# What is row reduced echelon form? How to row reduce this matrix?

I'm not being able to grasp the concept of row reduced echelon form. Please, explain how to row reduce one of the the following matrices.

$A = \begin{bmatrix} 1&3&4&5\\3&9&12&9\\1&3&4&1 \end{bmatrix}$

$B= \begin{bmatrix} 1&2&1&2\\0&1&0&1\\-1&2&0&3 \end{bmatrix}$

$\begin{array}{l} \begin{bmatrix} 1 & 3 & 4 & 5\\ 3 & 9 & 12 & 9\\ 1 & 3 & 4 & 1 \end{bmatrix}\overset{R_2:=3R_1 - R_2}{\to}\begin{bmatrix} 1 & 3 & 4&5\\0&0&0&6 \\ 1&3&4&1 \end{bmatrix}\overset{R_3:=R_1 - R_3}{\to}\begin{bmatrix} 1&3&4&5\\0&0&0&6\\0&0&0&4 \end{bmatrix}\overset{R_2:=\frac 16R_2}{\to}\begin{bmatrix}1&3&4&5\\ 0&0&0&1\\0&0&0&4\\\end{bmatrix}\\\overset{R_3:=4R_2-R_3}{\to} \begin{bmatrix}1&3&4&5\\0&0&0&1\\0&0&0&0\end{bmatrix}\overset{R_1:=R_1 - 5R_2}{\to}\begin{bmatrix}1&3&4&0\\ 0&0&0&1\\0&0&0&0 \end{bmatrix}. \end{array}$
You begin to put the matrix in echelon form, with the pivots equal to $1$, going downwards row after row. When that is done, all coefficients under a pivot (in the same column) are equal to $0$.Then you make the coefficients above the pivots equal to $0$, starting fom the last pivot and going upwards: \begin{align*} &\begin{bmatrix} 1&3&4&5\\3&9&12&9\\1&3&4&1 \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1&3&4&5\\0&0&0&-6\\0&0&0&-4\\ \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1&3&4&5\\0&0&0&1\\0&0&0&1 \end{bmatrix}\\ \rightsquigarrow &\begin{bmatrix} 1&3&4&5\\0&0&0&1\\0&0&0&0 \end{bmatrix}\rightsquigarrow \begin{bmatrix} 1&3&4&0\\0&0&0&1\\0&0&0&0 \end{bmatrix} \end{align*}