# solving homogeneous system of linear equations

 1 The system is written on the form Ax=0. I know this can be done by using, for example, solve([eq1==0,eq2==0],x1,x2) But this is somewhat complex. I wonder if the system can be directly solved by A.solve_right or some other simpler notation? asked Dec 14 '11 Anonymous Anonymous

 2 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. posted Dec 14 '11 kcrisman 7427 ● 17 ● 76 ● 166 the right_kernel is exactly what I want. Thanks very much for your answer and the resources provided!lainme (Dec 14 '11)

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