# How do you write the matrix [(1, -3, 0, -7), (-3, 10, 1, 23), (4, -10, 2, -24)] using the row echelon form?

Jun 7, 2017

The matrix in reduced row echelon form is $\left(\begin{matrix}1 & 0 & 3 & - 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0\end{matrix}\right)$

#### Explanation:

The matrix is

$\left(\begin{matrix}1 & - 3 & 0 & - 7 \\ - 3 & 10 & 1 & 23 \\ 4 & - 10 & 2 & - 24\end{matrix}\right)$

Replace ${L}_{2}$ by $\left({L}_{2} + 3 {L}_{1}\right)$, we get

$\left(\begin{matrix}1 & - 3 & 0 & - 7 \\ 0 & 1 & 1 & 2 \\ 4 & - 10 & 2 & - 24\end{matrix}\right)$

Replace ${L}_{3}$ by $\left({L}_{3} - 4 {L}_{1}\right)$, we get

$\left(\begin{matrix}1 & - 3 & 0 & - 7 \\ 0 & 1 & 1 & 2 \\ 0 & 2 & 2 & 4\end{matrix}\right)$

Replace ${L}_{3}$ by $\left({L}_{3} - 2 {L}_{2}\right)$, we get

$\left(\begin{matrix}1 & - 3 & 0 & - 7 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0\end{matrix}\right)$

Replace ${L}_{1}$ by $\left({L}_{1} + 3 {L}_{2}\right)$, we get

$\left(\begin{matrix}1 & 0 & 3 & - 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0\end{matrix}\right)$

This is the matrix in reduced row echelon form.

Jun 7, 2017

$\left[\begin{matrix}1 & - 3 & 0 & | & - 7 \\ 0 & 1 & 1 & | & 2 \\ 0 & 0 & 0 & | & 0\end{matrix}\right]$

#### Explanation:

Row echelon form has the form

$\left[\left(1 , {a}_{1 , 2} , {a}_{1 , 3} , | , {b}_{1}\right) , \left(0 , 1 , {a}_{2 , 3} , | , {b}_{2}\right) , \left(0 , 0 , 1 , | , {b}_{3}\right)\right]$

The available operations include

• Swap rows ($R$)
• Multiply/divide a row by a constant

Multiply Row 1 by $3$ and add to Row 2. In short hand:

$3 \times {R}_{1} + {R}_{2} \to {R}_{2}$ gives:
$\left[\begin{matrix}1 & - 3 & 0 & | & - 7 \\ 0 & 1 & 1 & | & 2 \\ 4 & - 10 & 2 & | & - 24\end{matrix}\right]$

$- 4 \times {R}_{1} + {R}_{3} \to {R}_{3}$ gives:
$\left[\begin{matrix}1 & - 3 & 0 & | & - 7 \\ 0 & 1 & 1 & | & 2 \\ 0 & 2 & 2 & | & 4\end{matrix}\right]$

Notice in this last step that that ${R}_{3} = 2 \times {R}_{2}$

$\frac{1}{2} {R}_{3} \to {R}_{3}$ gives:
$\left[\begin{matrix}1 & - 3 & 0 & | & - 7 \\ 0 & 1 & 1 & | & 2 \\ 0 & 1 & 1 & | & 2\end{matrix}\right]$

$- 1 \times {R}_{2} + {R}_{3} \to {R}_{3}$ gives:
$\left[\begin{matrix}1 & - 3 & 0 & | & - 7 \\ 0 & 1 & 1 & | & 2 \\ 0 & 0 & 0 & | & 0\end{matrix}\right]$

Because no further manipulation will produce the final form,

$\left[\left(1 , {a}_{1 , 2} , {a}_{1 , 3} , | , {b}_{1}\right) , \left(0 , 1 , {a}_{2 , 3} , | , {b}_{2}\right) , \left(0 , 0 , 1 , | , {b}_{3}\right)\right]$

We are finished.