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hw3sol - %Let's define the right hand side vector b...

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Problem 1 Matlab Code %Let's define a square matrix A=[10 2 -1; -3 -6 2; 1 1 5] %Determinant of A detA = det(A) %Inverse of A invA = inv(A) %Let's see if invA is really the inverse of A I1 = A*inv(A) I2 = inv(A)*A Output A = 10 2 -1 -3 -6 2 1 1 5 detA = -289 invA = 0.1107 0.0381 0.0069 -0.0588 -0.1765 0.0588 -0.0104 0.0277 0.1869 I1 = 1.0000 0.0000 0 0.0000 1.0000 0.0000 -0.0000 0.0000 1.0000
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I2 = 1.0000 0.0000 0.0000 -0.0000 1.0000 0.0000 0 0 1.0000
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Problem 2 Matlab Code: A = [-cos(pi/6) 0 cos(pi/3) 0 0 0; -sin(pi/6) 0 -sin(pi/3) 0 0 0; cos(pi/6) 1 0 1 0 0; sin(pi/6) 0 0 0 1 0; 0 -1 -cos(pi/3) 0 0 0; 0 0 sin(pi/3) 0 0 1]; b = [0; 1000; 0; 0; 0; 0]; forces = A\b Output forces = -500.0000 433.0127 -866.0254 0 250.0000 750.0000
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%Let's define the coefficient matrix A A=[-6 0 1 0 0; 3 -3 0 0 0; 0 1 -9 0 0; 0 1 8 -11 2; 3 1 0 0 -4]; c01=10;
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Unformatted text preview: %Let's define the right hand side vector b b=[-5*c01; 0; -160; 0; 0]; [L,U]=lu(A); z = L\b; c = U\z %part 2 of the problem c01=1:20; for i=c01 b=[-5*c01(i); 0; -160; 0; 0]; z = L\b; c = U\z; c2(i) = c(2); end plot(c01,c2);grid on 2 4 6 8 10 12 14 16 18 20 2 4 6 8 10 12 14 16 18 20 %problem 5 - Newton's method clear all ; ITERMAX = 2; Tol = 1e-4; % give the initial guess x = [0.75;0.75]; %note I use x(1)=x & x(2) = y iter = 0; while (1) xold = x; %define the Jacobian matrix & right hand side A = [1-x(2) -x(1);x(2) -1+x(1)]; b = -[(x(1)-x(1)*x(2)); (-x(2) +x(1)*x(2))]; %solve the system delx = A\b; iter = iter + 1; disp([ 'Iter : ' num2str(iter)]); %new root x = xold + delx %RE re = abs((x - xold)./x); if ((max(re) < Tol) | (iter == ITERMAX)) break ; end end OUPUT: Iter : 1 x = 1.1250 1.1250 Iter : 2 x = 1.0125 1.0125...
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