# MatLab Assignment 4.docx - Math 20D MatLab Assignment 4...

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Math 20D MatLab Assignment 4 Exercise 4.1 a. >> B= [1.2, 2.5; 4, .7] B = 1.2000 2.5000 4.0000 0.7000 b. >> [eigvec, eigval] = eig(B) eigvec = 0.6501 -0.5899 0.7599 0.8075 eigval = 4.1221 0 0 -2.2221 Exercise 4.2 a. >> A = [1,3; -1, -8] A = 1 3 -1 -8 b. >> [eigvec, eigval] = eig (A) eigvec = 0.9934 -0.3276 -0.1148 0.9448 eigval = 0.6533 0 0 -7.6533 c. v(t) = c 1 e .6533t (.9934 ) + c 2 e -7.6533 (-.3276 ) = (.9934c 1 e .6533 -.3276c 2 e -7.6533 ) (-.1148) (.9448 ) (-.1148c 1 e .6533 +.9448c 2 e -7.6533 ) As t approaches infinity, v(t) tends towards [+inf; -inf] in the direction [.9934; -.1148] for positive c 1 . This is the opposite for negat8ive c 1 . d. The plot support does support my answer, because the trend of v(t) approaches infinity when in the directions [.9934; -.1148] for positive c 1 and [-.9934; .1148] for negative c 1 .
Exercise 4.3 a. A = [2.7, -1; 4.1, 3.7] A = 2.7000 -1.0000 4.1000 3.7000 >> [eigvec, eigval] = eig(A) eigvec = -0.1093 + 0.4291i -0.1093 - 0.4291i 0.8966 + 0.0000i 0.8966 + 0.0000i eigval = 3.2000 + 1.9621i 0.0000 + 0.0000i 0.0000 + 0.0000i 3.2000 - 1.9621i b. v(t) = c 1 e (3.2+1.9621i)t (-.1093 + .4291i) + c 2 e (3.2-1.9621i)t (-.1093 - .4291i) (.8966 ) (.8966 ) c. The real component of the eigenvalue is what determines whether the solution tends to infinity or zero. If it is positive, it goes to infinity, if it is negative it tends towards zero.
Exercise 4.4 a. A = [1.25, -0.97, 4.6; -2.6, -5.2, -0.31; 1.18, -10.3, 1.12] [eigvec, eigval] = eig(A) A = 1.2500 -0.9700 4.6000 -2.6000 -5.2000 -0.3100 1.1800 -10.3000 1.1200 eigvec = 0.7351 + 0.0000i 0.4490 + 0.2591i 0.4490 - 0.2591i -0.1961 + 0.0000i -0.3375 + 0.2242i -0.3375 - 0.2242i 0.6490 + 0.0000i -0.7530 + 0.0000i -0.7530 + 0.0000i