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Oct22-11, 05:29 AM
Sci Advisor
PF Gold
P: 39,488
To add just a little bit, you are saying that for any vector in the solution space, [itex]<x_1, x_2, x_3, x_4>[/itex], we must have [itex]x_2= -x_1- x_3[/itex] and [itex]x_4= (1/3)x_3- (1/3)x_1[/itex]. That is, [itex]<x_1, x_2, x_3, x_4>= <x_1, -x_1- x_3, x_3, (1/3)x_3- (1/3)x_4>= x_1<1, -1, 0, -1/3>+ x_3<0, -1, 1, 1/3>[/itex] which makes it clear what a basis is.

Mark44 is saying that [itex]x_1= x_3- 3x_4[/itex] and [itex]x_2= -4x_3+ 3x_4[/itex] so that [itex]<x_1, x_2, x_3, x_4>= <x_3- 3x_4, -4x_3+ 3x_4, x_3, x_4>= x_3< 1, -4, 1, 0>+ x_4<-3, 3, 0, 1>[/itex]. That gives another basis for the same subspace.