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Is this what you mean? I'm not sure, because should the zero vector be a solution for you?

sage: A = matrix([[1,2],[2,4]])
sage: A.solve_right(vector([0,0]))
(0, 0)
sage: A\vector([1,2])
(1, 0)
sage: A*vector([1,0])
(1, 2)

Or maybe you wanted this.

sage: A.right_kernel()
Free module of degree 2 and rank 1 over Integer Ring
Echelon basis matrix:
[ 2 -1]

I hope I'm not misunderstanding something here. See this Linear Algebra quickref card for a lot more information.