Asked by Krzysztof
on 21 Mar 2014

The documentation for inv says:

A frequent misuse of inv arises when solving the system of linear equations Ax = b. One way to solve this is with x = inv(A)*b. A better way, from both an execution time and numerical accuracy standpoint, is to use the matrix division operator x = A\b.

The above taken for granted, is it nevertheless reasonable to precalculate invA = inv(A) in front of a loop containing (A\Xi), especially one that necessarily sequential?

Opportunities for recent engineering grads.

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