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

1. Mar 4, 2008

Hi,

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

Accordingly a system of 3 homogeneous linear congruences with 3 unknowns

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

possesses 3 solutions, their determinant has the value

$$H = \frac{a_{00}^3}{a_1 a_2 a_3}$$

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

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

$$a_{11} x_1 + a_{12} x_2 + a_{13} x_3 \equiv 0 \pmod{a_{00}}$$

$$a_{21} x_1 + a_{22} x_2 + a_{23} x_3 \equiv 0 \pmod{a_{00}}$$

$$a_{31} x_1 + a_{32} x_2 + a_{33} x_3 \equiv 0 \pmod a_{00}}$$

(where $$a_{ij}$$ 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

$$\frac{a_{00}^3}{a_1 a_2 a_3}$$

where

$$a_1 = \gcd(a_{00},e_1),$$

$$a_2 = \gcd(a_{00},e_2),$$

$$a_3 = \gcd(a_{00},e_3),$$

where

$$e_1, e_2, e_3$$

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.

Thank you.

Last edited: Mar 4, 2008