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Which set spans the same set as $\left\lbrace(1,2,−1),(0,1,1),(2,5,−1)\right\rbrace$ ?

The answers choices each give either a set of 3 vectors or 2 vectors. How will I go about to solve this question?

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If you provide the options, it will be easier to determine a way to spot the correct answer for this particular case. – Git Gud Jan 26 '14 at 22:54
like one answer choice is {(1,1,1),(1,1,2)(2,1,1)} – user123204 Jan 26 '14 at 23:00
That is not the correct answer as that set is linearly independent and the one in the question isn't. – Git Gud Jan 26 '14 at 23:02
so I basically check the answer choices that have a determinant of 0? – user123204 Jan 26 '14 at 23:04
That might not be enough. Depends on the choices. – Git Gud Jan 26 '14 at 23:22

A foolproof way to show that $A$ and $B$ span the same space is to show that each vector in $A$ can be made via combinations of vectors from $B$ and vice versa. This is essentially just setting up some simultaneous linear equations and solving them, for which there exist a variety of methods.

You can make this easier by first altering $A$ or $B$ (or both) to make a set with the same span but "simpler" elements. In this case, as the other answer points out, you can discard the third vector, since you can make it with the first two. You can also combine the vectors in different ways to make them simpler, e.g. you could replace the first vector with, say, $v_1 + v_2 = (1,3,0)$, since you can get back to the original first vector by subtracting the second. The zero in the third component is especially useful: now if $(x,y,z)$ is going to be a combination of $v_1 = (1,3,0)$ and $v_2 = (0,1,1)$, you can immediately deduce it must be $xv_1 + zv_2$ for the first and third coefficients to match, so you just need to check if the second does too.

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Take just the first two vectors. The last is $2v_{1}+v_{2}$.

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How do you whether this is one of the possible choices given in this special excercise? – user127.0.0.1 Jan 26 '14 at 22:58
That's a good point. – TheNumber23 Jan 26 '14 at 22:59

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