RREF of $$$\left[\begin{array}{ccc}4 & 8 & 16\\4 & 0 & 8\\-4 & -4 & -12\end{array}\right]$$$
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Find the reduced row echelon form of $$$\left[\begin{array}{ccc}4 & 8 & 16\\4 & 0 & 8\\-4 & -4 & -12\end{array}\right]$$$.
Solution
Divide row $$$1$$$ by $$$4$$$: $$$R_{1} = \frac{R_{1}}{4}$$$.
$$$\left[\begin{array}{ccc}1 & 2 & 4\\4 & 0 & 8\\-4 & -4 & -12\end{array}\right]$$$
Subtract row $$$1$$$ multiplied by $$$4$$$ from row $$$2$$$: $$$R_{2} = R_{2} - 4 R_{1}$$$.
$$$\left[\begin{array}{ccc}1 & 2 & 4\\0 & -8 & -8\\-4 & -4 & -12\end{array}\right]$$$
Add row $$$1$$$ multiplied by $$$4$$$ to row $$$3$$$: $$$R_{3} = R_{3} + 4 R_{1}$$$.
$$$\left[\begin{array}{ccc}1 & 2 & 4\\0 & -8 & -8\\0 & 4 & 4\end{array}\right]$$$
Divide row $$$2$$$ by $$$-8$$$: $$$R_{2} = - \frac{R_{2}}{8}$$$.
$$$\left[\begin{array}{ccc}1 & 2 & 4\\0 & 1 & 1\\0 & 4 & 4\end{array}\right]$$$
Subtract row $$$2$$$ multiplied by $$$2$$$ from row $$$1$$$: $$$R_{1} = R_{1} - 2 R_{2}$$$.
$$$\left[\begin{array}{ccc}1 & 0 & 2\\0 & 1 & 1\\0 & 4 & 4\end{array}\right]$$$
Subtract row $$$2$$$ multiplied by $$$4$$$ from row $$$3$$$: $$$R_{3} = R_{3} - 4 R_{2}$$$.
$$$\left[\begin{array}{ccc}1 & 0 & 2\\0 & 1 & 1\\0 & 0 & 0\end{array}\right]$$$
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 reduced row echelon form is $$$\left[\begin{array}{ccc}1 & 0 & 2\\0 & 1 & 1\\0 & 0 & 0\end{array}\right]$$$A.