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There are sixteen $ 2 $ by $ 2 $ matrices whose entries are $1$'s and $0$'s. How many are invertible?

I have already asked a similar question a while back; however, seeing as I am so terrible at counting, I need some more help. So, if I am not mistaken, $\det(A)=0$ if and only if the columns of $A$ are linearly dependent. So, I thought that counting the total number of ways of forming the first column and and ways that the second column could be a multiple of the first would solve this problem.

For the first column $(x,y)$, there are two possible values of $x$ and $y$, giving us four distinct vectors. Since there are only two possible multiples of this vector, this gives $4 \cdot 2 = 8$ total matrices that are not invertible and therefore $16-8=8$ that are. Sadly, the answer is $6$.

Where did I go wrong in my counting.

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    If the first column is $(0,0)$, then any second column makes the matrix non-invertible. You are correct for the other three possibilities. So it's $16 - (2\cdot 3 + 4) = 6$ – 2017-02-09
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    The first column can be any nonzero vector; with zeros and ones, $3$ possibilities. The second column can be any vector which is not a scalar multiple of the first. With entries restricted to zeros and ones, the only scalar multiples are the vector itself and the zero vector, so there are $4-2$ choices for the second column. The number of matrices is therefore $3\cdot2=6.$ – 2017-02-09

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