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Consider the matrix

$$ \left( \begin{matrix} 1 & 4 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{matrix} \right) $$ According to my textbook, this matrix is in reduced echelon form. However the rules of reduce echelon form state that all other elements in a column that contains a leading 1 are zero.

Would the 4 at position (1,2) not break that rule and make this matrix not be in reduced echelon form?

Thank you.

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    $\begingroup$ ??? That $4$ is not a "leading $1$". Look up the definition of "leading $1$"... $\endgroup$ Sep 25, 2018 at 14:53
  • $\begingroup$ Yup, just mixed it up $\endgroup$
    – GreenSaber
    Sep 25, 2018 at 14:58

2 Answers 2

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Well, what's the problem then? In every column that contains a leading $1$ all other elements are zeros. The $4$ is in the second column which has no leading $1$. Don't get confused between rows and columns.

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  • $\begingroup$ Thank you, I mixed up the row and column, I see how it works now! $\endgroup$
    – GreenSaber
    Sep 25, 2018 at 14:57
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Let us define "pivotal column" as the column containing the pivot.

Now the column having element 4 is NOT a pivotal column.

According to echelon form, only pivotal column has all elements (in the column) zero, except the pivot itself.

Since second column is not pivotal, it is perfectly a row-reduced echelon form.

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  • $\begingroup$ Thank you! I see how it's reduced now, that makes a lot of sense! $\endgroup$
    – GreenSaber
    Sep 25, 2018 at 14:58

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