Row Reduce Echelon Form on 3x4 Matrix I understand the rules for RREF are:
1) Each leading entry must be a 1 in each row
2) Each leading entry's column must be 0's other than the leading entry
3) In stair case order, the next element of the next row is suppose to be the next leading entry
So now, while RREFing we reach a value which is a 0 in staircase order. Since it is a 0, we move on to the next row.
But, the solution seems to be treating the next column entry in the same row there is a 0 value, and treating it as the leading value. I'm confused since how is this the stair case pattern?
Here is the question:


Here is the answer:


Here is what I think the stair case pattern is and the columns I need to 0 out other than the leading values


 A: Note that the algorithm as commonly defined asks you to get the current pivot point to be a nonzero number (by rowswapping with a row below it if necessary), then make that into a one (by multiplying by it's row by the pivot point's multiplicative inverse if necessary), and then zeroing out the rest of the entries in the pivot's column (by subtracting multiples of the pivot's row from the rest of the rows).
As in this case, once you pivot around the firstrow first column entry and zero out the rest of it's column, and attempt to shift your focus to the entry down and to the right to use as the new pivot point, you will find that it is a zero.  You will find further that you cannot swap with any rows below it to make it a nonzero (we do not want to swap with a row above it since that will damage our previous work in getting a pivot in the correct location in the upper row).  Our algorithm should have a clause written in it that if it is impossible to make our proposed pivot point nonzero by rowswapping with a row below it then shift your attention one space to the right and try to use that as the new pivot point for the row instead.

Remember the definition of a matrix to be in reduced row echelon form:


*

*The furthest left nonzero entry of each row is a 1

*The furthest left nonzero entry of each row has all entries down and/or to the left as zeroes

*If a row (or rows) of all zeroes occurs, it occurs (they occur) at the bottom

*The furthest left nonzero entry of each row has all other entries in its column as zeroes.


note, that the definition allows for some "stairs" to be "landings" instead of "steps" (metaphorically speaking).  In otherwords, we don't require that it be strictly diagonal.  $\begin{bmatrix}1 & 5 & 0\\ 0 & 0 & 1\end{bmatrix}$ is still considered to be in reduced row echelon form.

The algorithm (as I usually explain it)
From here out, refer to the entry that has our attention as $\color{blue}{\text{it}}$ and the row in which the entry is as $\color{blue}{\text{its row}}$ and the column in which the entry is as $\color{blue}{\text{its column}}$.  If rowswapping, $\color{blue}{\text{it}}$ will refer to the entry that takes $\color{blue}{\text{it}}$s place (i.e. refers to the same row/column entry).  Throughout the algorithm we will change focus from one $\color{blue}{\text{it}}$ to another.
Begin by setting $\color{blue}{\text{it}}$ as the first row first column entry.


*

*If $\color{blue}{\text{it}}$ ever refers to an entry outside of the dimension of the matrix, then exit the algorithm.

*$\color{red}{\star}$: If $\color{blue}{\text{it}}$ is zero, then if possible, rowswap $\color{blue}{\text{its row}}$ with a row below it (never above it) in order to make $\color{blue}{\text{it}}$ nonzero.  

*
*

*In the case that this is impossible (since all entries below $\color{blue}{\text{it}}$ in $\color{blue}{\text{its column}}$ are zero) then change your focus to a new $\color{blue}{\text{it}}$ one column to the right but in the same row and then return to $\color{red}{\star}$.


*If $\color{blue}{\text{it}}$ is not equal to $1$, then multiply all entries of $\color{blue}{\text{its row}}$ by $\frac{1}{\color{blue}{\text{it}}}$

*If any entry in $\color{blue}{\text{its column}}$ (other than $\color{blue}{\text{it}}$) are nonzero, then subtract an appropriate multiple of $\color{blue}{\text{its row}}$ from the row in question to make the entry zero.  Repeat this step until all other entries in $\color{blue}{\text{its column}}$ are zero.  Once done, shift your focus to a new $\color{blue}{\text{it}}$ one column to the right and one row down and then return to $\color{red}{\star}$


Following this algorithm for a matrix will always return the reduced row echelon form of the matrix.
A: The easiest way to possible go about this is just to try to eliminate the coefficient in each column so that there is only one 1 and all 0s.
The appearances of the 1s don't necessarily need to follow a "perfect" staircase pattern like you drew.  
