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GROUP NAME: BANGUL KHAN(F16BM05)
DEPARTMENT: BIOMEDICAL ENGINEERING(MUET)
PRESENTATION SUBJECT : LAAG.
PRESENTATION TOPIC : GUASS’S ELIMINATION METHOD AND
EXAMPLE.
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
Use eliminatory row operation on Augmented matrix to find
echelon form.
1 X X
0 1 X
0 0 1
ECHELON
MATRIX
Using backward substitution to find the answer of problem.
 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
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 .
ANTIGEN-A
ANTIGEN-B
ANTIGEN-C
x + y + z = 15
x + 2y + 3z = 30
x + 2y + 5z = 45
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
 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
 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

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
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
 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
2z = 15 z = 15/2
Z = 7.5
Y + 2z = 15 y + 2(7.5) = 15
y = 15/15y = 1
x + y + z = 15 x + 1 + 7.5 =15
x = 15 -8.5x = 6.5
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.
THANKS:

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GUASS’S ELIMINATION METHOD AND EXAMPLE

  • 1.
  • 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.