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RREF of $$$\left[\begin{array}{ccc}3 & 1 & 2\\-4 & 6 & 7\\2 & 8 & 9\end{array}\right]$$$

The calculator will find the reduced row echelon form of the $$$3$$$x$$$3$$$ matrix $$$\left[\begin{array}{ccc}3 & 1 & 2\\-4 & 6 & 7\\2 & 8 & 9\end{array}\right]$$$, with steps shown.

Related calculators: Gauss-Jordan Elimination Calculator, Matrix Inverse Calculator

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Your Input

Find the reduced row echelon form of $$$\left[\begin{array}{ccc}3 & 1 & 2\\-4 & 6 & 7\\2 & 8 & 9\end{array}\right]$$$.

Solution

Divide row $$$1$$$ by $$$3$$$: $$$R_{1} = \frac{R_{1}}{3}$$$.

$$$\left[\begin{array}{ccc}1 & \frac{1}{3} & \frac{2}{3}\\-4 & 6 & 7\\2 & 8 & 9\end{array}\right]$$$

Add row $$$1$$$ multiplied by $$$4$$$ to row $$$2$$$: $$$R_{2} = R_{2} + 4 R_{1}$$$.

$$$\left[\begin{array}{ccc}1 & \frac{1}{3} & \frac{2}{3}\\0 & \frac{22}{3} & \frac{29}{3}\\2 & 8 & 9\end{array}\right]$$$

Subtract row $$$1$$$ multiplied by $$$2$$$ from row $$$3$$$: $$$R_{3} = R_{3} - 2 R_{1}$$$.

$$$\left[\begin{array}{ccc}1 & \frac{1}{3} & \frac{2}{3}\\0 & \frac{22}{3} & \frac{29}{3}\\0 & \frac{22}{3} & \frac{23}{3}\end{array}\right]$$$

Multiply row $$$2$$$ by $$$\frac{3}{22}$$$: $$$R_{2} = \frac{3 R_{2}}{22}$$$.

$$$\left[\begin{array}{ccc}1 & \frac{1}{3} & \frac{2}{3}\\0 & 1 & \frac{29}{22}\\0 & \frac{22}{3} & \frac{23}{3}\end{array}\right]$$$

Subtract row $$$2$$$ multiplied by $$$\frac{1}{3}$$$ from row $$$1$$$: $$$R_{1} = R_{1} - \frac{R_{2}}{3}$$$.

$$$\left[\begin{array}{ccc}1 & 0 & \frac{5}{22}\\0 & 1 & \frac{29}{22}\\0 & \frac{22}{3} & \frac{23}{3}\end{array}\right]$$$

Subtract row $$$2$$$ multiplied by $$$\frac{22}{3}$$$ from row $$$3$$$: $$$R_{3} = R_{3} - \frac{22 R_{2}}{3}$$$.

$$$\left[\begin{array}{ccc}1 & 0 & \frac{5}{22}\\0 & 1 & \frac{29}{22}\\0 & 0 & -2\end{array}\right]$$$

Divide row $$$3$$$ by $$$-2$$$: $$$R_{3} = - \frac{R_{3}}{2}$$$.

$$$\left[\begin{array}{ccc}1 & 0 & \frac{5}{22}\\0 & 1 & \frac{29}{22}\\0 & 0 & 1\end{array}\right]$$$

Subtract row $$$3$$$ multiplied by $$$\frac{5}{22}$$$ from row $$$1$$$: $$$R_{1} = R_{1} - \frac{5 R_{3}}{22}$$$.

$$$\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & \frac{29}{22}\\0 & 0 & 1\end{array}\right]$$$

Subtract row $$$3$$$ multiplied by $$$\frac{29}{22}$$$ from row $$$2$$$: $$$R_{2} = R_{2} - \frac{29 R_{3}}{22}$$$.

$$$\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]$$$

Answer

The reduced row echelon form is $$$\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]$$$A.