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).
Hello! I have just one question!
We have the ring , which are all 2x2 matrices over the field . We need to find all the elements of ring (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?
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.
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 ...
@Deveno,
actually: matrix * , but this linear system doesn't give you all coefficients in so matrices with aren't invertible. So, all invertible matrices are those, which have the det(A) = 1.
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).)
(mod 3)
That matrix is invertible. More generally, in Zp, with p any prime number, every number has a multiplicative inverse.
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.