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Linear algebra: Solution space
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[QUOTE="_Bd_, post: 2968037, member: 249159"] [h2]Homework Statement [/h2] Find a basis for, and the dimension of, th esolution space of Ax = b A=[1 3 -2 4] [0 1 -1 0] [-2 -6 4 8] [h2]Homework Equations[/h2] Allowed to use calculator [h2]The Attempt at a Solution[/h2] getting the RRE form yields: [1 0 1 0] [0 1 -1 0] [0 0 0 1] which means its a rank 3 matrix (since there are no dependent vectors. . .right?) the nullity for a mxn matrix is n-r . . .so 4 - 3 = 1. . .that would be the nullity . . .which from what I see in the book is the same as the dimension? it says: so then using the RREf and setting up a system of equations I get x_1 + 0 + x_3 + 0 = 0 0 + x_2 - x_3 + 0 = 0 0 + 0 + 0 + x_4 = 0 so then x_4 = 0 | x_3 = t = x_2 and x_1 = -t so then [u]my basis would be [-1, 1, 1, 0] my dimension would be: 1 [/u] . . .the back of the book says the following: dimension = 2 basis: {[-1, 1, 1, 0], [ 2, -2, 0, 1]} thank you. [/QUOTE]
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Linear algebra: Solution space
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