The determinant of the solutions of a linear congruence

**perwiradua** · March 3rd 2008, 07:02 PM

Hi,

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

--> begin statement

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

$\sum_{i=1}^3 x_i \; \alpha_{ik} \equiv 0 \quad (a_0) \qquad (k = 1, 2, 3)$

possesses 3 solutions, their determinant has the value

$<br /> H = \frac{a_0^3}{a_1 \; a_2 \; a_3} <br />$

where a1, a2, a3 are the greatest common divisor the module(number?) a0 with the elementary divisors e1, e2, e3 respectively of the system \alpha.''

--> end statement

The matrix \alpha is a 3x3 matrix with integer entries. My understanding is
that the linear congruence:

$a_{11} x_1 + a_{12} x_2 + a_{13} x_3 \equiv 0 \mod a_0$

$a_{21} x_1 + a_{22} x_2 + a_{23} x_3 \equiv 0 \mod a_0$

$a_{31} x_1 + a_{32} x_2 + a_{33} x_3 \equiv 0 \mod a_0$

(where aij are entries of \alpha)

has three incongruent solutions (please prove it), and upon forming a 3x3 matrix of the solutions the determinant of the matrix is a0^3/(a1*a2*a3) where

$a_1 = \gcd(a_0,e_1),$

$a_2 = \gcd(a_0,e_2),$

$a_3 = \gcd(a_0,e_3),$

where e1, e2 and e3 are the elementary divisors of the matrix \alpha.

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

in the reply to this post.

Thank you.

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