% This code solves the three-loop circuit problem.
% Three methods are used.
% The resistors have the following values:
r1 = 1;
r2 = .1;
r3 = 3;
r4 = 2;
r5 = 4;
% The batteries have the following values:
E1 = 10;
E2 = 20;
E3 = 30;
d = [E1 -E2 0 0 -E3 0 0]';
A=[r1 0 0 0 0 -1 0;
0 r2 0 0 0 1 -1;
0 0 r3 0 0 -1 0;
0 0 0 r4 0 0 -1;
0 0 0 0 r5 0 1;
-1 1 -1 0 0 0 0;
0 -1 0 -1 1 0 0];
% Use Gauss elimination method.
A\d
% Use inverse matrix method.
inv(A)*d
% Use block LU factorization method.
A11 = A(1:5,1:5);
A12 = A(1:5,6:7);
A21 = A(6:7,1:5);
A22 = A(6:7,6:7);
d1 = d(1:5);
d2 = d(6:7);
A22hat = A22 - A21*inv(A11)*A12
d2hat = d2 - A21*inv(A11)*d1
x2 = A22hat\d2hat
x1 = A11\(d1 - A12*x2)