# Thread: Solving a system of Z_2

1. ## Solving a system of Z_2

I'm having difficulty applying modular arithmetic in solving basic systems over finite fields. If I think of it in terms of time, the arithmetic makes a little more sense. For example, $23(\bmod 12)=11$ means that $23=11$ in $Z_{12}$ right?

Suppose I want to solve the system in $Z_2={0,1}$ that has the augmented matrix:

$\left(\begin{array}{cccc}1&1&1&0\\1&0&1&1\end{arra y}\right)$

So whenever I perform a row-op, I have to apply the definitions of modular addition and multiplication on each element. So I could right the matrix obtained by the row-op $R_1\rightarrow R_1+R_2$

$\left(\begin{array}{cccc}1\oplus 1 &1\oplus 0 &1\oplus 1&0\oplus 1\\1&0&1&1\end{array}\right)$

$=\left(\begin{array}{cccc}2(\bmod 2) &1(\bmod 2) & 2(\bmod 2)&1(\bmod 2)\\1&0&1&1\end{array}\right)$

$=\left(\begin{array}{cccc}0&-1&0&-1\\1&0&1&1\end{array}\right)$

This is where I get confused because $-1$ isn't an element of $Z_2$. So do I just use the fact that $1=-1(\bmod 2)$, which means that $-1=1$ in $Z_2$? This would give $y=1$.

Is my reasoning correct so far?

Thanks

2. Actually, I think I was wrong about that last bit. Since $Z_2$ is a field, then every element must have an additive inverse, by the field axioms.

I'm new to this kind of math.

So I guess I could write:

$-y=-1$

Then I would just multiply both sides by 1 to obtain y=1, since 1=-1 in this field.