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Thread: Solving a system of Z_2

  1. #1
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    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, $\displaystyle 23(\bmod 12)=11$ means that $\displaystyle 23=11$ in $\displaystyle Z_{12}$ right?

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

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

    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 $\displaystyle R_1\rightarrow R_1+R_2$

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

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

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

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

    Is my reasoning correct so far?

    Thanks
    Last edited by adkinsjr; Mar 18th 2010 at 11:00 PM.
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  2. #2
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    Actually, I think I was wrong about that last bit. Since $\displaystyle Z_2$ 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:

    $\displaystyle -y=-1$

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