Consider the linear system $Ax=b$ where
$$A=\begin{bmatrix} 1&2&3&5\\ 2&4&8&12\\ 3&6&7&13 \end{bmatrix}\tag{1}$$
$$b=\begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix}\tag{2}$$
Let's use the augmented matrix
$$M=\begin{bmatrix} 1&2&3&5&b_1\\ 2&4&8&12&b_2\\ 3&6&7&13&b_3 \end{bmatrix}\tag{3}$$
to perform row operations.
I did the calculations that follow in the program Maple.
If we perform Gaussian elimination we obtain the matrix
$$U=\begin{bmatrix} 1&2&3&5&b_1\\ 0&0&2&2&b_2-2b_1\\ 0&0&0&0&b_3-5b_1+b_2\end{bmatrix}\tag{4}$$
My question regards what the reduced row echelon form of this matrix is. Maple tells me it is
$$R=\begin{bmatrix} 1&2&0&2&0\\ 0&0&1&1&0\\ 0&0&0&0&1\end{bmatrix}\tag{5}$$
I can't quite understand the last column. When I do the calculations by hand I get the following as the last column (starting from (4), divide the second row by $2$, then subtract $3$ times the second row from the first row)
$$\begin{bmatrix} b_1-3\left (\frac{b_2}{2}-b_1\right )\\ \frac{b_2}{2}-b_1\\ b_3-5b_1+b_2\end{bmatrix}\tag{6}$$
What am I missing?
User @jgd1729 provided the answer in the comments, and I want to add some information here about what I did in Maple to obtain the expected results.
I was using the command
but this reduced the entire augmented matrix including the last column representing $b$.
To obtain the $b$ that would be associated with the reduced row echelon system I did the following
This command provides the matrices in the factorization
$$A=P\cdot L\cdot U1\cdot R$$
where $P$ is a permutation matrix representing row exchanges, $L$ is the lower triangular inverse of the elimination matrix used to perform Gaussian elimination, $U1\cdot R$ is the Gaussian elimination reduced matrix, and $R$ by itself is the reduced row echelon form of $A$.
We can write
$$R=(P\cdot L\cdot U1)^{-1}A$$
Thus, when we go from $Ax=b$ to $Rx=d$, $d$ is
$$d=(P\cdot L\cdot U1)^{-1}b$$
In Maple, this is
which gives the result I got by hand