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All of the examples I can find on youtube show how to check if three $3\times 1$ vectors are independent by combining them into a matrix and checking if the determinant is $0$. However, since a determinant only exists for square matrices, how can you check if only two vectors are linearly independent (procedurally rather than by inspection)?

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    You don't always have to use the determinant, you could row reduce and check the number of pivot columns, though for two vectors it's genuinely very straightforward to see if one is a multiple of the other. You could also use the dot product as linearly independent vectors lie on the same line and are parallel/antiparallel – 2017-01-20
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    Would you please elaborate a little bit more on how to determine this using the dot product? – 2017-01-20
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    Well if the vectors are parallel then the dot product is exactly equal to the product of their magnitudes, anything less and they do not lie on the same line – 2017-01-20
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    Thanks that makes sense after looking up what the actual geometric interpretation of linear independence is – 2017-01-20
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    It's one of many useful tools besides gaussian elimination – 2017-01-20

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Two nonzero vectors $v$ and $w$ are linearly independent if and only if they are not collinear, i.e., not of the form $w=\lambda v$ for nonzero $\lambda$. This is much easier than to compute a determinant, of course.

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    But how can you determine if there exists a lambda that would make this equality true? (rather than just being comfortable enough with linear algebra to be able to eye it up, for example say I wanted an algorithm that I can do by hand to check) – 2017-01-20
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    If there is such a $\lambda$, then you have $v_k=\lambda w_k$ for every dimension $k$. So do $\lambda = v_x/w_x$, and then see if that $\lambda$ works for the other dimensions. (obviously, pick a dimension in which $w_x \ne 0$ – 2017-01-20
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For two $3\times 1$ vectors $u$ and $v$, you can compute their cross-product $w=u\times v$.

$u$ and $v$ are linearly dependent iff $w=0$

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    Does this always work or only in particular cases? – 2017-01-20
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    This works for all vectors in the 3-dimensional case; other numbers of dimensions do not admit a cross product, so it doesn't work outside of 3 dimensions. – 2017-01-20