REF of $$$\left[\begin{array}{ccc}1 & 2 & 3\\2 & 5 & 4\\1 & 1 & 5\end{array}\right]$$$
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Find the row echelon form of $$$\left[\begin{array}{ccc}1 & 2 & 3\\2 & 5 & 4\\1 & 1 & 5\end{array}\right]$$$.
Solution
Subtract row $$$1$$$ multiplied by $$$2$$$ from row $$$2$$$: $$$R_{2} = R_{2} - 2 R_{1}$$$.
$$$\left[\begin{array}{ccc}1 & 2 & 3\\0 & 1 & -2\\1 & 1 & 5\end{array}\right]$$$
Subtract row $$$1$$$ from row $$$3$$$: $$$R_{3} = R_{3} - R_{1}$$$.
$$$\left[\begin{array}{ccc}1 & 2 & 3\\0 & 1 & -2\\0 & -1 & 2\end{array}\right]$$$
Add row $$$2$$$ to row $$$3$$$: $$$R_{3} = R_{3} + R_{2}$$$.
$$$\left[\begin{array}{ccc}1 & 2 & 3\\0 & 1 & -2\\0 & 0 & 0\end{array}\right]$$$
Since the element at row $$$3$$$ and column $$$3$$$ (pivot element) equals $$$0$$$, we need to swap the rows.
Find the first nonzero element in column $$$3$$$ under the pivot entry.
As can be seen, there are no such entries.
Answer
The row echelon form is $$$\left[\begin{array}{ccc}1 & 2 & 3\\0 & 1 & -2\\0 & 0 & 0\end{array}\right]$$$A.