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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.

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

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

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In order to obtain the reduced row echelon form (rref) of a matrix, we apply some row operations. According to this article, for the first case we have:

$\begin{array}{l} \begin{bmatrix} 1 & 3 & 4 & 5\\ 3 & 9 & 12 & 9\\ 1 & 3 & 4 & 1 \end{bmatrix}\overset{R_2:=3R_1 - R_2}{\to}\begin{bmatrix} 1 & 3 & 4&5\\0&0&0&6 \\ 1&3&4&1 \end{bmatrix}\overset{R_3:=R_1 - R_3}{\to}\begin{bmatrix} 1&3&4&5\\0&0&0&6\\0&0&0&4 \end{bmatrix}\overset{R_2:=\frac 16R_2}{\to}\begin{bmatrix}1&3&4&5\\ 0&0&0&1\\0&0&0&4\\\end{bmatrix}\\\overset{R_3:=4R_2-R_3}{\to} \begin{bmatrix}1&3&4&5\\0&0&0&1\\0&0&0&0\end{bmatrix}\overset{R_1:=R_1 - 5R_2}{\to}\begin{bmatrix}1&3&4&0\\ 0&0&0&1\\0&0&0&0 \end{bmatrix}. \end{array}$

The last matrix satisfies all the conditions of the reduced row echelon form of a matrix.

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Reduced row echelon form is a specific echelon form of a matrix that can be obtained by modifying a matrix with basic row operations, such that every leading coefficient is 1 and is the only nonzero entry in its column.

Useful: http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi?c=rref

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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*}

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