2130_Chapter7_Notes

1 10 11 exa 2 slide 4 1 1 1 1 1 1 e2t 1

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Unformatted text preview: A − λ1 I = A − 3I = = −2 1 4 −2 Remember: λ is an eigenvalue ⇐⇒ det(A − λI) = 0. Use this to check your work: det(A − 3I) = −2 1 4 −2 =4−4=0 =⇒ λ1 = 3 is an eigenvalue. −2 1 4 −2 (A − 3I)v = 0 =⇒ v1 v2 = 0 0 =⇒ −2v1 + v2 = 0 (first equation) c 1 =c , c=0 2c 2 =⇒ v2 = 2v1 =⇒ v(1) = Again, check your work: (A − 3I)v(1) = −2 1 4 −2 1 2 = 0 0 . 5 / 11 3 Exa 1 (slide 3) λ2 = −1 =⇒ A − λ2 I = A + I = Check #1: det(A + I) = 1+1 1 4 1+1 = 21 42 21 = 4 − 4 = 0. 42 (A + I)v = 0 =⇒ 21 42 v1 0 = v2 0 =⇒ 2v1 + v2 = 0 (first equation) =⇒ v2 = −2v1 =⇒ v(2) = Check #2: (A + I)v(2) = 21 42 c 1 =c , c=0 −2c −2 1 0 = . −2 0 6 / 11 Exa 1 (slide 3) λ1 = 3, v(1) = λ2 = −1, v(2) = 1 2 =⇒ x(1) = 1 −2 1 3t e 2 =⇒ x(2) = 1 −t e −2 The solutions x(1) , x(2) are linearly independent because v(1) , v(2) are linearly independent =⇒ x = c1 x(1) + c2 x(2) is the general solution of x′ = Ax. General solution: x = c1 1 3t 1...
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