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Does this matrix have 1 or 2 pivot columns?

\begin{bmatrix}1& 1 &3 &-7 &-2\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0 \end{bmatrix}

I think it's 1 but I don't know if the second 1 counts in the second column.

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If you take it column-wise you could take any element as pivot, depending on your definition. –  user127.0.0.1 Feb 27 at 7:40
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Pivot is defined to be the first $1$ in each row. So in your matrix, there is only one nonezero row (the first row), and there is only one pivot, and there is only one pivot column (the first column).

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I do not agree, you are free to define the pivot nearly as you want to. eg in numerical algorithms you would usually take the $-7$ here, as it is the element with max abs-value // However, I think the OP has to clarify the question. –  user127.0.0.1 Feb 27 at 7:42
    
I am thinking about the pivot in the rref form. In Gilbert Strang's Linear Algebra, pivot in the rref form is defined to be the first $1$ in each row, and the pivot column is the column which contains pivot. But I think others have some other ways to define pivot, as you said. –  Paul Feb 27 at 7:49
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