SOLUTION OF LINEAR EQUATION BY LU FACTORIZATION METHOD.

1. LU FACTORIZATION
METHOD

2. 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]

3. [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.

4.  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
)

5.  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)

6. 
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

7.  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 .

8. THANK YOU