# How do you solve using gaussian elimination or gauss-jordan elimination, 5x + y + 5z = 3 , 4x − y + 5z = 13 , 5x + 2y + 2z = 2?

May 14, 2017

$x = 1 \frac{32}{99}$,
$y = - 5 \frac{131}{198}$,
$z = \frac{9}{22}$

#### Explanation:

To solve the system of linear equations

$\left\{\begin{matrix}5 x + y + 5 z = 3 \\ 4 x - y + 5 z = 13 \\ 5 x + 2 y + 2 z = 2\end{matrix}\right.$

First convert it to the augmented matrix form

$R i g h t a r r o w \left[\begin{matrix}5 & 1 & 5 & | & 3 \\ 4 & - 1 & 5 & | & 13 \\ 5 & 2 & 2 & | & 2\end{matrix}\right]$

The goal is to write this in row-echelon form, which is triangular form where the leading coefficient is 1. The three row operations we can use are:

1. Switch rows
2. Multiply a row by a constant
3. Add a multiple of a row to another

Multiply the first row by $\frac{1}{5}$

$R i g h t a r r o w \left[\begin{matrix}1 & \frac{1}{5} & 1 & | & \frac{3}{5} \\ 4 & - 1 & 5 & | & 13 \\ 5 & 2 & 2 & | & 2\end{matrix}\right]$

Multiply first row by -4 and add to second row

$R i g h t a r r o w \left[\begin{matrix}1 & \frac{1}{5} & 1 & | & \frac{3}{5} \\ 0 & - \frac{9}{5} & 1 & | & \frac{53}{5} \\ 5 & 2 & 2 & | & 2\end{matrix}\right]$

Multiply first row by -5 and add to the third row

$R i g h t a r r o w \left[\begin{matrix}1 & \frac{1}{5} & 1 & | & \frac{3}{5} \\ 0 & - \frac{9}{5} & 1 & | & \frac{53}{5} \\ 0 & 1 & - 3 & | & - 1\end{matrix}\right]$

Multiply row 2 by $- 5 / 9$

$R i g h t a r r o w \left[\begin{matrix}1 & \frac{1}{5} & 1 & | & \frac{3}{5} \\ 0 & 1 & - \frac{5}{9} & | & - \frac{53}{9} \\ 0 & 1 & - 3 & | & - 1\end{matrix}\right]$

Multiply row 2 by -1 and add to row 3

$R i g h t a r r o w \left[\begin{matrix}1 & \frac{1}{5} & 1 & | & \frac{3}{5} \\ 0 & 1 & - \frac{5}{9} & | & - \frac{53}{9} \\ 0 & 0 & - \frac{22}{9} & | & - 1\end{matrix}\right]$

Multiply the last row by -9/22

$R i g h t a r r o w \left[\begin{matrix}1 & \frac{1}{5} & 1 & | & \frac{3}{5} \\ 0 & 1 & - \frac{5}{9} & | & - \frac{53}{9} \\ 0 & 0 & 1 & | & \frac{9}{22}\end{matrix}\right]$

Plugging these values back into formulas gives

$\left\{\begin{matrix}x + \frac{1}{5} y + z = \frac{3}{5} \\ y - \frac{5}{9} z = - \frac{53}{9} \\ z = \frac{9}{22}\end{matrix}\right.$

Using backwards substitution we get the remaining answers:
$x = \frac{131}{99} = 1 \frac{32}{99}$,
$y = - \frac{1121}{198} = - 5 \frac{131}{198}$,
$z = \frac{9}{22}$