RREF of $$$\left[\begin{array}{cc}4 & 2\\2 & 1\end{array}\right]$$$

Find the reduced row echelon form of $$$\left[\begin{array}{cc}4 & 2\\2 & 1\end{array}\right]$$$.

Divide row $$$1$$$ by $$$4$$$: $$$R_{1} = \frac{R_{1}}{4}$$$.

$$$\left[\begin{array}{cc}1 & \frac{1}{2}\\2 & 1\end{array}\right]$$$

Subtract row $$$1$$$ multiplied by $$$2$$$ from row $$$2$$$: $$$R_{2} = R_{2} - 2 R_{1}$$$.

$$$\left[\begin{array}{cc}1 & \frac{1}{2}\\0 & 0\end{array}\right]$$$

Since the element at row $$$2$$$ and column $$$2$$$ (pivot element) equals $$$0$$$, we need to swap the rows.

Find the first nonzero element in column $$$2$$$ under the pivot entry.

As can be seen, there are no such entries.

The reduced row echelon form is $$$\left[\begin{array}{cc}1 & \frac{1}{2}\\0 & 0\end{array}\right] = \left[\begin{array}{cc}1 & 0.5\\0 & 0\end{array}\right].$$$A