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Home/ Questions/Q 8771839
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T17:53:28+00:00 2026-06-13T17:53:28+00:00

How does one solve a large system of linear equations efficiently when only a

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How does one solve a large system of linear equations efficiently when only a few of the constant terms change. For example:

I currently have the system Ax= b. I compute the inverse of A once, store it in a matrix and each time any entry updates in b perform a matrix-vector multiplication A^-1(b) to recompute x.

This is inefficient as only a couple of entries would have update in b. Are there more efficient ways of solving this system when A-1 remains constant but specific known values change in b?

I use uBlas and Eigen, but not aware of solutions that would address this problem of selective recalculation. Thanks for any guidance.

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  1. Editorial Team
    Editorial Team
    2026-06-13T17:53:29+00:00Added an answer on June 13, 2026 at 5:53 pm

    Compute A^-1. If b_i is the ith component of b, then examine d/db_i A^-1 b (the derivative of A^-1 with respect to the ith component of b) — it equals a column of A^-1 (in particular, the ith column). And derivatives of linear functions are constant over their domain. So if you have b and b', and they differ only in the ith component, then A^-1 b - A^-1 b' = [d/db_i A^-1] * (b-b')_i. For multiple components, just add them up (as A^-1 is linear).

    Or, in short, you can calculate A^-1 (b'-b) with some optimizations for input components that are zero (which, if only some components change, will be most of the components). A^-1 b' = A^-1 (b'-b) + A^-1 (b). And if you know that only some components will change, you can take a copy of the appropriate column of A^-1, then multiply it by the change in that component of b.

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