RREF of $$$\left[\begin{array}{ccc}2 & -1 & 2\\4 & -3 & h\end{array}\right]$$$

Find the reduced row echelon form of $$$\left[\begin{array}{ccc}2 & -1 & 2\\4 & -3 & h\end{array}\right]$$$.

Divide row $$$1$$$ by $$$2$$$: $$$R_{1} = \frac{R_{1}}{2}$$$.

$$$\left[\begin{array}{ccc}1 & - \frac{1}{2} & 1\\4 & -3 & h\end{array}\right]$$$

Subtract row $$$1$$$ multiplied by $$$4$$$ from row $$$2$$$: $$$R_{2} = R_{2} - 4 R_{1}$$$.

$$$\left[\begin{array}{ccc}1 & - \frac{1}{2} & 1\\0 & -1 & h - 4\end{array}\right]$$$

Multiply row $$$2$$$ by $$$-1$$$: $$$R_{2} = - R_{2}$$$.

$$$\left[\begin{array}{ccc}1 & - \frac{1}{2} & 1\\0 & 1 & 4 - h\end{array}\right]$$$

Add row $$$2$$$ multiplied by $$$\frac{1}{2}$$$ to row $$$1$$$: $$$R_{1} = R_{1} + \frac{R_{2}}{2}$$$.

$$$\left[\begin{array}{ccc}1 & 0 & 3 - \frac{h}{2}\\0 & 1 & 4 - h\end{array}\right]$$$

The reduced row echelon form is $$$\left[\begin{array}{ccc}1 & 0 & 3 - \frac{h}{2}\\0 & 1 & 4 - h\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 3 - 0.5 h\\0 & 1 & 4 - h\end{array}\right].$$$A