The document presents methods for solving systems of linear equations, specifically focusing on the LU factorization method and the Gauss elimination method. It describes the process of using triangular matrices to find solutions and illustrates the techniques with examples. The conclusion highlights the efficiency of triangular systems over more complex methods like matrix inversion.

1. NUMERICALSOLUTIONOF
SYSTEMOF LINEAREQUATIONS
NUMERICAL METHODS 1

2. NUMERICAL METHODS 2
REPRESENTED BY:-
NAME
ARKA ROY
ARPANDIP MONDAL
ARANYA MONDAL
APURBA MONDAL
AMIYA SING
AMIT SENA

3.  CONTENTS :
 SOLUTION OF LINEAR EQUATIONS BY USING LU FACTORIZATION METHOD.
 SOLUTION OF LINEAR EQUATIONS BY USING GAUSS ELEMINATION METHOD.
 INVERSSION OF MATRIX USING GAUSS ELEMINATION METHOD.
 DATE :02/11/2018NUMERICAL METHODS 3

4. LU FACTORIZATION
METHOD
NUMERICAL METHODS 4
DATE :02/11/2018

5. LU Factorization Method :
This method is based on the fact that every square matrix can be represented as a
product of lower and upper triangular matrix; i.e. [A]=[L]*[U]
[L]=Lower triangular matrix; [U]=Upper triangular matrix
Provided all the principle minors are non singular .
Let us consider we have : 𝑎11 𝑥 + 𝑎12 𝑦 + 𝑎13 𝑧 = 𝑏1
𝑎21 𝑥 + 𝑎22 𝑦 + 𝑎23 𝑧 = 𝑏2
𝑎31 𝑥 + 𝑎32 𝑦 + 𝑎33 𝑧 = 𝑏3
the system of equation can be represented as : [A]*[X]=[B]
now matrix A can be represented as [A]=[L]*[U]
DATE :02/11/2018NUMERICAL METHODS 5

6. [A]=[L]*[U]=
𝑙11 0 0
𝑙21 𝑙22 0
𝑙31 𝑙32 𝑙33
*
1 𝑢12 𝑢13
0 1 𝑢23
0 0 1
From here we have to find the values of the elements of lower & upper triangular
matrix.
Then : [A]*[X]=[B];
[L]*[U]*[X]=[B]; As : [A]=[L]*[U]
Now we have to consider that [U]*[X]=[Y] ----------(i)
So that [L]*[Y]=[B]---------------------(ii)
Now using the (ii) equ. We have to find the elements of matrix [Y];where [L],[B]
Is known to us.
Then using the matrix [Y] in equ. (i) we can find the unknown variable matrix
[X] & get the set of solution of the given equation.
DATE :02/11/2018
NUMERICAL METHODS 6

7.  For example we have three equation given bellow and find the sol. By LU method :
8𝑥 − 3𝑦 + 2𝑧 = 20
4𝑥 + 11𝑦 − 𝑧 = 33
6x + 3𝑦 + 12𝑧 = 36
The given equ. Can be represented as : [A]*[X]=[B];
A=
8 −3 2
4 11 −1
6 3 12
,B=
20
33
36
,X=
𝑥
𝑦
𝑧
 A=L*U;

𝑙11 0 0
𝑙21 𝑙22 0
𝑙31 𝑙32 𝑙33
*
1 𝑢12 𝑢13
0 1 𝑢23
0 0 1
=
8 −3 2
4 11 −1
6 3 12
 Leading to : 𝑙11 = 8 ; 𝑙11 𝑢12 = −3 𝑢12 = −(
3
8
)
DATE :02/11/2018
NUMERICAL METHODS 7

8.  𝑙11 𝑢13 = 2 𝑢13 =
1
4
;
 𝑙21 = 4; 𝑙21 𝑢12 + 𝑙22 = 11 𝑙21 = 11 − 𝑙21 𝑢12 = (
25
2
)
 𝑙21 𝑢13 + 𝑙22 𝑢23 = −1 𝑢23 = −
4
5
 𝑙31 = 6; 𝑙31 𝑢12 + 𝑙32 = 3 6 ∗ −
3
8
+ 𝑙32 = 3 𝑙32 = (
21
4
)
 𝑙31 𝑢13 + 𝑙22 𝑢23 + 𝑙33 = 12
𝑙33= (
567
50
) [using the values ]
 Hence [L]=
8 0 0
4 (
25
2
) 0
6 (
21
4
) (
567
30
)
& [U]=
1 −(
3
8
) (
1
4
)
0 1 (−
4
25
)
0 0 1
 Thus the equation A*X=B i.e. L*U*X=B
 Now consider L*Y=B ------------(I)
 where U*X=Y-------------(II)
 DATE :02/11/2018
NUMERICAL METHODS 8

9. 
8 0 0
4 (
25
2
) 0
6 (
21
4
) (
567
30
)
*
𝑌1
𝑌2
𝑌3
=
20
33
36
 So that : 8𝑌1 = 20 𝑌1 = (
5
2
)
 4𝑌1 +
25
2
∗ 𝑌2 = 33 𝑌2 = 33 − 4 ∗
5
2
∗
2
25
= (
46
25
)
 6𝑌1 +
21
4
𝑌2 +
567
50
𝑌3 = 36 𝑌3 =
567
50
∗ 36 − 15 − 21 ∗
23
50
= 1
 Now putting the values on equ. (ii) we get :

1 −(
3
8
) (
1
4
)
0 1 (−
4
25
)
0 0 1
*
𝑥
𝑦
𝑧
=
(
5
2
)
(
46
25
)
1
 which gives
 z= 1,
 𝑦 −
4
25
𝑧 = (
46
25
) y = (
46
25
) +
4
25
)=2
NUMERICAL METHODS
9

10.  𝑥 −
3
8
𝑦 +
𝑧
4
= (
5
2
) x = (
5
2
+
6
8
−
1
4
)=3

 Hence the set of solution is given as :
 𝑥 = 3 ; 𝑦 = 2 ; 𝑧 = 1 .
DATE :02/11/2018NUMERICAL METHODS 10

11.  CONCLUSION :
 The benefit of this approach is that we only
ever need to solve triangular systems. The
cost is that we have to solve two of them.
 It is one of the easiest way of solve the
system of equation.
 But it is also a time consuming one.
NUMERICAL METHODS 11

12. GAUSS-ELEMINATION METHOD
NUMERICAL METHODS 12

13. GAUSS ELEMINATION METHOD :
 The method is based on the idea of reducing the given system of equations Ax = B; to
an upper triangular system of equations Ux = z, using elementary row operations
.That is, the solutions of both the systems are identical. Here 𝑥 =
𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒′ 𝑠 𝑚𝑎𝑡𝑟𝑖𝑥
 We illustrate the method using the 3 × 3 system
 𝑎11 𝑥 + 𝑎12 𝑦 + 𝑎13 𝑧 = 𝑏1 ∶ R1
 𝑎21 𝑥 + 𝑎22 𝑦 + 𝑎23 𝑧 = 𝑏2 ∶ 𝑅2
 𝑎31 𝑥 + 𝑎32 𝑦 + 𝑎33 𝑧 = 𝑏3 ∶ 𝑅3
 Now to make the patter of equation upper triangular matrix we have to use the
operation :
 𝑅𝑖 = 𝑅𝑖 −
𝑎 𝑖𝑘
𝑎 𝑘𝑘
∗ 𝑅 𝑘
 Where i =(k+1),(k+2),……n ; k= column number
 DATE :02/11/2018NUMERICAL METHODS 13

14. So after applying the operation the equation becomes :
𝑎11 𝑥 + 𝑎12 𝑦 + 𝑎13 𝑧 = 𝑏1
0+𝑎′12 𝑦 + 𝑎′13 𝑧 = 𝑏′2 [𝑅′2 = 𝑅2 −
𝑎21
𝑎11
∗ 𝑅1 ]
0+𝑎′12 𝑦 + 𝑎′13 𝑧 = 𝑏′3 [ 𝑅′3 = 𝑅3 −
𝑎31
𝑎11
∗ 𝑅1 ]
then after that we have to apply the same process to make that pattern
to a upper triangular matrix.
𝑎11 𝑥 + 𝑎12 𝑦 + 𝑎13 𝑧 = 𝑏1
0+𝑎′22 𝑦 + 𝑎′23 𝑧 = 𝑏′2
0 + 0 + 𝑎′′33 𝑧 = 𝑏′′3 [ 𝑅′′3 = 𝑅′3 −
𝑎32
𝑎22
∗ 𝑅2 ]
 the successively we can find the values Z, Y, X.
 DATE :02/11/2018NUMERICAL METHODS 14

15.  For example we take three equations:
𝑥 − 2𝑦 + 9𝑧 = 8
3𝑥 + 𝑦 − 𝑧 = 3
2x − 8𝑦 + 𝑧 = −5
Now we have to make it a upper triangular matrix using the operation:
𝑅2 = 𝑅2 −
𝑎21
𝑎11
∗ 𝑅1 𝑅2 = 𝑅2 − 3/1 ∗ 𝑅1
They become : 𝑥 − 2𝑦 + 9𝑧 = 8
0 + 7𝑦 − 28𝑧 = −21
0 − 4𝑦 − 17𝑧 = −21
To get the pattern of upper triangular matrix we again do the operation:
DATE :02/11/2018
NUMERICAL METHODS 15

16. The operation is : 𝑅3 = 𝑅3 − −4/7 ∗ 𝑅2
Now they become : : 𝑥 − 2𝑦 + 9𝑧 = 8
0 + 7𝑦 − 28𝑧 = −21
0 + 0 − 33𝑧 = −33
So that : 𝑧 = 1;
𝑦 =
−21+28𝑧
7
= 1;
𝑥 = 8 − 9𝑧 + 2𝑦 = 1 .
The set of solution is : X=1; Y=1; Z=1.
DATE :02/11/2018NUMERICAL METHODS 16

17.  CONCLUSION :
 In case of matrix inversion method the process of computation will be
lengthy if the number of equation is as well as number of variables are more.
But by using the gauss elimination method we just have to use the formula to
get the upper triangular matrix & compute the values of the variable.
NUMERICAL METHODS 17

18. INVERSION OF MATRIX
NUMERICAL METHODS 18

19. INVERSSION OF MATRIX :
Let us consider we have a matrix [A] & we have to find the inverse of that
matrix . Let [x] be the inverse of matrix [A].
we know that : A ∗ 𝐴−1 = [𝐼]
now : [A]*[X]=[I] as we considered that : [X]= 𝐴−1
Then we convert the system into an augmented matrix
in the form of [A|I ].
then we have to use elementary row operation on the
augmented matrix to get upper triangular matrix by using
Gauss Elimination Method.
DATE :02/11/2018NUMERICAL METHODS 19

20. Let us consider we have a matrix like this :
A=
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
& the inverse of [A] is [x] (say).
Let x is : x=
𝑥11 𝑥12 𝑥13
𝑥21 𝑥22 𝑥23
𝑥31 𝑥32 𝑥33
: A ∗ 𝐴−1 = [𝐼]
now : [A]*[X]=[I] as we considered that : [X]= 𝐴−1
The augmented matrix looks like : [A| I ]
=
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
1 0 0
0 1 0
0 0 1
=
𝑎11 𝑎12 𝑎13
0 𝑎′22 𝑎′23
0 𝑎′32 𝑎′33
1 0 0
𝑔21 𝑔22 0
𝑔31 0 1
=
𝑎11 𝑎12 𝑎13
0 𝑎′22 𝑎′23
0 0 𝑎′′33
1 0 0
𝑔21 𝑔22 0
𝑔′31 𝑔′32 1
DATE :02/11/2018
NUMERICAL METHODS 20

21. Then we can get the equivalent system of three equ.
𝑎11 𝑎12 𝑎13
0 𝑎′22 𝑎′23
0 0 𝑎′′33
*
𝑥11
𝑥21
𝑥31
=
1
𝑔21
𝑔′31
from here we can calculate the values of 𝑥11; 𝑥21; 𝑥31 .
Similarly :
𝑎11 𝑎12 𝑎13
0 𝑎′22 𝑎′23
0 0 𝑎′′33
*
𝑥12
𝑥22
𝑥32
=
0
𝑔22
𝑔′32
From here we can calculate the values of 𝑥12; 𝑥22 ; 𝑥32
Similarly :
𝑎11 𝑎12 𝑎13
0 𝑎′22 𝑎′23
0 0 𝑎′′33
*
𝑥13
𝑥23
𝑥33
=
0
𝑔22
𝑔′32
from here we can calculate the values of 𝑥13; 𝑥23 ; 𝑥33.
so that we have the elements of inverse matrix [X] ; which is our required solution.
DATE :02/11/2018
NUMERICAL METHODS 21

22.  For example we take a matrix and try to find it’s inverse :
A =
2 −2 4
2 3 2
−1 4 −1
consider the augmented matrix is :
2 −2 4
2 3 2
−1 4 −1
: 1 0 0
: 0 1 0
: 0 0 1
=
2 −2 4
0 5 −2
0 3 1
: 1 0 0
: −1 1 0
: (
1
2
) 0 1
[using 𝑅′2 = 𝑅2 −
2
2
∗ 𝑅1; 𝑅′
3 = 𝑅3 − (
1
2
)𝑅1]
DATE :02/11/2018NUMERICAL METHODS 22

23.  =
2 −2 4
0 5 −2
0 0 (
11
5
)
: 1 0 0
: −1 1 0
: (
11
10
) −(
3
5
) 1
[using 𝑅′′3 = 𝑅3 −
3
5
∗ 𝑅2]
 The inverse matrix is given as : X=
𝑥11 𝑥12 𝑥13
𝑥21 𝑥22 𝑥23
𝑥31 𝑥32 𝑥33
 Thus we have an equivalent system of three equation :

2 −2 4
0 5 −2
0 0 (
11
5
)
*
𝑥11
𝑥21
𝑥31
=
1
−1
(
11
10
)
 which gives :2𝑥 − 2𝑦 + 4𝑧 = 1;
 5𝑦 − 2𝑧 = −1;

11
5
∗ 𝑧 = (
11
10
)
 Solving by back substitution method we get ;
 𝑥 = −
1
2
; 𝑦 = 0; 𝑧 = (
1
2
)
 DATE :02/11/2018NUMERICAL METHODS 23

24. 
2 −2 4
0 5 −2
0 0 (
11
5
)
*
𝑥12
𝑥22
𝑥32
=
0
1
(−
3
5
)
 similarly by solving these we get :
 𝑥12 = (
7
11
) ; 𝑥22= (
1
11
); 𝑥32= (−
3
11
) .


2 −2 4
0 5 −2
0 0 (
11
5
)
*
𝑥13
𝑥23
𝑥33
=
0
0
1
 similarly by solving these we get :
 𝑥13 = (
−8
11
) ; 𝑥22= (
2
11
); 𝑥32= (
5
11
) .
 The required solution is :
 𝐴−1 =
(−
1
2
) (
7
11
) (−
8
11
)
0 (
1
11
) (
2
11
)
(
1
2
) (−
3
11
) (
5
11
)
 DATE :02/11/2018
NUMERICAL METHODS 24

25. CONCLUSION :
1. In case of matrix inversion method the process of computation will be
lengthy if the number of equation is as well as number of variables are more.
But by using the gauss elimination method we just have to use the formula to
get the upper triangular matrix & compute the inverse matrix.
NUMERICAL METHODS 25

26. REFERENCES
 CLASS NOTES
 NUMERICAL METHODS:- S.A. MOLLAH
NUMERICAL METHODS 26

27. THANK YOU
NUMERICAL METHODS 27