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The problem at hand is specifically to find the inverse of a matrix using Gaussian elimination modulo 29. I am familiar with the process of regular Gaussian elimination and modular arithmetic but not the combination of the two. The matrix you are given is:

$$A=\begin{pmatrix} 12 & 3 & 23 \\ 28 & 1 & 2 \\ 11 & 0 & 7 \end{pmatrix}$$

I first start out the way I would when doing a normal Gaussian elimination to find the inverse by writing out the following matrix:

$$A=\begin{pmatrix} 12 & 3 & 23 & 1 & 0 & 0 \\ 28 & 1 & 2 & 0 & 1 & 0\\ 11 & 0 & 7 & 0 & 0 & 1 \end{pmatrix}$$

From here, I first tried solving it how I would normally and then taking the elements modulo 29 at the end but unfortunately you end up with fractions at the end. After that I've tried working with only integers and tried to reduce numbers using modulo 29 as I go but haven't found success and I'm not sure if that is how you're supposed to do it. My question more generally would be how you go about Gaussian elimination modulo p for any prime number. Thanks in advance!

Frozty
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  • See here and here for modular fraction arithmetic. Using that classical Gaussian elimination works $!\bmod p.,$ Indeed, it works over any field (here $\Bbb Z_p).\ \ $ – Bill Dubuque Mar 05 '22 at 08:25

1 Answers1

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You do it exactly the same way except that all operations are done modulo $p$ ($p=29$ here).

The only difficult one is division. You make use of the fact that, for every integer $n$ that is not a multiple of $p$ there is an integer $m$ such that $mn \equiv 1 \bmod p$ so that $1/n \equiv m \bmod p$.

Then $a/n\equiv am \bmod p$.

marty cohen
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    Also, you should be careful not to divide by a multiple of $29$. This may sometimes require more swapping rows than you'd usually do. – Misha Lavrov Mar 05 '22 at 00:37
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    Perhaps it should be remarked that given $n$ to find $m$ so that $mn$ such that $mn-1$ is divisible by $p$, one needs to use the Extended GCD algorithm. – Kapil Mar 05 '22 at 03:05