The document discusses the Gaussian elimination method for solving homogeneous and non-homogeneous linear systems of equations. It describes how to form the augmented matrix and use elementary row operations to reduce it to row echelon form. This allows the equations to be solved using back-substitution. For homogeneous systems, the solutions are either the trivial solution where all variables are zero, or there are infinite solutions when the determinant of the coefficient matrix is zero.

1. Gaussian Elimination Method &
Homogeneous Linear Equation
Prepared By:
130110120007 – Ravi
130110120008 – Hemdeep Bhavsar
130110120009 – Nayan Chauhan
130110120010 – Chintan Katerecha
130110120011 – Chitt Kakadiya
130110120012 – Yashraj Chudasama
Faculty :-
VRS Sir

2. Gaussian elimination method
• Gaussian elimination method is used for
solving the non- homogeneous and
homogeneous linear equation.
• In the Gaussian elimination method we have
to make an augmented matrix of given m
linear equation in n variables.

3. • A system of m homogeneous or non
homogeneous linear equations in n
variables x1, x2, …,xn or simply a
linear system is a set of m linear
equation, each in n variables.

4. • A linear equation is represented by
• Writing this equation in matrix form,
Ax = B

5. • Where
called coefficient of
matrix of order m*n.
is any vector of order .
is any vector of order .

6. • To solve a system of linear equation,
elementary transformation are used to
reduce the augmented matrix to either
row echelon form or reduced row echelon
form.
• Reducing the augmented matrix to row
echelon form is called Gaussian elimination
method.

7. • The Gaussian elimination method for solving:
1. Write the augmented matrix.
2. Obtain the row echelon form of the
augmented matrix by using elementary row
operations.
3. Write the corresponding linear system of
equation from row echelon form.
4. Solve the corresponding linear system of
equation by back solution.

8. Forward Elimination
• The goal of forward elimination is to transform
the coefficient matrix into an upper triangular
matrix































2.279
2.177
8.106
112144
1864
1525
3
2
1
x
x
x
































735.0
21.96
8.106
7.000
56.18.40
1525
3
2
1
x
x
x

9. Forward Elimination
• A set of n equations and n unknowns
11313212111 ... bxaxaxaxa nn 
22323222121 ... bxaxaxaxa nn 
. .
. .
. .
nnnnnnn bxaxaxaxa  ...332211

10. Forward Elimination
• Subtract the result from Equation 2.
or
22323222121 ... bxaxaxaxa nn 
1
11
21
1
11
21
212
11
21
121 ... b
a
a
xa
a
a
xa
a
a
xa nn 
1
11
21
21
11
21
2212
11
21
22 ... b
a
a
bxa
a
a
axa
a
a
a nnn 












'
2
'
22
'
22 ... bxaxa nn 

11. Forward Elimination
• At the end of (n-1) Forward Elimination steps,
the system of equations will look like
11313212111 ... bxaxaxaxa nn 
'
2
'
23
'
232
'
22 ... bxaxaxa nn 
"
3
"
33
"
33 ... bxaxa nn 
   11 

n
nn
n
nn bxa

12. Matrix Form at End of Forward
Elimination

















































 )(n-
n
"
'
n
)(n
nn
"
n
"
'
n
''
n
b
b
b
b
x
x
x
x
a
aa
aaa
aaaa
1
3
2
1
3
2
1
1
333
22322
1131211
0000
00
0





13. Back Substitution
• Solve each equation starting from the last
equation
Example of a system of 3 equations
































735.0
21.96
8.106
7.000
56.18.40
1525
3
2
1
x
x
x

14. Back Substitution Starting Equation
11313212111 ... bxaxaxaxa nn 
'
2
'
23
'
232
'
22 ... bxaxaxa nn 
"
3
"
3
"
33 ... bxaxa nn 
. .
. .
. .
   11 

n
nn
n
nn bxa

15. Back Substitution
• Start with the last equation because it has
only one unknown
)1(
)1(


 n
nn
n
n
n
a
b
x

16. Back Substitution
)1(
)1(


 n
nn
n
n
n
a
b
x
       
  1,...,1for
...
1
1
,2
1
2,1
1
1,
1


 







ni
a
xaxaxab
x i
ii
n
i
nii
i
iii
i
ii
i
i
i
   
  1,...,1for1
1
11


 


ni
a
xab
x i
ii
n
ij
j
i
ij
i
i
i

17. 1 . x + y + 2z = 9
2x + 4y – 3z = 1
3x + 6y – 5z = 0
The matrix form of the equation is
the augmented matrix of the system

































0
1
9
563
342
211
z
y
x

18. • Reducing the augmented matrix to row echelon
form,
R2 – 2R1, R3 – 3R1
(1/2) R2

19. R3 – 3R2
(-2)R3

20. • The corresponding system of equation
Solving this equations,
x = 1, y = 2
Hence, x =1, y=2, z=3 is the solution of the
system.

21. System of homogeneous linear
equation
• A system of m homogeneous linear equation
in n variables x1, x2, x3,……,xn or simply a linear
system, is a set of m linear equation in n
variables.
• A linear system is represented by,
0... 1313212111  nn xaxaxaxa
0... 2323222121  nn xaxaxaxa
0...332211  nnnnnn xaxaxaxa
. .
. .
. .

22. • Writing the equation in matrix form,
Ax = 0
• Where A is any matrix of order , x
is a vector of order and 0 is a null
vector of order .
• The matrix A is called co efficient of the
system of equation.

23. Solutions of system of linear
equations
• For a system of m linear equations in n
variables, there are two possibilities of the
solution to the system.
1. The system has exactly one solution, i.e. x1=0,
x2=0…….xn =0. This solution is called Trivial
solution.
2. The system has infinite solutions.
3. System has a non trivial solution if det(A) = 0.

24. Example:
3x – y – z = 0
x + y + 2z = 0
5x + y + 3z = 0
• The augmented matrix of the system is,

25. • Reducing the augmented matrix to reduced
row echelon form,
R12
R2- 3R1, R3 – 5R1

26. (-1/4) R2, (-1/4) R3,
R3 – R2

27. R3 – R2
The corresponding system of equation is

28. • Assigning the free variables z an arbitrary
value t,
x = (-1/4)t y = (-7/4)t
• Hence , x = (-1/4)t, y= (-7/4)t is the non trivial
solution of the matrix.