Why is this matrix not in reduced row echelon form? See matrix G:
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
 A: For a matrix to be in Reduced Row Echelon Form it must satisfy the following conditions:


*

*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.

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

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