RREF of $$$\left[\begin{array}{cccc}\frac{5}{2} & 120 & \frac{6}{5} & \frac{37}{10}\\5 & 240 & \frac{12}{5} & \frac{37}{5}\\3 & 180 & \frac{9}{5} & \frac{11}{2}\end{array}\right]$$$

The calculator will find the reduced row echelon form of the $$$3$$$x$$$4$$$ matrix $$$\left[\begin{array}{cccc}\frac{5}{2} & 120 & \frac{6}{5} & \frac{37}{10}\\5 & 240 & \frac{12}{5} & \frac{37}{5}\\3 & 180 & \frac{9}{5} & \frac{11}{2}\end{array}\right]$$$, with steps shown.

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

$$$\times$$$
A

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please contact us.

Your Input

Find the reduced row echelon form of $$$\left[\begin{array}{cccc}\frac{5}{2} & 120 & \frac{6}{5} & \frac{37}{10}\\5 & 240 & \frac{12}{5} & \frac{37}{5}\\3 & 180 & \frac{9}{5} & \frac{11}{2}\end{array}\right]$$$.

Solution

Multiply row $$$1$$$ by $$$\frac{2}{5}$$$: $$$R_{1} = \frac{2 R_{1}}{5}$$$.

$$$\left[\begin{array}{cccc}1 & 48 & \frac{12}{25} & \frac{37}{25}\\5 & 240 & \frac{12}{5} & \frac{37}{5}\\3 & 180 & \frac{9}{5} & \frac{11}{2}\end{array}\right]$$$

Subtract row $$$1$$$ multiplied by $$$5$$$ from row $$$2$$$: $$$R_{2} = R_{2} - 5 R_{1}$$$.

$$$\left[\begin{array}{cccc}1 & 48 & \frac{12}{25} & \frac{37}{25}\\0 & 0 & 0 & 0\\3 & 180 & \frac{9}{5} & \frac{11}{2}\end{array}\right]$$$

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

$$$\left[\begin{array}{cccc}1 & 48 & \frac{12}{25} & \frac{37}{25}\\0 & 0 & 0 & 0\\0 & 36 & \frac{9}{25} & \frac{53}{50}\end{array}\right]$$$

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.

The first nonzero element is at row $$$3$$$.

Swap the rows $$$2$$$ and $$$3$$$:

$$$\left[\begin{array}{cccc}1 & 48 & \frac{12}{25} & \frac{37}{25}\\0 & 36 & \frac{9}{25} & \frac{53}{50}\\0 & 0 & 0 & 0\end{array}\right]$$$

Divide row $$$2$$$ by $$$36$$$: $$$R_{2} = \frac{R_{2}}{36}$$$.

$$$\left[\begin{array}{cccc}1 & 48 & \frac{12}{25} & \frac{37}{25}\\0 & 1 & \frac{1}{100} & \frac{53}{1800}\\0 & 0 & 0 & 0\end{array}\right]$$$

Subtract row $$$2$$$ multiplied by $$$48$$$ from row $$$1$$$: $$$R_{1} = R_{1} - 48 R_{2}$$$.

$$$\left[\begin{array}{cccc}1 & 0 & 0 & \frac{1}{15}\\0 & 1 & \frac{1}{100} & \frac{53}{1800}\\0 & 0 & 0 & 0\end{array}\right]$$$

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. Move to the next column.

Since the element at row $$$3$$$ and column $$$4$$$ (pivot element) equals $$$0$$$, we need to swap the rows.

Find the first nonzero element in column $$$4$$$ under the pivot entry.

As can be seen, there are no such entries.

Answer

The reduced row echelon form is $$$\left[\begin{array}{cccc}1 & 0 & 0 & \frac{1}{15}\\0 & 1 & \frac{1}{100} & \frac{53}{1800}\\0 & 0 & 0 & 0\end{array}\right]\approx \left[\begin{array}{cccc}1 & 0 & 0 & 0.066666666666667\\0 & 1 & 0.01 & 0.029444444444444\\0 & 0 & 0 & 0\end{array}\right].$$$A


Please try a new game Rotatly