+2353 Morning!
Consider matrix $$A = \begin{pmatrix}
1 & 1 & 1 & 1 & 1\\
0 & 1 & 3 & 2 & 0\\
1 & 1 & 0 & 5 & 1\\
0 & 0 & 1 & -4 & 2\\
\end{pmatrix}$$
I want to find the row echolon form of A.
According to the solutions this is the first step toward that form.
$$\begin{pmatrix}
1 & 1 & 1 & 1 & 1\\
0 & 1 & 3 & 2 & 0\\
1 & 1 & 0 & 5 & 1\\
0 & 0 & 1 & -4 & 2\\
\end{pmatrix} \sim \begin{pmatrix}
1 & 1 & 1 & 1 & 1\\
0 & 1 & 3 & 2 & 0\\
0 & 0 & 1 & -4 & 2\\
0 & 0 & -1 & 4 & 0\\
\end{pmatrix}$$
It appear like $$R_3-R_1 \mbox{ has been applied, but } R_3-R_1 \mbox{ is } \begin{pmatrix}
1&1&0&5&1
\end{pmatrix}-\begin{pmatrix}
1&1&1&1&1
\end{pmatrix} = \begin{pmatrix}
0&0&-1&4&0
\end{pmatrix}.$$
What have I been missing here?
