sln3 - 7224 Homework Solution 3 Problem 1 (a) For a 2 2...

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Unformatted text preview: 7224 Homework Solution 3 Problem 1 (a) For a 2 2 real matrix A = a 11 a 12 a 21 a 22 ! The eigenvalues of the matrix A are given by = 2- 4 2 where = trace ( A ) and = det ( A ) = a 11 a 22- a 12 a 21 . If > , < 0 and 2- 4 0 then there are two real eigenvalues, which are negative. If > , < 0 and 2- 4 < 0 then there are two complex eigenvalues with negative real parts. Each discussion above is invertible. So they are necessary and sufficient conditions for A to have eigenvalues with negative real parts. (b) For a linear dynamical system d u dt = A u When det ( A ) > 0 and trace ( A ) < 0, there exist an invertible 2 2matrix P such that the matrix B = P- 1 AP has one of the following forms B = 1 2 ! , B = 1 ! or B = a- b b a ! Where = a + ib is the complex eigenvalue in the third form. The system becomes d v dt = B v where v = P u is the linear transform of u . The eigenvalues ( s ) of v are also eigenvalues of u . 1 The solution of system could be written as u ( t ) = e Bt u where e Bt has one of the following forms e Bt = e 1 e 2 !...
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sln3 - 7224 Homework Solution 3 Problem 1 (a) For a 2 2...

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