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 Math Emeritus Sci Advisor Thanks PF Gold P: 39,563 To add just a little bit, you are saying that for any vector in the solution space, , we must have $x_2= -x_1- x_3$ and $x_4= (1/3)x_3- (1/3)x_1$. That is, $= = x_1<1, -1, 0, -1/3>+ x_3<0, -1, 1, 1/3>$ which makes it clear what a basis is. Mark44 is saying that $x_1= x_3- 3x_4$ and $x_2= -4x_3+ 3x_4$ so that $= = x_3< 1, -4, 1, 0>+ x_4<-3, 3, 0, 1>$. That gives another basis for the same subspace.