Matrix determinants (modulo)

Apr 2008
748
159
When calculating the inverse of a normal matrix, for \(\displaystyle M = \begin{pmatrix}
x & y \\
z & v
\end{pmatrix}\) for example, you would simply use \(\displaystyle \frac{1}{xv-yz}\begin{pmatrix}
v & -y \\
-z & x
\end{pmatrix}\).

However, if your matrix was \(\displaystyle \begin{pmatrix}
\bar{1} & \bar{2} \\
\bar{2} & \bar{0}
\end{pmatrix}\) in \(\displaystyle \mathbb{Z}_3\), the inverse would be:

\(\displaystyle (-\bar{4})^{-1}\begin{pmatrix}
\bar{0} & \bar{-2} \\
\bar{-2} & \bar{1}
\end{pmatrix} = (\bar{2})^{-1}\begin{pmatrix}
\bar{0} & \bar{1} \\
\bar{1} & \bar{1}
\end{pmatrix}\).

Ok up to here, just forgotten how to calculate \(\displaystyle (\bar{2})^{-1}\)

Thanks in advance
 
Dec 2009
411
131
I suppose with \(\displaystyle \mathbb{Z}_3\) you mean the invertible elements of \(\displaystyle \mathbb{Z}/3\mathbb{Z}\)? (I've seen different notation for that).

Anyway, \(\displaystyle (\overline{2})^{-1}=2\) since \(\displaystyle 2\cdot 2 = 4 \equiv 1\) mod 3
 
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Apr 2008
748
159
Yeh that was what I meant, cheers.
 

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Mar 2010
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For finding an arbitrary modular inverse (if it exists), there is an algorithm using the extended Euclidean algorithm, described here. Also, if you choose the method involving direct exponentiation using Euler's theorem, there are algorithms for fast modular exponentiation, one of which is described in this thread, posts #5 and #7.