# Thread: The ring M2(Z3) - how to find the multiplicative inverses?

1. ## The ring M2(Z3) - how to find the multiplicative inverses?

Hello! I have just one question!

We have the ring $M_{2}(\mathbb{Z}_{3})$, 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 $A \in K^{*}$ if the $det(A) \neq 0$. But here in $\mathbb{Z}_{3}$ we just have $(det(A) = 1)$ or $(det(A) = 2)$.

2. ## Re: The ring M2(Z3) - how to find the multiplicative inverses?

Here is one way you might proceed: we know that a matrix $A$ will be invertible if and only if its columns form a basis for $\Bbb Z_3 \times \Bbb Z_3$.

So that means that we can choose any non-zero vector $(a,b)$ for the first column (here, it might be more profitable to view $\Bbb Z_3$ as the congruence classes of $-1,0,1$, instead of $0,1,2$).

That means we have 8 choices for our first vector.

Now the second vector will form a linearly dependent set with the first if (and only if):

a) The second vector is 0
b) the second vector is the same vector as the first
c) the second vector is the negative of the first

So for each of the possible 8 vectors we chose, we have 6 possible choices for the 2nd vector, giving 48 in all.

You should, knowing this, be able to explicitly write down all 48 matrices. Why must exactly half of these have determinant 1?

3. ## Re: The ring M2(Z3) - how to find the multiplicative inverses?

Hey, maybe was a little bit misunderstanding... hehe, I don't need to write down ALL the matrices. Just find in implicit way.

If we have this K = $M_{2}(\mathbb{Z}_{3})$. Matrix A is invertible if $det(A) \neq 0$. So matrix A is invertible if det(A) = 1. I'm asking: is matrix A invertible even if det(A) = 2?

So, is the solution: $K^{*} = \left\{A \in K | where (det(A) = 1 || det(A) = 2)\}\right}$? Is det(A) = 2 a solution?

4. ## Re: The ring M2(Z3) - how to find the multiplicative inverses?

Yes any matrix with a non-zero determinant is invertible.