Liner Algebra and Analytical method

2. GROUP NAME: BANGUL KHAN(F16BM05)
DEPARTMENT: BIOMEDICAL ENGINEERING(MUET)
PRESENTATION SUBJECT : LAAG.
PRESENTATION TOPIC : GUASS’S ELIMINATION METHOD AND
EXAMPLE.

3. GUASS’S ELIMINATION METHOD:
It is method which is used to solve
the non homogenous system of linear equation with square.
 This method has four steps:
Convert the system of linear equation into matrix form.
Ax=b
Construct the Augmented matrix which is denoted by Ab or A/B

4. Use eliminatory row operation on Augmented matrix to find
echelon form.
1 X X
0 1 X
0 0 1
ECHELON
MATRIX

5. Using backward substitution to find the answer of problem.

6.  Example :
A lab technician needs 15% amount of Antigen-A, 30% of Antigen-B
and 45% of Antigen-C to test the blood Group. There is three sources
( x y z ) where these antigen can be obtained and composition of Antigen-A ,
Antigen-B and Antigen-C according to sources are given below.
ANTIGEN SOURCE x SOURCE y SOURCE z
A 1 1 1
B 1 2 3
C 1 2 5

7. Using Guass elimination method to determine the required amount of
Antigen-A , Antigen-B and Antigen-C that must be satisfied from the tabular
sources in order to meet the lab technician need?
SOLUTION:
Let x, y, z be the sources of each antigen so according to question
first we made linear equation of each antigen then we apply Guass’s elimination
method .

8. ANTIGEN-A
ANTIGEN-B
ANTIGEN-C
x + y + z = 15
x + 2y + 3z = 30
x + 2y + 5z = 45

9. So , Apply steps of Guass elimination method to find the answer.
 Step no 1:
Convert the system of linear equation into matrix form.
Ax = b
1 1 1 X 15
1 2 3 y = 30
1 2 5 z 45

10.  Step no 2:
Construct the Augmented matrix which is denoted by Ab or A/B
1 1 1 . 15
Ab = 1 2 4 . 60
1 2 5 . 45
Augmented
matrix

11.  Step no 3:
Apply eliminatory row operation on Augmented matrix to find
echelon matrix.
1 1 1 . 15
Ab = 1 2 3 . 30 R2 – R1
1 2 5 . 45
R2 = 1 2 3 30
(-)R1 = 1 1 1 15
0 1 2 15

12. 
1 1 1 . 15
Ab = 0 1 2 . 15 R3 – R1
1 2 5 . 45
R3 = 1 2 5 45
(-) R1 = 1 1 1 15
0 1 4 30
1 1 1 . 15
Ab = 0 1 2 . 15 R3 – R2
0 1 4 . 30

13. R3 = 0 1 4 30
(-) R2 = 0 1 2 15
0 0 2 15
1 1 1 . 15
Ab = 0 1 2 . 15
0 0 2 . 15
So finally echelon form
acheived

14.  Step no 4:
Using backward substitution on echelon matrix to find the answer of
problem.
1 1 1 . 15
Ab = 0 1 2 . 15
0 0 2 . 15
x + y+ z = 15
y + 2z = 15
2z = 15

15. 2z = 15 z = 15/2
Z = 7.5

16. Y + 2z = 15 y + 2(7.5) = 15
y = 15/15y = 1

17. x + y + z = 15 x + 1 + 7.5 =15
x = 15 -8.5x = 6.5

18. x = 7.5
Y = 1
Z = 6.5
These are the amount of
sources to meet the demand of
antigens to test the blood
group by lab technician.

19. THANKS: