Well, let's see what happens if you don't do anything fancy and solve this by substitution. I'm going to do this little step by little step as I get screwed up when I skip steps. I'll do this for Z_7.

2x + 4y + 5z = 1

2x + y + z = 3

5x + 4y = 0

Solving the bottom equation we get:

5x + 4y = 0

5x + 4y + 3y = 0 + 3y

5x = 3y

In Z_7 the multiplicative inverse of 5 is 3, so:

3(5x) = 3(3y)

x = 2y

Put x = 2y into the remaining equations:

2(2y) + 4y + 5z = 1 ==> 4y + 4y + 5z = 1

2(2y) + y + z = 3 ==> 4y + y + z = 3

Which leads to

y + 5z = 1

5y + z = 3

I'm going to solve the bottom equation for z.

5y + z = 3

2y + 5y + z = 2y + 3

z = 2y + 3

Put this into the top equation:

y + 5(2y + 3) = 1

y + 3y + 1 = 1

4y + 1 = 1

4y + 1 + 6 = 1 + 6

4y = 0

Since Z_7 has no zero divisors, I get that y = 0.

z = 2(0) + 3 = 3

x = 2(0) = 0

So x = 0, y = 0, z = 3.

-Dan