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GROUPS:

The LinearSolve command instead of Solve

Posted 12 years ago
POSTED BY: peter B

Umm, what did you try? LinearSolve works like the documentation states. For example:

A = {{1, 2}, {3, 0}};
B = {{11}, {3}};
LinearSolve[A, B] 

(* Out[186]= {{1}, {5}} *)

This agrees with multiplying on the left by the inverse matrix.

Inverse[A].B

(* Out[187]= {{1}, {5}} *)
POSTED BY: Daniel Lichtblau
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