# Gaussian Elimination

Gaussian elimination is a technique to change the augmented matrix into a row echelon form.There are many echelon forms, but subsequently, we must find the reduced row echelon form. The reduced row echelon form can can be achieved through another technique called the Gauss-Jordan elimination technique.

### Augmented Matrix To Row Echelon Form

Given an augmented matrix that represents a system of linear equations with n equations and m variables. We want to reduce the matrix to a simple form called the echelon form and solve the unknowns.

$5x + y + z = 10$
$2x + y - 3z = -5$
$x - 2y + 2z = 3$

These are the steps to achieve the row echelon form.

• Begin with column 1
• Obtain leading 1 for each row and
• Change 0 below all leading 1s.

#### Example

In this example we will convert the following augmented matrix into a row echelon form by series of row operations.

Step 1

The augmented matrix first row starts with a leading 1 and all other entries in the first column are zeros. If the first row starts with a 0 , then do row exchange with second or third row and bring a non-zero value.

To obtain leading 1 in the first row do following row operation.

$R_{1} = \frac{R_{1}}{5}$

Our resultant matrix is as follows.

$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\ & 1 & -3 & -5\ & -2 & 2 & 3\end{bmatrix}$

Step 2

Perform following row operations to make rest of the entries 0 in the first column under leading 1.

$R_{2} = R_{2} - 2R_{1}$
$R_{3} = R_{3} - R_{1}$

Now, our matrix look like the following.

$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\
& \frac{3}{5} & \frac{-17}{5} & -9\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & \frac{3}{5} & \frac{-17}{5} & -9\\0 & \frac{-11}{5} & \frac{9}{5} & 1\end{bmatrix}$
& \frac{-11}{5} & \frac{9}{5} & 1\end{bmatrix}' title='\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & \frac{3}{5} & \frac{-17}{5} & -9\\0 & \frac{-11}{5} & \frac{9}{5} & 1\end{bmatrix}$
& \frac{3}{5} & \frac{-17}{5} & -9\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & \frac{3}{5} & \frac{-17}{5} & -9\\0 & \frac{-11}{5} & \frac{9}{5} & 1\end{bmatrix}$
& \frac{-11}{5} & \frac{9}{5} & 1\end{bmatrix}' class='latex' />

We must repeat the same steps for Group 2.

Step 3

In the group 2 which is column 2 , first change the value to leading 1.

$R_{2} = R_{2} * 5$
$R_{2} = \frac{R_{2}}{3}$

The result of the row operation is given below.

$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\
& 1 & \frac{-17}{3} & -15\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & 1 & \frac{-17}{3} & -15\\0 & \frac{-11}{5} & \frac{9}{5} & 1\end{bmatrix}$
& \frac{-11}{5} & \frac{9}{5} & 1\end{bmatrix}' title='\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & 1 & \frac{-17}{3} & -15\\0 & \frac{-11}{5} & \frac{9}{5} & 1\end{bmatrix}$
& 1 & \frac{-17}{3} & -15\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & 1 & \frac{-17}{3} & -15\\0 & \frac{-11}{5} & \frac{9}{5} & 1\end{bmatrix}$
& \frac{-11}{5} & \frac{9}{5} & 1\end{bmatrix}' class='latex' />

Step 4

All entries below leading 1 in second row must be changed to 0. Therefore, perform following row operations.

$R_{3} = R_{3} + \frac{11}{5}R_{2}$

The result is as follows. The second row leading 1 has a 0 below it in the third row.

$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\
& 1 & \frac{-17}{3} & -15\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & 1 & \frac{-17}{3} & -15\\0 & 0 & \frac{-32}{3} & -32\end{bmatrix}$
& 0 & \frac{-32}{3} & -32\end{bmatrix}' title='\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & 1 & \frac{-17}{3} & -15\\0 & 0 & \frac{-32}{3} & -32\end{bmatrix}$
& 1 & \frac{-17}{3} & -15\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & 1 & \frac{-17}{3} & -15\\0 & 0 & \frac{-32}{3} & -32\end{bmatrix}$
& 0 & \frac{-32}{3} & -32\end{bmatrix}' class='latex' />

Step 5

Now, we must find the leading 1 for row 3. Do the following row operations on

$R_{3} = \frac{R_{3}}{-32}$
$R_{3} = R_{3} * -3$

The augmented matrix is in row echelon form and it is easy to find a solution to the matrix.

$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\
& 1 & \frac{-17}{3} & -15\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & 1 & \frac{-17}{3} & -15\\0 & 0 & 1 & 3\end{bmatrix}$
& 0 & 1 & 3\end{bmatrix}' title='\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & 1 & \frac{-17}{3} & -15\\0 & 0 & 1 & 3\end{bmatrix}$
& 1 & \frac{-17}{3} & -15\
$\begin{bmatrix} 1 & \frac{1}{5} & \frac{1}{5} & 2\\0 & 1 & \frac{-17}{3} & -15\\0 & 0 & 1 & 3\end{bmatrix}$
& 0 & 1 & 3\end{bmatrix}' class='latex' />

### Back Substitution

The echelon form gives us the following simplified system of linear equations.

$x + \frac{y}{5} + \frac{z}{5} = 2$
$y + \frac{-17}{3} = -15$
$z = 3$

Using value of $z = 3$ and substituting in other equations,we get

$x + \frac{2}{5} + \frac{3}{5} = 2$
$x = 2 - \frac{2}{5} - \frac{3}{5}$
$x = \frac{10 - 2 - 3}{5}$
$x = 1$

$y - 17 = -15$
$y = 17-15 = 2$

$z = 3$

Therefore, the solution to the equation is $x = 1, y =2, z = 3$.