# Homework Help: Matrix groups

1. Jul 31, 2010

### roam

1. The problem statement, all variables and given/known data

[PLAIN]http://img844.imageshack.us/img844/2975/31668142.gif [Broken]

3. The attempt at a solution

I know the inverse would be of the form:

$$M^{-1}_{a,p}=\frac{1}{1-(p-1)(10 \bmod\ p)} \begin{bmatrix} 1&-(a \bmod\ p)\\ -(p-1)& 1 \end{bmatrix}$$

Where $$1-(p-1)(10 \bmod\ p)$$ is the determinant. Whatever this equals to, I must find its inverse.

But how does one evaluate $$10 \bmod\ p$$ and $$(p-1)$$ when $$p \in \{ 3,7 \}$$? Do we take '10 mod 7' or '10 mod 3'? I appreciate if anyone could show me that.

Last edited by a moderator: May 4, 2017
2. Aug 1, 2010

### losiu99

I believe you are supposed to compute two inverses, for p = 3 and for p = 7 (with a = 10 in both cases).

3. Aug 2, 2010

### roam

Are you sure? Because the question apparently asks for one inverse...

But in the case we need 2 inverses

10 mod 3 = 1 & (p-1)=2
10 mod 7 = 3 & (p-1)=6

The inverses will be

$$M^{-1}_{10,3}=1^{-1} \begin{bmatrix} 1&-1\\ -2& 1 \end{bmatrix}$$

$$M^{-1}_{10,7}=3^{-1} \begin{bmatrix} 1&-3\\ -6& 1 \end{bmatrix}$$

Inverse of 1 is 1, and the inverse of 3 is 5 [since (3.5) mod 7 = 1]. When the arithmetic is done modulo 3 and modulo 7 respectively the inverse matrices are:

$$M^{-1}_{10,3}=\begin{bmatrix} 1&2\\ 1& 1 \end{bmatrix}$$

$$M^{-1}_{10,3}=\begin{bmatrix} 5.1&5.4\\ 5.1& 5.1 \end{bmatrix}$$$$=\begin{bmatrix} 5&6\\ 5& 5 \end{bmatrix}$$

Is this correct then?