# REF of $\left[\begin{array}{ccc}1 & 2 & 3\\2 & 5 & 7\\4 & 9 & 13\end{array}\right]$

The calculator will find the row echelon form of the $3$x$3$ matrix $\left[\begin{array}{ccc}1 & 2 & 3\\2 & 5 & 7\\4 & 9 & 13\end{array}\right]$, with steps shown.

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

$\times$

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

Find the row echelon form of $\left[\begin{array}{ccc}1 & 2 & 3\\2 & 5 & 7\\4 & 9 & 13\end{array}\right]$.

### Solution

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

$\left[\begin{array}{ccc}1 & 2 & 3\\0 & 1 & 1\\4 & 9 & 13\end{array}\right]$

Subtract row $1$ multiplied by $4$ from row $3$: $R_{3} = R_{3} - 4 R_{1}$.

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

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

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

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.

The row echelon form is $\left[\begin{array}{ccc}1 & 2 & 3\\0 & 1 & 1\\0 & 0 & 0\end{array}\right]$A.