Hint :
Well it's clear from the hint I gave you that every vector in your subspace can be written as a linear combination of the two vectors above. So the two vectors form a basis if and only if they are linearly independent. It's very easy to check for linear independence when you have only two vectors; two vectors are linearly dependent if and only if one is a multiple of the other. It's clearly not the case here! Your method is also valid (though I'm not sure why you added a third zero column to the matrix you reduced.)
Thank you that makes sense and actually helps me a lot about understanding these.
One quick (and kind of a stupid question) but could you please quickly explain what you mean by "one is a multiple of the other."
I added the extra columns of zero's cause when i plug the matrix into my calculator to get the rref it needed extra columns of zeros in order to calculate it without my calculator saying error
Thanks again