q8 - w gives v = ( A-I ) w as a desired member of the...

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MA366 Sathaye Quiz 8 Name Sec. Row 1. Find the Special Fundamental Matrix for the system: X 0 = - 3 - 2 8 5 ! X . Be sure to show all work. Hint: Try the substitution X = e t Y . Answer: The substitution gives a new equation Y 0 = - 4 - 2 8 4 ! Y = BY where B 2 = 0. Hence a Special Fundamental Matrix for the new system is I + Bt. The corresponding matrix for the original system is then e t ( I + Bt ) . Using the book method, you see that the original matrix A = - 3 - 2 8 5 ! has eigenvalue 1 with multiplicity 2. There is only one independent eigenvector v = 1 - 2 ! . You need a generalized eigenvector w satisfying ( A - I ) w = v . It is easy to see that w = 0 - 1 / 2 ! works. This gives Ψ( t ) = e t ± v tv + w ² = 1 t - 2 - 2 t - 1 / 2 ! To get the desired Special Matrix, you take Φ( t ) = Ψ( t )Ψ(0) - 1 = Ψ( t ) 1 0 - 2 - 1 / 2 ! - 1 . Note. A better way to find v,w is to start with w as any convenient vector for which ( A - I ) w = v is non zero. For example, we could have chosen w = 1 0 ! . Then we use v = ( A - I ) w = - 4 - 2 8 4 ! 1 0 ! = - 4 8 ! . The point is that since ( A - I ) 2 = 0 any non zero vector in the column space of ( A - I ) is an eigenvector belonging to 1. And any choice of
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Unformatted text preview: w gives v = ( A-I ) w as a desired member of the column space, as long as it is non zero. 2. You are given that: T = 1 1-2-1 ! , A = 1 2-4-5 ! , and T-1 AT =-3-1 ! . Use this information to nd a particular solution to X = AX +-1 1 ! . Hint: Use the substitution X = TY to nd the Special Fundamental Matrix for the given system and then use it. Answer: We know that the Special Fundamental Matrix is given by the formula: ( t ) = T e-3 t e-t ! T-1 and variation of parameters gives the answer: X = ( t ) u where u = -1 ( t )-1 1 ! = T e 3 t e t ! T-1-1 1 ! We thus solve: u = T-e t ! and get u = T-e t ! . Thus, a particular solution is T e-3 t e-t ! T-1 T-e t ! = T-1 ! =-1 1 ! . 2...
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q8 - w gives v = ( A-I ) w as a desired member of the...

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