients $b_{th}$.

$mathbf x_{th}=x_{1m}-x_{2n}$

$mathbf b_{p}=mathbf{B_{TH}}^{-1} mathbf x_{th}$

begin{align}
mathbf b_{th}=begin{bmatrix}0&cdots&b_{p}&cdots&-b_{p}&cdots&0
end{bmatrix}^T
end{align}

The $mathbf b_{th}$ matrix only has nonzero elements at $m$ an $N/2 +n$. Now we can finally find the solution $x_n$ for the whole system:
begin{align}
mathbf x_n=begin{bmatrix}x_1\x_2
end{bmatrix}-begin{bmatrix}B_1&0\0&B_2
end{bmatrix}begin{bmatrix}b_{th}
end{bmatrix}
end{align}

I'm more used to the engineering notation with $Z, I, U$ and so on, so excuse for non-standard symbol usage.

It is an open important mathematical question

The Wikipedia page https://en.wikipedia.org/wiki/Matrix_multiplication_algorithm#Sub-cubic_algorithms shows how the fast known asymptotic algorithm has evolved through time:

This begs the question: how close can we get to O(n^2), which is of course a lower bound since we have to read 2 * n^2 inputs at least once?

Proving the upper lower bound, even non constructively, would make you instantly famous. Wikipedia lists it as an "Unsolved problem in computer science".

The constant factor for the better algorithms is so large though that it makes them impractical for most matrix sizes found in practice today, and I doubt this will change if new algorithms are found. Therefore any practical real world answer will come down to optimizing against a given machine model.

Personal guess about solving for single variables: I don't think you can reduce complexity like that in general, since the entire system is coupled. How would you know that your solution for some variables also satisfies the entire global solution?

Unless your matrix is sparse or structured (e.g. Vandermonde, Hankel, or those other named matrix families that admit a fast solution method), there is not much hope of doing things better than $O(n^3)$ effort. Even if one were to restrict himself to solving for just one of the 50,000 variables, Cramer will demand computing two determinants for your answer, and the effort for computing a determinant is at least as much as decomposing/inverting a matrix to begin with.