Finding multiplicative inverse modulo n using matrix method

Question

According to this video (15:17 onwards), there is a "matrix method" to find the multiplicative inverse of $a$ mod $n$ by row reducing $$\begin{bmatrix} a & 1\\ n & 0 \end{bmatrix}$$

In the author's case, to find $97^{-1} mod\ n$, he did row operations: $$ \left[ \begin{array}{ccc} 97 & 1\\ 224 & 0 \end{array} \right] \sim \left[ \begin{array}{ccc} 97 & 1\\ 30 & -2 \end{array} \right] \sim \left[ \begin{array}{ccc} 7 & 7\\ 30 & -2 \end{array} \right] \sim \left[ \begin{array}{ccc} 7 & 7\\ 2 & -30 \end{array} \right] \sim \left[ \begin{array}{ccc} 1 & 97\\ 2 & -30 \end{array} \right] $$ and concluded that $97^{-1}=97$.

My question is how exactly does this method work and why didn't he row reduce traditionally into row echelon form?

This method really just solves the two simultaneous equations $97x = 1$ (true for some specific $x = 97^{-1}$) and $224x = 0$ (true for any $x$, so in particular for $x = 97^{-1}$) without using division. It ends up with $1x = 97$ (and $2x = -30$). — Billy, Jul 21 '13 at 05:26
It looks like the video does row reduction but stops when it gets 1 in the upper left entry --- no need to continue on to row echelon form, since you can read off the answer from what you get to. — Gerry Myerson, Jul 21 '13 at 06:45

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

0

In short, this is a compact form of the extended Euclidean algorithm (the typical way of finding modular inverses). I think the best way to see what is going on is to compare the matrix calculations to the calculations that you get coming out of the extended Euclidean algorithm.

If you get stuck, this answer by Marc van Leeuwen explains what's happening.

edited Apr 13 '17 at 12:21

Community

1

answered Jul 23 '13 at 20:21

davidlowryduda

91,687

Finding multiplicative inverse modulo n using matrix method

1 Answers1