Loop hole of finding rank using Minor method

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According to definition of Rank,

Rank is the number of dimensions in column space of a matrix. So rank can be found from reduced echelon form.

But in another definition, that is the reduction of the matrix to Minors, the size of the largest non-zero minor is the rank.

Taking this example,

\begin{bmatrix}0&1&0\\0&0&0\\0&0&1\end{bmatrix}

Applying the first definition , it can be seen the column 2 and column 3 are linearly independent and hence rank is 2.

But upon applying the second defination all the 2x2 minors are zero, which by definition gives a rank 1.

I know rank 2 is the correct answer. So what is it that I'm missing ? Are both these definitions correct ?

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Delete the first column and the second row. The resulting $2\times2$ minor is $1$.