MATH_20D#4 - Exercise 4.1(a Code >> B=[1.2 2.5 4 0.7 Output...

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Exercise 4.1 (a) Code: >> B=[1.2, 2.5; 4, 0.7] Output: B = 1.2000 2.5000 4.0000 0.7000 (b) Code: >> [eigvec,eigval]=eig(B) Output: eigvec = 0.6501 -0.5899 0.7599 0.8075 eigval = 4.1221 0 0 -2.2221 Exercise 4.2 (a)
(b) Code: >> A=[1,3; -1,-8] Output: A = 1 3 -1 -8 Code: >> [eigvec, eigval] = eig(A) Output: eigvec = 0.9934 -0.3276 -0.1148 0.9448 eigval = 0.6533 0 0 -7.6533 (c) (d)
x ' = x + 3 y y ' = - x - 8 y -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 x y It matches my answer in (c) Exercise 4.3 (a) Code: >> A=[2.7, -1; 4.2, 3.5] Output: A = 2.7000 -1.0000 4.2000 3.5000 Code: >> [eigvec, eigval] = eig(A) Output: eigvec =
-0.0856 + 0.4301i -0.0856 - 0.4301i 0.8987 0.8987 eigval = 3.1000 + 2.0100i 0 0 3.1000 - 2.0100i (b) (c) x' = 2.7 x - y y' = 4.2 x + 3.5 y -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 x y Since the real part of eigenvalue is positive, which is 3.1, the solution becomes unbounded as t goes to infinity. Exercise 4.4 (a) Code: >> A = [1.25, -.97, 4.6; -2.6, -5.2, -.31; 1.18, -10.3, 1.12]
Output: A = 1.2500 -0.9700 4.6000 -2.6000 -5.2000 -0.3100 1.1800 -10.3000 1.1200 Code: >> [eigvec, eigval] = eig(A) Output: eigvec = 0.7351 -0.4490 - 0.2591i -0.4490 + 0.2591i -0.1961 0.3375 - 0.2242i 0.3375 + 0.2242i 0.6490 0.7530 0.7530 eigval = 5.5698 0 0 0 -4.1999 + 2.6606i 0 0 0 -4.1999 - 2.6606i (b) The system is not stable. The three eigenvalues the system has are one real number (5.5698) and two complex numbers (-4.1999 + 2.6606i and -4.1999 - 2.6606i). The real eigenvalue is positive, which means the solution goes to infinity as t becomes to infinity. Since the complex eigenvalues contain real part, which is -4.1999. This indicates that as t goes to

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