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# 1b 35 a 4 tra a 1 det a 3 a so tra 2 det a

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Unformatted text preview: en b = 0, the normal mo des are e t � , for c a real constant. When b = 0, the 0 normal � o des � re et v for any vector v. When b � 0, we must solve (A − � 1 I )w = v1 , that is, m a� � = � � 0b 1 0 w= . The solution is w = , so the extra solution is u2 = e�1 t (tv1 + w) = 00 0 1/b � � t t e . 1/b 35. (a) [4] trA = a − 1, det A = 3 − a, so trA = 2 − det A. det A = 0 when a = 3. trA = 0 � when a = 1. det A = (trA)2 /4 when a2 + 2a − 11 = 0 or a = −1 ± 2 3, i.e. a � −4.4641 and a = 2.4641. � (c) [4] Diagram showing: a < −1 − 2 3—stable no de = no dal sink � a = −1� 2 3—defective stable no de = defective no dal sink − −1 − 2 3 < a < 1—counterclo ckwise stable spiral = spiral sink a = 1—counterclockwise center � 1 < a < −1 + 2 3–counterclockwise unstable spiral = spiral source � a = 1� 2 3—unstable defective node = defective...
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## This note was uploaded on 01/22/2014 for the course MATH 18.03 taught by Professor Vogan during the Spring '09 term at MIT.

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