Gaussian Elimination technique by matlab
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Hello every body , i am trying to solve an (nxn) system equations by Gaussian Elimination method using Matlab , for example the system below :
x1 + 2x2 - x3 = 3
2x1 + x2 - 2x3 = 3
-3x1 + x2 + x3 = -6
C = [ 1 2 -1 ; 2 1 -2 ; -3 1 1 ]
b= [ 3 3 -6 ]
by using this code :
% Matlab Program to solve (nxn) system equation
% by using Gaussian Elimination method
clear ; clc ; close all
n = input('Please Enter the size of the equation system n = ') ;
C = input('Please Enter the elements of the Matrix C ' ) ;
b = input('Please Enter the elements of the Matrix b ' ) ;
dett = det(C)
if dett == 0
print('This system unsolvable because det(C) = 0 ')
else
b = b'
A = [ C b ]
for j = 1:(n-1)
for i= (j+1) : n
mult = A(i,j)/A(j,j) ;
for k= j:n+1
A(i,k) = A(i,k) - mult*A(j,k) ;
A
end
end
end
for p = n:-1:1
for r = p+1:n
x(p) = A(p,r)/A(p,r-1)
end
end
end
everything is good but i need help to do the back substitution to find and print the matrix x ( which is contains the solutions of this system ) , could any one help me to do that ? i will thankful in advance Razi
9 commentaires
John D'Errico
le 14 Mai 2017
Modifié(e) : John D'Errico
le 14 Mai 2017
Note that testing to see if det(C)==0 is a terribly bad idea, since it will essentially NEVER be zero, even for a singular matrix. Yes, I know they taught you that in class. But "they" are wrong here.
For example:
A = rand(3,4);
A = [A;A(1,:)];
det(A)
ans =
-1.5101e-19
Here A has an EXACT duplicate row. Is det(A)==0?
How about this one, here A has rank 2, in a 10x10 matrix.
A = randn(10,2)*randn(2,10);
det(A)
ans =
-5.4818e-125
Not zero. How about this one?
A = randn(10,9)*rand(9,10)*100;
det(A)
ans =
7.1118e+06
Still not zero. In fact, this one had a pretty large determinant for a known to be singular matrix. So you cannot even test to see if det is a small number, since it can easily be quite large yet the matrix is still singular.
rank(A)
ans =
9
Use tools like rank or cond to decide if a matrix is singular. NOT det. Anyone who tells you to use det using floating point arithmetic is flat out wrong.
det is a liar.
Razi Naji
le 15 Mai 2017
Tanzina Nasrin Tania
le 24 Déc 2019
I don't understand that how much amounts of size in the equation system n?So please give me answer.
Robby Ching
le 26 Mai 2020
Hello! I have question. What does n:-1:1 mean? Thank you
AKSHAY KUMAR
le 2 Nov 2020
if n=3, then for p = 3:-1:1 will generate 3,2,1.
It simply means that p will go from 3 to 1 decreasing by 1
Robby Ching
le 3 Nov 2020
Thank you so much!
jealryn obordo
le 20 Fév 2022
can I ask , what is the value of n?
Muntasir
le 15 Juin 2022
Jealryn obordo
'n' is the number of variables or equations. Here n=3
shrawani
le 5 Mar 2024
what is A(i,j) and how, why it is used
Réponse acceptée
Plus de réponses (3)
M Waqar Nadeem
le 11 Mar 2021
C = [1 2 -1; 2 1 -2; -3 1 1]
b= [3 3 -6]'
A = [C b]; %Augmented Matrix
n= size(A,1); %number of eqns/variables
x = zeros(n,1); %variable matrix [x1 x2 ... xn] coulmn
for i=1:n-1
for j=i+1:n
m = A(j,i)/A(i,i)
A(j,:) = A(j,:) - m*A(i,:)
end
end
x(n) = A(n,n+1)/A(n,n)
for i=n-1:-1:1
summ = 0
for j=i+1:n
summ = summ + A(i,j)*x(j,:)
x(i,:) = (A(i,n+1) - summ)/A(i,i)
end
end
3 commentaires
Belkacem Faraheddine
le 25 Mar 2022
Big thanks
Ahmad
le 30 Avr 2022
Hi Nadeem
seems it does not work for:
C=[0 0 0 5 ; 4 0 2 5; 1 3 0 2; 3 4 2 0]
b=[16 10 13 -1]'
Mourad
le 30 Oct 2022
it does work only if the elements of the diagonal are different from zero.
pss CHARAN
le 2 Déc 2020
Modifié(e) : Walter Roberson
le 8 Mar 2024
clc
n=input(‘Enter number of variables’);
for i=1:1:n
for j=1:1:n
A(i,j)=input(‘Enter Coefficient”);
end
end
for i=1:1:n
for j=1:1:n
B(i,1)=input(‘Enter Constant’);
end
end
A
B
E=[AB]
for i=1:1:n
E(i,:)=E(i,:)+E(i+1,:);
end
E(n,1)=E(n,1)+E(1,1);
q=1;
for i=q:1:n
for j=1:1:n
if E(j,i)==0;
E(j,:)=E(j,:);
else
E(j,:)=E(j,:)/E(j,i);
end
end
for k=i+1:1:n
E(k,:)=E(k,:)-E(i,:);
end
q=i+1;
end
q=n;
for i=n:-1:1
for j=1:1:q
if E(j,i)==0;
E(j,:)=E(j,:);
else
E(j,:)=E(j,:)/E(j,i);
end
end
for k=1:1:i-1
E(k,:)=E(k,:)-E(i,:);
end
q=q-1;
end
X=E(:,n+1);
fprintf(‘****The Solution is****\n’)X
3 commentaires
Razi Naji
le 8 Fév 2021
Martin
le 17 Mai 2023
The lack of tabs/formatting broke my brain :D
DGM
le 18 Mai 2023
No documentation, no formatting, invalid characters, improper indexing. To add insult to injury, you harass the user by forcing them to blindly enter matrices using input() without any explanation of how the inputs should be oriented-- and then you throw it away and force them to do it again n times.
Why post something that's not even working code? It's not an answer, so you're not helping anyone else. It's not a question, so you're not helping yourself either. What's the point of this bizarre ritual?
I was going to fix the formatting, but it belongs like this.
Vishwa Lawliet
le 3 Sep 2022
Modifié(e) : DGM
le 18 Mai 2023
disp('bX = c')
%4
n = input('Enter the size of matrix')
%[2 1 -1 2; 4 5 -3 6; -2 5 -2 6; 4 11 -4 8]
b = input('Enter the elements of the Matrix b ' )
%[5;9;4;2]
c = input('Enter the elements of the Matrix c ' )
dett = det(b);
if dett == 0
disp('This system unsolvable because det(b) = 0 ')
else
a=[b c];
for i = 0:n-2
for j = 0:n-2-i
a(i+j+2:i+j+2,i+1:n+1)=(a(i+j+2:i+j+2,i+1:n+1).*(a(i+1,i+1)/a(i+j+2,i+1)))-a(i+1:i+1,i+1:n+1);
disp(a)
end
end
X=c';
for i = 0:n-1
X(n-i)=(a(n-i,n+1)-sum(a(n-i:n-i,1:n).*X)+a(n-i,n-i)*X(n-i))/a(n-i,n-i);
end
X=X';
disp(X)
end
1 commentaire
Deepak
le 6 Juin 2023
dett = det(b);
This is not valid as b is a row matrix
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