Gauss Jordan Elimination Calculator calculates the variables of given system of linear equations.

This method is also known as back substitution method. Here in this method all we have to do is converting the given matrix into diagonal matrix.
Gauss Jordon elimination Calculator solves the value of variables x, y and z if those elements are entered in the given block.Consider the equations of the form:

$a_{1} x + b_{1} y + c_{1} z$ = $d_{1}$

$a_{2} x + b_{2} y + c_{2}$ = $d_{2}$

$a_{3} x + b_{3} y + c_{3}$ = $d_{3}$

We can get easily the value of $x, y$ and $z$ if we enter all these variables in their corresponding blocks.

## Gauss Jordon Elimination Examples

Steps for solving Gauss Jordon Elimination:
1) Read the problem  and write the linear system of equations in augmented form.

2) By applying row operations convert the given matrix in reduced row echelon form

3) Find the value of variables using the echelon form of matrix.
Below are given some problems based on Gauss Jordon elimination which may be helpful for you.

### Solved Example

Question: Solve the following equations by Gauss Jordon elimination:
2x + z = -12
y - 2z = 7
6x + 3y = 2
Solution:

Given linear system of equations are
2x + z = -12
y - 2z = 7
6x + 3y = 2

Take first row as R1, second row as R2 and third row as R3.
$\begin{bmatrix} 2 & 0 & 1 & | -12\\ 0 & 1 & -2 & | 7\\ 6 & 3 & 0 & | 2 \end{bmatrix}$

Apply row operations, we get
R3 ~ $\frac{1}{3}$ R3
$\begin{bmatrix} 2 & 0 & 1 & | -12\\ 0 & 1 & -2 & | 7\\ 2 & 1 & 0 & | \frac{2}{3} \end{bmatrix}$

R3 ~ R3 - R1
$\begin{bmatrix} 2 & 0 & 1 & | -12\\ 0 & 1 & -2 & | 7\\ 0 & 1 & -1 & | \frac{38}{3} \end{bmatrix}$

R3 ~ R3 - R2
$\begin{bmatrix} 2 & 0 & 1 & | -12\\ 0 & 1 & -2 & | 7\\ 0 & 0 & 1 & | \frac{17}{3} \end{bmatrix}$

R2 ~ $\frac{1}{2}$ R2
$\begin{bmatrix} 2 & 0 & 1 & | -12\\ 0 & \frac{1}{2} & -1 & | \frac{7}{2}\\ 0 & 0 & 1 & | \frac{17}{3} \end{bmatrix}$

Applying R1 ~ R1 - R2 and R2 ~ R2 + R3
$\begin{bmatrix} 2 & 0 & 0 & | - \frac{53}{3}\\ 0 & \frac{1}{2} & -1 & | \frac{55}{6}\\ 0 & 0 & 1 & | \frac{17}{3} \end{bmatrix}$

Solving the matrix we get
2x = - $\frac{53}{3}$ => x = - 8.83
$\frac{y}{2}$ = $\frac{55}{6}$ => y = 18.33
z = $\frac{17}{3}$ => z = 5.667.

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