Question

(Use MATLAB) Use Gaussian elimination with backward substitution to solve the following linear system. For this problem you will have to do scaled partial pivoting. The matrix A and the vector b are in the Matlab code shown below

A=[3 -13 9 3;-6 4 1 -18;6 -2 2 4;12 -8 6 10];
display(A);
b=[-19;-34;16;26];
display(b);

Answer 1

Please find the required MATLAB script as the following:

%==================================================

clear all;
clc;

% Given data
A=[3 -13 9 3;-6 4 1 -18;6 -2 2 4;12 -8 6 10];
display(A);
b=[-19;-34;16;26];
display(b);

% Preliminary calculation
n=length(A(:,1));
A=[A,b];
x=zeros(n,1);
flag=0;
NROW=1:n;

for i=1:n-1
    if flag==0
        M=max(abs(A(NROW(i:n),i)));
        if M==0
            flag=1;
        else
            I=find(abs(A(NROW(i:n),i))==M);
            p=I(1)+i-1;
            if p>i
                NCOPY=NROW(i);
                NROW(i)=NROW(p);
                NROW(p)=NCOPY;
            end
        end
        if flag==0
            for j=i+1:n
                m=A(NROW(j),i)/A(NROW(i),i);
                A(NROW(j),i:n+1)=A(NROW(j),i:n+1)-m*A(NROW(i),i:n+1);
                clear m;
            end
        end
    end
end
if flag==0
   if A(NROW(n),n)==0
      flag=1;
   end
end
if flag==0
    x(n,1)=A(NROW(n),n+1)/A(NROW(n),n);
    for i=n-1:-1:1
        x(i,1)=(A(NROW(i),n+1)-sum(x(i+1:n,1)'.*A(NROW(i),i+1:n)))/A(NROW(i),i);
    end
else
    x='no unique solution may exist';
end
display('The solution is: ');
display(x);

%==================================================

output:

A- 3 13 9 3 1 18 6 2 24 12 8 6 10 -6 4 -19 -34 LES 26 The solution is: 3.0000 L. -2.0000 L.