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
[Math] Why is this matrix not in reduced row echelon form
linear algebra
Related Solutions
When the matrix is in echelon form, for each non-zero row $R_i$, you can divide the row by its leading non-zero value. that makes the leading value $1$. Next for each row $R_n$ above $R_i$, you can subtract $R_i$ multiplied by the entry in row $R_n$ in the same column as that leading $1$ from $R_n$. This results in $R_n$ having $0$ in that column. If you follow this procedure starting with the first row and going down, by the time you are done, the matrix will be reduced echelon form, and guess what! Those leading $1$s that define the pivot points are exactly the locations of the leading non-zero values in each row back before it was reduced.
I.e., When a matrix is in echelon form, the pivot points are exactly the leading non-zero values in each row. Quite frankly, if I had written the definition, that's how I would have defined it, since the two are equivalent, and you need to know them before you get in reduced echelon form.
For example, in your matrix, I marked the leading non-zero entries in red:
$$\begin{bmatrix} \color{red}1 &4 &5 &-9 &7 \\ 0 &\color{red}2 &4 &-6 &-6 \\ 0 &0 &0 &\color{red}{-5} &0 \\ 0 &0 &0 &0 &0 \\ \end{bmatrix}$$ First divide each row by its leading non-zero value $$\begin{bmatrix} \color{red}1 &4 &5 &-9 &7 \\ 0 &\color{red}1 &2 &-3 &-3 \\ 0 &0 &0 &\color{red}1 &0 \\ 0 &0 &0 &0 &0 \\ \end{bmatrix}$$ Subtract $4$ times Row 2 from Row 1: $$\begin{bmatrix} \color{red}1 &0 &-3 &3 &19 \\ 0 &\color{red}1 &2 &-3 &-3 \\ 0 &0 &0 &\color{red}1 &0 \\ 0 &0 &0 &0 &0 \\ \end{bmatrix}$$ Subtract $3$ times row 3 from row 1, and add 3 times row 3 to row 2: $$\begin{bmatrix} \color{red}1 &0 &-3 &0 &19 \\ 0 &\color{red}1 &2 &0 &-3 \\ 0 &0 &0 &\color{red}1 &0 \\ 0 &0 &0 &0 &0 \\ \end{bmatrix}$$ And now, we are in reduced echelon form. See that the the pivot points from the definition are in the same locations as the echelon from leading non-zero values.
When you perform elementary row operations, then linear relations between columns don't change. More precisely, if $A=[a_1\;a_2\;\dots\;a_n]$ is a matrix (given with its columns $a_i$) and $B$ is obtained by $A$ by performing elementary row operations, if $1\le i_1,i_2,\dots,i_j,p\le n$ are column indices, then $$ b_p=\alpha_1b_{i_1}+\alpha_2b_{i_2}+\dots+\alpha_jb_{i_j} \quad\text{if and only if}\quad a_p=\alpha_1a_{i_1}+\alpha_2a_{i_2}+\dots+\alpha_ja_{i_j} $$ This follows from the fact that performing row operations is the same as multiplying $A$ on the left by invertible matrices. So $B=FA$ for some invertible matrix $F$ and proving the above results is easy.
In particular, a set of columns of $B$ is linearly independent if and only if the corresponding set of columns of $A$ is linearly independent.
If $U=[u_1\;u_2\;\dots\;u_n]$ is a reduced row echelon form for $A$, let $u_{i_1},u_{i_2},\dots,u_{i_j}$ be the pivot columns, with $i_1<i_2<\dots<i_j$.
Note that if $i_j<p<i_{j+1}$, then $u_p$ is a linear combination of $u_{i_1},\dots,u_{i_j}$ and the coefficients of the linear combination are determined by $u_p$: if $$ u_p=\begin{bmatrix}\alpha_1\\\alpha_2\\\vdots\\\alpha_m\end{bmatrix} $$ then $\alpha_i=0$ if $i>i_j$ and $$ u_p=\alpha_1u_{i_1}+\alpha_2u_{i_2}+\dots+\alpha_{i_j}u_{i_j} $$ Therefore $$ a_p=\alpha_1a_{i_1}+\alpha_2a_{i_2}+\dots+\alpha_{i_j}a_{i_j} $$ and this linear combination is unique, because the set of columns $$ \{a_{i_1},a_{i_2},\dots,a_{i_j}\} $$ is linearly independent.
Thus the position of the pivot columns in $U$ is uniquely determined by the columns of $A$ and the coefficients on the non-pivot columns are likewise determined by the linear relations between the columns of $A$.
Best Answer
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.
$$G=\left[ \begin{array} 11&0&0&20\\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0\\ \end{array} \right] $$