02_hwc1SolnsODDA

02_hwc1SolnsODDA - Adding the terms on the left and adding...

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Adding the terms on the left and adding the terms on the right, we cancel out y and find 3 x = u + v , or x = ( u + v ) / 3 . From the first equation above we have y = x - u , and so y = ( u + v ) / 3 - u = ( v - 2 u ) / 3 . Thus, with ± x y ² = 1 3 ± u + v v - 2 u ² , we have f ³± x y ²´ = ± u v ² . Since ± u v ² is an arbitrary vector in IR 2 , this shows that f is onto. Since, as we have seen in solving for x and y , this is the only vector with f ³± x y ²´ = ± u v ² , f is one to one. Since it is both one to one and onto, it is invertible. In fact, we have already found the formula for the inverse: f - 1 ³± u v ²´ = 1 3 ± u + v v - 2 u ² . The key thing to note is how the solution of all three parts revolves around solving the equation ( * ). That is the point of the question. 1.7 Consider the function from IR 2 to IR 2 defined by f µh x y = h x 2 - y 2 x - 2 i . (a) Find all solutions of f µh x y = h 0 0 i . (b) Is f invertible? If so find a formula for the inverse. If not, explain why not.
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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