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I know a matrix can have different basis but here I am not sure how to efficiently check which answer is correct. We have to be able to answer this in under 3 min so I can't check each one individualy? Do you have a trick to go faster than row reducing A?

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    I know that every basis of a subspace has the same dimention. So if I just look at the dimention of my basis for col, nul .. etc? – 2017-01-21
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    Well, you can go through the definition of a basis: just check that the given sets are linearly independent and that every element in the claim vector space can be written as a linear combination of the elements in the claimed basis – 2017-01-21
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    @Danielbut that would take more than 3 min to do if you have to check for each basis. – 2017-01-21

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So I followed @daniel's advice and I check if a corresponding element in A would be a lin combination and you realise pretty quickly that the 1st and last answer fail to be consistent.

Plus the 3rd answer does not make sense since col (A)^T is in R4 but the base is given in R3 so the only one left is Nul(A) which is easy to check by solving A(basisNUL) = 0 which works so we can confirm that the 2nd answer is correct!

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    In the 3rd answer, there's no transpose; that $\perp$ sign means the space orthogonal to the column space, and that is a subspace of ${\bf R}^3$. So, you have to find some other way to deal with the 3rd answer. – 2017-01-21
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    Yes sorry. By multiplying a column of A by a column of that basis I should get 0 then – 2017-01-21