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Inverse matrices are close iff matrices are close

Consider this problem:

$Ax = b$

I want to solve it/find x and the matrix A is ill-conditioned. Why is the fact "A is ill-conditioned" a "bad" thing?

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marked as duplicate by Did, Listing, J. M., Ilya, t.b. Oct 28 '11 at 14:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2 Answers 2

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Because if one knows the coefficients of $A$ only up to a given precision, a small variation in them will cause a huge variation in the coefficients of $A^{-1}$, hence, presumably, in the solution $x=A^{-1}b$. Alternatively, even if $A^{-1}$ is known with an absolute precision, if one knows the coefficients of $b$ only up to a given precision, a small variation in them will cause a notable variation in the coefficients of the solution $x=A^{-1}b$ since some coefficients of $A^{-1}$ are large. In real life, both effects are often conspiring. See the definition of the condition number of a matrix.

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Do you plan to ask over and over the same question here? –  Did Oct 28 '11 at 8:43

A matrix is ill-conditioned if the condition number is too large .

Condition number is the ratio C of the largest to smallest singular value in the singular value decomposition of a matrix.

The base-b logarithm of C is an estimate of how many base-b digits are lost in solving a linear system with that matrix. In other words, it estimates worst-case loss of precision.

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