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Unformatted text preview: . Then the Jacobian of a is the 2 3 matrix b J ( , x ) = s , (1 ) I + J ( x ) , where J ( x ) is the matrix from Problem 1. (b) In order for the function to be transversal to zero, the matrix b J ( , x ) must be full rank (i.e., rank-2) at every point [0 , 1), x, y ( , ). The matrix J ( x ) has two eigenvalues call them 1 and 2 . The matrix K = (1 ) I + J ( x ) has eigenvalues (1 )+ i , so it is singular only if = 1 / (1 1 ) or = 1 / (1 2 ). Even if that happens, it is likely that the vector s will point in a dierent direction, making the rank of J ( , x ) equal to 2....
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- Fall '11