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ch 9 rudin

ch 9 rudin - Math 118C Homework 3 Solutions Charles Martin...

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Math 118C Homework 3 Solutions Charles Martin April 21, 2009 9.19 Show that the system of equations 3 x + y - z + u 2 = 0 x - y + 2 z + u = 0 2 x + 2 y - 3 z + 2 u = 0 can be solved in terms of x , in terms of y , in terms of z , but not in terms of u . For k = 1 , 2 , 3 define f k : R 4 R by f 1 ( x, y, z, u ) = 3 x + y - z + u 2 f 2 ( x, y, z, u ) = x - y + 2 z + u f 3 ( x, y, z, u ) = 2 x + 2 y - 3 z + 2 u. Let f : R 4 R 3 be the map given by f = ( f 1 , f 2 , f 3 ). Clearly f is smooth and f (0 , 0 , 0 , 0) = 0. We can compute f (0 , 0 , 0 , 0); in the standard basis the Jacobian matrix is 3 1 - 1 0 1 - 1 2 1 2 2 - 3 2 Consider the problem of solving for y, z, u in terms of x in the equation f ( x, y, z, u ) = 0. By the implicit function theorem, this is possible since the determinant of the corresponding submatrix is nonzero: det 1 - 1 0 - 1 2 1 2 - 3 2 = 3 = 0 Similarly, the equation f ( x, y, z, u ) = 0 can be solved for x, z, u in terms of y since the corresponding submatrix has nonzero determinant: det 3 - 1 0 1 2 1 2 - 3 2 = 21 = 0 . Also, the same equation can be solved for x, y, u in terms of z because the corresponding submatrix is invertible: det 3 1 0 1 - 1 1 2 2 2 = - 12 = 0 . Finally, suppose we try to solve f ( x, y, z, u ) = 0 for u . Treating u as a parameter, we can use techniques of linear algebra and attempt to solve the system via Gaussian elimination. 3 x + y - z = - u 2 x - y + 2 z = - u 2 x + 2 y - 3 z = - 2 u = 3 1 - 1 - u 2 1 - 1 2 - u 2 2 - 3 - 2 u = 1 - 1 2 - u 0 4 - 7 - u 2 + 3 u 0 0 0 u 2 - 3 u We see that the equation is solvable only when u ∈ { 0 , 3 } . Thus we cannot solve for

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