# Math Help - Multiplicative elements in ring M<sub>2</sub>(Z<sub>3</sub>) ...

1. ## Multiplicative elements in ring M2(Z3) ...

Hello! I have just one question!

We have the ring $K = M_{2}(\mathbb{Z}_{3})$, which are all 2x2 matrices over the field $\mathbb{Z}_{3}$. We need to find all the elements of ring $K*$ (the set of multiplicative inverses).

I know that A € K* if the det(A) != 0. But here in Z3 we just have (det(A) == 1) || (det(A) == 2). Just want to know why det(A) == 2 is not the right solution?

2. ## Re: Multiplicative elements in ring M<sub>2</sub>(Z<sub>3</sub>) ...

Not sure what you are asking, here. The matrix:

$\begin{bmatrix}2&0\\0&1 \end{bmatrix}$ is invertible, as is the 2x2 identity matrix (the first has det 2, the second det 1).

3. ## Re: Multiplicative elements in ring M<sub>2</sub>(Z<sub>3</sub>) ...

Ok, thanks ... so in ring $M_{2}(\mathbb{Z}_{3})$ are invertible elements the matrixes with det(A) = 1 and det(A) = 2.
How can I get the $|{M_{2}(\mathbb{Z}_{3})}|$?

4. ## Re: Multiplicative elements in ring M<sub>2</sub>(Z<sub>3</sub>) ...

If you want the number of 2x2 matrices, that is fairly easy: you have 3 choices for every entry, and 4 possible entries, for a total of $3^4 = 81$ matrices.

If you want the number of invertible matrices, see my answer in another thread. And do try not to double-post in the future.

5. ## Re: Multiplicative elements in ring M<sub>2</sub>(Z<sub>3</sub>) ...

Actually, my niece is a student and has the exam of Algebra 2 on Friday, and he sent me some exercises for the exam. I had learned Algebra 2 4 years ago, so I forgot some things, so I asked for the solution ...

6. ## Re: Multiplicative elements in ring M<sub>2</sub>(Z<sub>3</sub>) ...

@Deveno,

actually: matrix $\begin{bmatrix}2&0\\0&1\end{bmatrix}$ * $\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}2a&2b\\c&d\end{bmatrix}= \begin{bmatrix}1&0\\0&1\end{bmatrix}=I$, but this linear system doesn't give you all coefficients in $\mathbb{Z}$so matrices with $det(A)=2$ aren't invertible. So, all invertible matrices are those, which have the det(A) = 1.

7. ## Re: Multiplicative elements in ring M<sub>2</sub>(Z<sub>3</sub>) ...

You need to practice with Z3 more! 2(2)= 4= 3+1= 1 (mod3) so that the multiplicative inverse 2 in Z3 is 2 itself. (We could say that 1/2= 2 (mod 3).)

$\begin{bmatrix}2 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}$ (mod 3)

That matrix is invertible. More generally, in Zp, with p any prime number, every number has a multiplicative inverse.

8. ## Re: Multiplicative elements in ring M<sub>2</sub>(Z<sub>3</sub>) ...

Ok .... Thanks! hehe, I said that this isn't for me, it's for my niece who has the exam Algebra 2 on Friday. Just I forgot these things in Algebra ... I learned this 4 years ago.

9. ## Re: Multiplicative elements in ring M<sub>2</sub>(Z<sub>3</sub>) ...

It is not hard to show that:

$\begin{bmatrix}a&b\\c&d \end{bmatrix}^{-1} = \begin{bmatrix}(ad - bc)^{-1}d&(ad - bc)^{-1}(-b)\\(ad - bc)^{-1}(-c)&(ad - bc)^{-1}a \end{bmatrix}$

Even when we are in the field $\Bbb Z_3$. Of course, the expression $(ad - bc)^{-1}$ is only defined when $ad - bc \neq 0$.

As I remarked elsewhere, since -1 = 2 (mod 3), it is more fruitful in this situation to think of $\Bbb Z_3$ as consisting of the elements $\{-1,0,1\}$, which makes multiplying a bit easier.