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Math 2280 Quiz 7 1. Find the general solution to the system ~x′ = 6 1 −7 −2 ~x. First we must find the eigenvalues of the coefficient matrix. To do this, we solve the characteristic equation 6−λ −7 0 = = (6 − λ)(−2 − λ) + 7 = −12 − 6λ + 2λ + λ2 + 7 1 −2 − λ = λ2 − 4λ − 5 ⇒ λ2 − 4λ + 4 = 5 + 4 ⇒ (λ − 2)2 = 9 ⇒ λ = 2 ± 3 = 5 or − 1. Now we need the eigenvectors corresponding to these eigenvalues. For λ = 5 we have 1 −7 1 −7 A − 5I = ∼ ⇒ a = 7b. 1 −7 0 0 If we choose b = 1, then an eigenvector corresponding to λ = 5 is 7 ~v1 = . 1 For λ = −1, we have A+I = 7 −7 1 −1 ∼ 1 −1 0 0 ⇒ a = b. 1 1 If we choose b = 1, then ~v2 = 1 1 . Our general solution is ~x(t) = e 5t 7 1 +e −t . 2. Find the general solution to the system 1 0 0 ~x′ = 1 3 0 ~x. −2 −4 1 Since A is lower triangular, it is clear that its eigenvalues are λ = 1, 1, and 3. Since 3 is not a repeated eigenvalue, let’s deal with it first. For λ = 3 we have 1 0 0 1 0 0 −2 0 0 1 0 ∼ 0 1 21 ⇒ a = 0, b = − c. 0 0 ∼ 0 0 A − λI = 1 2 0 −4 −2 0 0 0 −2 −4 −2 If we choose c = 2, then an eigenvecto for λ = 3 is 0 ~v1 = −1 . 2 For λ = 1 we have 0 0 0 1 1 2 0 0 A − λI = −2 −4 0 0 1 2 0 0 0 ⇒ a = 2b, 0 0 so if we choose b = 0 and c = 1, then we get 0 ~v2 = 0 . 1 If instead we choose b = 1 and c = 0, we get −2 ~v3 = 1 , 0 so even though λ = 1 is repeated with algebraic multiplicity 2, we were able to find two linearly independent eigenvectors. Thus the geometric multiplicity of λ = 1 is 2 and its defect is zero. Our general solution is ~x(t) = c1 e3t~v1 + c2 et~v2 + c3 et~v3 . 2