# Math 20D Matlab 4.docx - Section C05 TA Garrett Williams...

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Section: C05TA: Garrett WilliamsAssignment: Math 20D, Matlab 4Exercise 4.1a.B = [1.2 2.5; 4 0.7]B = 1.2000 2.5000 4.0000 0.7000b.[eigvec, eigval] = eig(B)eigvec = 0.6501 -0.5899 0.7599 0.8075 eigval = 4.1221 0 0 -2.2221Exercise 4.2a.d.b.A = [1 3; -1 -8]A = 1 3 -1 -8 [eigvec, eigval] = eig(A) eigvec = 0.9934 -0.3276 -0.1148 0.9448 eigval = 0.6533 0 0 -7.6533c.v(t)=c1e0.6533t[0.99340.1148]+c2e7.6533t[0.32760.9448]1
Section: C05TA: Garrett WilliamsAssignment: Math 20D, Matlab 4Exercise 4.3
c.Exercise 4.4
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Section: C05TA: Garrett WilliamsAssignment: Math 20D, Matlab 4A = 1.2500 -0.9700 4.6000 -2.6000 -5.2000 -0.3100 1.1800 -10.3000 1.1200 eig(A)ans = 5.5698 + 0.0000i -4.1999 + 2.6606i -4.1999 - 2.6606ib.The system is unstable because its eigenvalues include real, nonzero values. The solutions are not bounded.Exercise 4.5a.eig(A)ans = -0.0329 + 0.9467i -0.0329 - 0.9467i -0.5627 + 0.0000i -0.0073 + 0.0000ib.x’=Axis an asymptotically stable system. Its eigenvalues have real and negative components; hence the solution tends to zero.c.[eigvec, eigval] = eig(A)eigvec = Columns 1 through 3 0.1994 - 0.1063i 0.1994 + 0.1063i -0.0172 + 0.0000i -0.0780 - 0.1333i
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