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    The Jacobian Matrix in the Implicit Function Theorem

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    In the attached problem, I am having trouble showing that the determinants are in the same form in the Lagrangian as the one in the implicit function theorem.

    © BrainMass Inc. brainmass.com October 10, 2019, 7:49 am ad1c9bdddf
    https://brainmass.com/math/calculus-and-analysis/jacobian-matrix-implicit-function-theorem-596680

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    Let us start with a general discussion. Given a function G: R^nto R^m, denote by g_1,...,g_m: R^nto R the component
    functions of G, meaning that for X=(x_1,...,x_n)in R^n,
    G({X})=(g_1({X}),...,g_m({X})).
    Suppose that all the partial derivatives of g_1,...,g_m exist at a point {X}in R^n. We define the {Jacobian
    matrix} of G at {X}, denoted by G_{{X}}({X}) and also by DG({X}) or G'({X}), by
    the following expression:
    G_{{X}}({X})={pmatrix}
    pd{g_1({X})}{x_1} ... pd{g_1({X})}{x_n}
    vdots ddots vdots
    pd{g_m({X})}{x_1} ... pd{g_m({X})}{x_n}
    {pmatrix}.

    The Jacobian matrix of a function from R^n to R^m is an mtimes n matrix, that is, it has m rows and n columns. The
    number of rows equals the number component functions, or equivalently, equals the exponent of R where the image of G is
    contained, in this case m; the number of columns equals the number of variables, in this case the vector {X} has n
    variables.

    Note that the i^{th}-row of the Jacobian matrix corresponds to the (transpose) of the gradient vector nabla g_i,
    nabla g_i({X})={pmatrix}
    pd{g_i({X})}{x_1}
    pd{g_i({X})}{x_2}
    vdots
    pd{g_1({X})}{x_n}
    {pmatrix},

    so (nabla g_i)^T=(pd{g_i}{x_1},pd{g_i}{x_2},...,pd{g_i}{x_n}). Here the superscript T denotes transposition. We ...

    Solution Summary

    In the proof of the Theorem of Lagrange multipliers via the Implicit Function Theorem, we need to verify that a certain Jacobian matrix is non-singular. We explain how to calculate such a Jacobian matrix and why it is non-singular.

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