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The LinearSolve command instead of Solve

Posted 10 years ago

lets say I have something easy AX=B (matrix)

x+2y=10 3x+0y=3

A={{1,2},{3,0}};B={{11},{3}};X={{x}},{y}};Inverse[A].B

This solves the system, if i have a bigger system how does the command LinearSolve works, i dont find any example

Is that a better solution?

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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