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## Main Question or Discussion Point

Hello. First, I'd like to apologize because I don't know where to go ask for homework on linear algebra on the forums so if anyone could please let me know, that would be appreciated.

so the matrix A is

( 2 4 -2 3)

(-2 -2 2 -4)

(1 3 -1 1)

and the book puts it in reduced row echelon form which comes out to

( 1 0 -1 5/2 )

( 0 1 0 -1/2)

(0 0 0 0 )

Thus a basis for the subspace is { (1, 0, -1, 5/2), (0, 1, 0, -1/2) }

Is this the only answer you can get or can I get different answers.

Say, if I used paramters "a" and "b" for column x3 and x4 respectively, then I can get

x1 = a - (5/2)b

x2 = (1/2)b

x3 = a

x4 = b

then if I let a = 0 and b =1, I'd get:

(-5/2, 1/2, 0, 1)

then if I let a = 1 and b = 0, I'd get:

(1, 0, 1, 0)

thus a basis for the subspace is { (-5/2, 1/2, 0, 1), (1, 0, 1, 0) }

Could that be another answer?

Also, I thought for it to be a basis, it has to be linearly independent. Are they assuming that by inspection it's linearly independent so that's why the book isn't testing for independence?

Thank you so much for any guidance.

**Here's the question:**Find a basis for the subspace of R^4 spanned by the given vectors**Here's the answer at the end of the textbook:**so the matrix A is

( 2 4 -2 3)

(-2 -2 2 -4)

(1 3 -1 1)

and the book puts it in reduced row echelon form which comes out to

( 1 0 -1 5/2 )

( 0 1 0 -1/2)

(0 0 0 0 )

Thus a basis for the subspace is { (1, 0, -1, 5/2), (0, 1, 0, -1/2) }

**My question is:**Is this the only answer you can get or can I get different answers.

Say, if I used paramters "a" and "b" for column x3 and x4 respectively, then I can get

x1 = a - (5/2)b

x2 = (1/2)b

x3 = a

x4 = b

then if I let a = 0 and b =1, I'd get:

(-5/2, 1/2, 0, 1)

then if I let a = 1 and b = 0, I'd get:

(1, 0, 1, 0)

thus a basis for the subspace is { (-5/2, 1/2, 0, 1), (1, 0, 1, 0) }

Could that be another answer?

Also, I thought for it to be a basis, it has to be linearly independent. Are they assuming that by inspection it's linearly independent so that's why the book isn't testing for independence?

Thank you so much for any guidance.