See matrix G:enter image description here

Why does the textbook say matrix G is not in reduced row echelon form? The zero row is at the bottom, and the other rows have leading 1s in appropriate locations. Thanks

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    $\begingroup$ Based on Wikipedia, reduced row echelon form requires the leading $1$'s to be the only non-zero numbers in their columns (i.e. there must be zeros above them, not just below). This is how it differs from the row echelon form (without the "reduced" specifier). $\endgroup$ – Peter Košinár Jan 7 '14 at 0:12
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    $\begingroup$ This is an answer. $\endgroup$ – Berci Jan 7 '14 at 0:22

For a matrix to be in Reduced Row Echelon Form it must satisfy the following conditions:

  1. The first non-zero entry in any row is the number $1$. These are called pivots. This implies that every row has a $0$/$1$ pivot. Also, the first non-zero element of any non-zero row appears in the later column (furthest to the right) than the first non-zero element of the preceding row.

  2. The pivot is the only non-zero entry in the column. This implies that each column can have a $0$/$1$ pivot.

  3. The rows are ordered so that any rows consisting of all $0$'s are at the bottom of the matrix, i.e. all non-zero rows precede zero rows.

While the matrix $G$ satisfies condition $1$, since the pivot of each row is the number $1$ and each pivot appears in a later column to the right and also satisfies condition $3$, since the row consisting pure $0$'s is at the bottom (the none-zero rows precede the zero row)...

$G$ however, does not satisfy condition $2$, as both columns $3$ and $4$ have non-zero entries besides their pivots.

The below matrix is $G$ in reduced row echelon form, notice that in all columns with leading ones, the number $1$ (the pivot), is only non-zero entry in the column.

$$G=\left[ \begin{array} 11&0&0&20\\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0\\ \end{array} \right] $$

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