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Unformatted text preview: + x 2 y + xy-( y-a 2 ) . Then the Jacobian of a is the 2 3 matrix 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 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 diFerent direction, making the rank of J ( , x ) equal to 2. 2...
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- Fall '06