1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
|
//Example No. 7_03
// Gauss Elimination using partial pivoting
// Pg No. 220
clear ; close ; clc ;
A = [ 2 2 1 ; 4 2 3 ; 1 -1 1];
B = [ 6 ; 4 ; 0 ];
[ ar , ac ] = size(A);
Aug = [ 2 2 1 6 ; 4 2 3 4 ; 1 -1 1 0 ];
for i = 1 : ar-1
[ p , m ] = max(abs(Aug(i:ar,i)))
Aug(i:ar,:) = Aug([i+m-1 i+1:i+m-2 i i+m:ar],:);
disp(Aug)
for k = i+1 : ar
Aug(k,i:ar+1) = Aug(k,i:ar+1) - (Aug(k,i)/Aug(i,i))*Aug(i,i:ar+1);
end
disp(Aug)
end
//Back Substitution
X(ar,1) = Aug(ar,ar+1)/Aug(ar,ar)
for i = ar-1 : -1 : 1
X(i,1) = Aug(i,ar+1);
for j = ar : -1 : i+1
X(i,1) = X(i,1) - X(j,1)*Aug(i,j);
end
X(i,1) = X(i,1)/Aug(i,i);
end
printf('\n The solution can be obtained by back substitution \n x1 = %i \n x2 = %i \n x3 = %i \n',X(1,1),X(2,1),X(3,1))
|
