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Proof required: The determinant of the solutions of a linear homogeneous congruence

  1. Mar 4, 2008 #1

    I came across a statement by Frobenius, in German, and online translator says,

    ``Accordingly a system of 3 homogeneous linear congruences with 3 unknowns

    [tex] \sum_{i=1}^3 x_i \; \alpha_{ik} \equiv 0 \quad (a_{00}) \qquad (k = 1, 2, 3) [/tex]

    possesses 3 solutions, their determinant has the value

    [tex] H = \frac{a_{00}^3}{a_1 a_2 a_3} [/tex]

    where [tex] a_1, a_2, a_3 [/tex] are the greatest common divisor the module(number?) [tex] a_{00} [/tex]
    with the elementary divisors [tex] e_1, e_2, e_3 [/tex] respectively of the system [tex] \alpha [/tex].''

    The matrix [tex] \alpha [/tex] is a [tex] 3 \times 3 [/tex] matrix with integer entries. My understanding is
    that the linear congruence:

    [tex] a_{11} x_1 + a_{12} x_2 + a_{13} x_3 \equiv 0 \pmod{a_{00}} [/tex]

    [tex] a_{21} x_1 + a_{22} x_2 + a_{23} x_3 \equiv 0 \pmod{a_{00}} [/tex]

    [tex] a_{31} x_1 + a_{32} x_2 + a_{33} x_3 \equiv 0 \pmod a_{00}} [/tex]

    (where [tex] a_{ij} [/tex] are entries of [tex] \alpha [/tex])

    has three incongruent solutions (please prove it), and upon forming a 3x3 matrix
    of the solutions the determinant of the matrix is

    [tex] \frac{a_{00}^3}{a_1 a_2 a_3} [/tex]


    [tex] a_1 = \gcd(a_{00},e_1), [/tex]

    [tex] a_2 = \gcd(a_{00},e_2), [/tex]

    [tex] a_3 = \gcd(a_{00},e_3), [/tex]


    [tex] e_1, e_2, e_3 [/tex]

    are the elementary divisors of the matrix [tex] \alpha [/tex].

    I would appreciate it if
    (1) a proof of the statement is found,
    (2) an example illustrating the fact of the statement is provided.

    Thank you.
    Last edited: Mar 4, 2008
  2. jcsd
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