Gauss elimination and Gauss-Jordan elimination are methods for solving systems of linear equations. Gauss elimination puts the matrix of coefficients into row-echelon form using elementary row operations, while Gauss-Jordan elimination further reduces the matrix to reduced row-echelon form. These methods can also be used to find the rank of a matrix, calculate the determinant, and compute the inverse of an invertible square matrix. Examples demonstrate applying the methods to solve systems of 3 equations with 3 unknowns.

1. Gauss Jorden and
Gauss Elimination method
TO SOLVE SYSTEM OF LINEAR EQUATIONS

2. Define and History
ALI SHAN 19011598-044

3. System of linear equation
 If all equations in a system are linear, the
system is a system of linear equation or
linear system.
Example:
2y + z = -8
x -2y -3z = 0
-x+y+2z = 3
 There are two methods to solve these equations:
Direct and Iterative method

4. Gauss Elimination
 Define:-
It is also known as row reduction, is an algorithm in linear algebra for solving
a system of linear equations. It is usually understood as a sequence of operations
performed on the corresponding matrix of coefficients.
This method can also be used to find the rank of a matrix, to calculate the determinant of
a matrix, and to calculate the inverse of an invertible square matrix.
 History :-
The method is named after Carl Friedrich Gauss (1777–1855). Some special cases of
the method - albeit presented without proof - were known to Chinese mathematicians as
early as circa 179 CE.

5. Gauss Jorden Elimination
 Gauss–Jordan elimination to refer to the procedure which ends in reduced echelon
form. The name is used because it is a variation of Gaussian elimination as described
by Wilhelm Jordan in 1888.
 It is a further calculation of gauss elimination method.
 For Guass elimination: [A]=
1 ∗ ∗
0 1 ∗
0 0 1
∗
∗
∗
 For guass Jordan Elimination: [A]=
1 0 0
0 1 0
0 0 1
∗
∗
∗

6. Why we need these methods?
 It is usually understood as a sequence of operations performed on the
corresponding matrix of coefficients. This method can also be used to
find the rank of a matrix, to calculate the determinant of a matrix, and
to calculate the inverse of an invertible square matrix.
 Gaussian Elimination helps to put a matrix in row echelon form,
while Gauss-Jordan Elimination puts a matrix in reduced row echelon
form. For small systems (or by hand), it is usually more convenient to
use Gauss-Jordan elimination and explicitly solve for each variable
represented in the matrix system.

7. Techniques of Gauss
Elimination and GaussJorden
method
RAMEEN IFTIKHAR 19011598-042

8. Guass Elimination method
 First consider the system of linear equations as;
AX=B
a11x1+ a12x2+ a13x3= b1
a21x1+ a22x2+ a23x3= b2
a31x1+ a32x2+ a33x3= b3
 Find the augmented matric for the given system of equations as;
C=[A;B]
C=
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
𝑏1
𝑏2
𝑏3
 Use row operation to transform the augmented matrix into the Row Echelon Form.

9.  An elementary row operation is;
 Interchange the two rows.
 Multiply a row by non-zero constant.
 Add a multiple of the row to another row.
 Now check the resulting matrix and re-interpret it as a system of linear
equations.
 The resulting matrix will be;
1 𝑎12 𝑎13
0 1 𝑎23
0 0 1
𝑏1
𝑏2
𝑏3
 Find the solution of the equations by interpreting the equations.

10. Guass Jordan method
 First consider the system of linear equations as;
AX=B
a11x1+ a12x2+ a13x3= b1
a21x1+ a22x2+ a23x3= b2
a31x1+ a32x2+ a33x3= b3
 Find the augmented matric for the given system of equations as;
C=[A;B]
C=
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
𝑏1
𝑏2
𝑏3
 Use row operation to transform the augmented matrix into the Reduced Row
Echelon Form.

11.  An elementary row operation is;
 Interchange the two rows.
 Multiply a row by non-zero constant.
 Add a multiple of the row to another row.
 Now check the resulting matrix and re-interpret it as a system of linear
equations.
 The resulting matrix will be;
1 0 0
0 1 0
0 0 1
𝑏1
𝑏2
𝑏3
 Find the solution of the equations.

12. Examples of Gauss
Elimination method
AHSAN MEHBOOB 19011598-023

13. EXAMPLE.1:
The equation is:
x−y+z=8
x+3y−z=−2
3x−2y−9z=9
In augmented form
=
1 −1 1
1 3 −1
3 −2 −9
8
−2
9
=
1 −1 1
0 5 −3
0 1 −12
8
−18
−15
−2R1+R2=R2 and −3R1+R3=R3
1 ∗ ∗
0 1 ∗
0 0 1
∗
∗
∗

14. =
1 −1 1
0 1 −12
0 5 −3
8
−15
−18
Interchange R2 and R3
=
1 −1 1
0 1 −12
0 0 57
8
−15
57
−5R2+R3=R3
=
1 −1 1
0 1 −12
0 0 1
8
−15
1
−1/57R3=R3
Using back-substitution to get the values:
Z=1
x−y+z=8
y−12z=−15
Hence
The solution set is (4,-3,1)

15. Example.2:
 The linear equation is:
2x+3y+2z=-3
x+y+z=0
-x+2y-3z=-1
Solution:
In augmented form=
2 3 2
1 1 1
−1 2 −3
−3
0
−1
=
1 1 1
2 3 2
−1 2 −3
0
−3
−1
Interchanging R1 and R2

16. =
1 1 1
0 1 0
0 3 −2
0
−3
−1
R2=-2R1+R2 & R3=R1+R3
=
1 1 1
0 1 0
0 0 −2
0
−3
8
R3=-3R2+R3
=
1 1 1
0 1 0
0 0 1
0
−3
−4
R3=-
1
2
R3
Here
z=-4
y=-3
x+y+z=0
X-3-4=0
X=7
S.S={(7,-3,-4)}

17. Examples of Gauss Jordan
Elimination method
AMINA ZUBAIR 19011598-035

18. Example.1:
 1.Solve the system of linear equations by using Gauss-Jordan Eliminating Method:
2y + z = -8
x -2y -3z = 0
-x+y+2z = 3
Solution:
The augmented matrix of the system is:
0 2 1
1 −2 −3
−1 1 2
−8
0
3
Interchanging R1 and R2
1 −2 −3
0 2 1
−1 1 2
0
−8
3

19. R3=R3+R1
1 −2 −3
0 2 1
0 −1 −1
0
−8
3
Interchanging R2 and R3
1 −2 −3
0 −1 −1
0 2 1
0
3
−8
R3=R3+2R2
1 −2 −3
0 −1 −1
0 0 −1
0
3
−2
R1=R1-2R2
1 0 −1
0 −1 −1
0 0 −1
−6
3
−2
R2=R2-R3 & R1=R1-R3
1 0 0
0 −1 0
0 0 −1
−4
5
−2
Multiple R2&R3 both by (-1)
1 0 0
0 1 0
0 0 1
−4
−5
2
Therefore x=-4,y=-5&z=2.
Thus the solution of the given equations is (-4,-5,2)

20. Example.2:
 2.Solve the system of linear equations by using Gauss-Jordan Eliminating Method:
x+ y + z = 2
6x -4y+5z = 31
5x+2y+2z=13
Solution:
The augmented matrix of the system is:
1 1 1
6 −4 5
5 2 2
2
31
13
R2=R2-6R1
1 1 1
0 −10 −1
5 2 2
2
19
13

21. R3=R3-5R1
1 1 1
0 −10 −1
0 −3 −3
2
19
3
R2=R2/-10
1 1 1
0 1 1
10
0 −3 −3
2
−19
10
3
R1=R1-R2
1 0 9
10
0 1 1
10
0 −3 −3
39
10
−19
10
3
R3=R3+3R2
1 0 9
10
0 1 1
10
0 0 −27
10
39
10
−19
10
−27
10
R3=-(10/27)R3
1 0 9
10
0 1 1
10
0 0 1
39
10
−19
10
1

22. R1=R1-(9/10)R3
1 0 0
0 1 1
10
0 0 1
3
−19
10
1
R2=R2-(1/10)R3
1 0 0
0 1 0
0 0 1
3
−2
1
Therefore x=3,y=-2&z=1.
Thus the solution of the given equations
x+ y + z = 2
6x -4y+5z = 31
5x+2y+2z=13
Are
x=3
y=-2
z=1.

23. Applications

24. Applications:
 Computing determinants
By using row operations of Gaussian elimination we can find out the determinant
of any square matrix
 Finding the inverse of a matrix
A variant of Gaussian elimination called Gauss–Jordan elimination can be used for
finding the inverse of a matrix.
 Computing ranks and bases

25. Thanks for all
HAVE A NICE DAY