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//Example 8.6
//Block Jacobi Method
//Page no. 281
clc;clear;close;
A=[10,1,0,0,0,-1;1,10,1,0,0,0;2,0,20,1,0,0;0,0,1,10,-1,0;0,3,0,0,30,3;0,0,0,2,-2,20]; //equation matrix
B=[5;10;10;0;0;5] //solution matrix
disp(B,'B=',A,'A=')
for i=1:3
for j=1:3
A11(i,j)=A(i,j);
end
B1(i,1)=B(i,1);
end
for i=1:3
for j=1:3
A12(i,j)=A(i,j+3);
end
end
for i=1:3
for j=1:3
A21(i,j)=A(i+3,j);
end
end
for i=1:3
for j=1:3
A22(i,j)=A(i+3,j+3);
end
B2(i,1)=B(i+3,1);
end
disp(B2,'B2=',B1,'B1=',A22,'A22=',A21,'A21=',A12,'A12=',A11,'A11=');
A11_1=inv(A11);A22_1=inv(A22);
disp(A22_1,'Inverse of A22=',A11_1,'Inverse of A11=')
for i=1:3
X1(i,1)=0;
X2(i,1)=0;
end
for r=1:2
X11=A11_1*(-1*A12*X2+B1);
X22=A22_1*(-1*A21*X1+B2);
X1=X11;
X2=X22;
disp(X1,'X1=')
disp(X2,'X2=')
end
for i=1:6
if(i<4)
X(i,1)=X1(i,1);
else
X(i,1)=X2(i-3,1);
end
end
disp(X,'X=')
printf('\n\n\nNote : There is a computation error in calculation of X1(2)')
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