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107 MAE Spring 2007 HW 4 Problem 1 Contents The terms can be collected to give: The terms can be collected to give: A = [0 -7 5;0 4 7;-4 3 -7] b = [50;-30;40] x = A\b AT = A' AI = inv(A) % transpose % inverse A = 0 0 -4 -7 4 3 5 7 -7 b = 50 -30 40 x = -15.1812 -7.2464 -0.1449 AT = 0 -7 5 0 4 7 -4 3 -7 AI = -0.1775 -0.1232 -0.2500 -0.1014 0.0580 0.0725 0.1014 0 0 Published with MATLAB 7.4 MAE 107 Spring 2007 HW 4 Problem 2 Contents Function to multiply two matrices Test mat_mult for matrices from Prob 8.4 Function to multiply two matrices type mat_mult.m % function to multiply two matrices function X=mat_mult(Y,Z) if nargin < 2 disp('at least 2 input arguments required'); return end [m,n]=size(Y); [n2,p]=size(Z); if n ~= n2 disp('Inner matrix dimensions must agree.'); return end for i = 1:m for j = 1:p s=0; for k=1:n s=s+Y(i,k)*Z(k,j); end X(i,j)=s; end end end Test mat_mult for matrices from Prob 8.4 A = [6 -1;12 8;-5 4] B = [4 0;0.5 2] C = [2 -2;-3 1] mat_mult(A,B) mat_mult(A,C) mat_mult(B,C) mat_mult(C,B) mat_mult(B,A) mat_mult(C,A) A = 6 12 -5 -1 8 4 B = 4.0000 0.5000 0 2.0000 C = 2 -3 -2 1 ans = 23.5000 52.0000 -18.0000 -2.0000 16.0000 8.0000 ans = 15 0 -22 -13 -16 14 ans = 8 -5 -8 1 ans = 7.0000 -11.5000 -4.0000 2.0000 Inner matrix dimensions must agree. Inner matrix dimensions must agree. Published with MATLAB 7.3 MAE 107 Spring 2007 HW 4 Problem 3 Contents Part (a) Part (b) Part (a) The determinant can be computed as follows: D = 0*det([2 -1; -2 0]) - (-3)*det([1 -1; 5 0]) + 7*det([1 2; 5 -2]) D = -69 Part (b) Cramer's rule x1 = det([2 -3 7; 3 2 -1; 2 -2 0]) / D x2 = det([0 2 7; 1 3 -1; 5 2 0]) / D x3 = det([0 -3 2; 1 2 3; 5 -2 2]) / D x1 = 0.9855 x2 = 1.4638 x3 = 0.9130 Published with MATLAB 7.3 HW 4 Problem 3 Part (c) Pivoting is necessary, switch so first and third rows 5x1 - 2x2 = 2 x1 + 2x2 + x3 = 3 -3x2 + 7x3 = 2 Multiply pivot row 1 by 1/5 and subtract the result from the second row to eliminate the a21 term. 5x1 - 2x2 = 2 2.4x2 - x3 = 2.6 -3x2 + 7x3 = 2 Pivoting is necessary so switch the second and third row, 5x1 - 2x2 = 2 -3x2 + 7x3 = 2 2.4x2 - x3 = 2.6 Multiply pivot row 2 by 2.4/(3) and subtract the result from the third row to eliminate the a32 term. 5x1 - 2x2 = 2 -3x2 + 7x3 = 2 4.6x3 = 4.2 By back substituion, x3 = and, x2 = 1.463768, x1 = 0.985507 4.2 = 0.913043 4.6 Part (d) Substituing back in original equations, we see that they are satisfied, -3(1.463768) + 7(0.913043) = 2 0.985507 + 2(1.463768) - 0.913043 = 3 5(0.985507) - 2(1.463768) = 2 1 HW 4 Problem 4 Discretization of problem Discretizing the continuous ODE using central finite difference, 0=D Collecting terms, -(D + 0.5U x)ci-1 + (2D + kx2 )ci - (D - 0.5U x)ci+1 = 0 Assuming x = 1, and substituting parameters, -2.5ci-1 + 4.2ci - 1.5ci+1 = 0 ci+1 - 2ci + ci-1 c-+1 - ci-1 -U - kci 2 x 2x Matlab output A=[4.2 -1.5 0 0 0 0 0 0 0 -2.5 4.2 -1.5 0 0 0 0 0 0 0 -2.5 4.2 -1.5 0 0 0 0 0 0 0 -2.5 4.2 -1.5 0 0 0 0 0 0 0 -2.5 4.2 -1.5 0 0 0 0 0 0 0 -2.5 4.2 -1.5 0 0 0 0 0 0 0 -2.5 4.2 -1.5 0 0 0 0 0 0 0 -2.5 4.2 -1.5 0 0 0 0 0 0 0 -2.5 4.2] b=[200 0 0 0 0 0 0 0 30]' c=A\b A = 4.2000 -2.5000 0 0 0 0 0 0 0 -1.5000 4.2000 -2.5000 0 0 0 0 0 0 0 -1.5000 4.2000 -2.5000 0 0 0 0 0 0 0 -1.5000 4.2000 -2.5000 0 0 0 0 0 0 0 -1.5000 4.2000 -2.5000 0 0 0 0 0 0 0 -1.5000 4.2000 -2.5000 0 0 0 0 0 0 0 -1.5000 4.2000 -2.5000 0 0 0 0 0 0 0 -1.5000 4.2000 -2.5000 0 0 0 0 0 0 0 -1.5000 4.2000 b = 200 1 0 0 0 0 0 0 0 30 c = 68.6755 58.9581 50.6236 43.4825 37.3783 32.1884 27.8305 24.2779 21.5940 Plotting plot(0:10, [80 c' 20]) 2 80 70 60 50 40 30 20 0 1 2 3 4 5 6 7 8 9 10 3

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