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$GL(2,\mathbb{Z}_{11})$ is a general linear group whose elements are $2\times 2$ matrices where every element of matrices belongs to the set $\mathbb{Z}_{11}$ (so modulo 11 arithmetic would take place) and determinant should non-zero for matrices.

So I have been given a matrix $$ A = \begin{pmatrix} 2 & 6 \\ 3 & 5 \end{pmatrix}, $$

where A belongs to $GL(2,\mathbb{Z}_{11})$ and I have to find it's inverse. So what confused me is the way the inverse would be solved ? What is process to find it's inverse ?

Stefan4024
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Aashish Loknath Panigrahi
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    You pretty much find the inverse exactly as if this was done with real numbers. The only difference is you do everything modulo 11. – Hamed Apr 06 '18 at 22:57

1 Answers1

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Especially for a $2\times2$ matrix, there is an easy formula for the inverse: $$ \left( \matrix{a & b \\ c& d} \right )^{-1} = \frac{1}{ad-bc}\left( \matrix{d & -b \\ -c& a} \right ) $$ In your case, that gives us $$ \left( \matrix{2 & 6 \\ 3& 5} \right )^{-1} = \frac{1}{2\cdot5-3\cdot6}\left( \matrix{5 & 5 \\ 8 & 2} \right ) = \frac{1}{3}\left( \matrix{5 & 5 \\ 8 & 2} \right ) = 4\cdot\left( \matrix{5 & 5 \\ 8 & 2} \right ) = \left( \matrix{9 & 9 \\ 10 & 8} \right ) $$ You can check it by just multiplying them.

To address a very valid point raised in the comments, let me explain why I write $\tfrac13=4$. It is important to note that by "$3$" and "$4$" I do not mean the natural numbers, but their residue class in $\Bbb Z_{11}$. And since $3\cdot 4 = 12=1 + 11$, we can see that $3\cdot4=1$ holds in $\Bbb Z_{11}$. Since $\Bbb Z_{11}$ is a field, we may divide by any nonzero element and the formula $\tfrac13=4$ is justified. Now, how to figure out that $\tfrac13=4$ holds in $\Bbb Z_{11}$? Since $11$ is rather small, in this case I did it by trial and error; just multiply $3$ with a couple of numbers and see if the residue modulo $11$ is $1$. When the prime number in question is much larger, the extended Euclidean algorithm is your best way to find the inverse of an element, however I feel that this may be a topic for another question.

Jesko Hüttenhain
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    That $\frac 13 = 4$ in $\Bbb Z_{11}$ is a tricky point for many starting out with finite fields, so it may bear further clarification. – Ben Grossmann Apr 07 '18 at 01:59
  • Can you please explain me how can it find y when I have been given 1/x where x is in integer . @Omnomnomnom – Aashish Loknath Panigrahi Apr 07 '18 at 20:21
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    @AashishLoknathPanigrahi for instance: to find that $\frac 13 = 4$, we'd like to solve $$ 3x \equiv 1 \pmod{11} $$ which has solution $x \equiv 4 \pmod{11}$. For low primes like $11$ trial and error is a straightforward approach, but these inverses can be found systematically with the so-called "reverse Euclidean algorithm" – Ben Grossmann Apr 07 '18 at 22:03