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Let $p$ be a prime number and let $\mathbb{Z}_p$ denote the field of integers modulo $p$. Find the number of $2 × 2$ invertible matrices with entries from this field.

an I get some help how to solve the problem. thanks.

Michael Albanese
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dutyr
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  • Hint: Search function. – Martin Brandenburg Dec 23 '13 at 10:03
  • Probably not worth saying at this late date, but: what did you try? – solitaireartist May 13 '22 at 14:46
  • Due to my lack of reputation I can't reply to: "Sorry for the very late question, but why does this only work for $p$ is prime?" from @user423841. I believe the answer to this question should be: If $p$ wasn't prime, then $\mathbb{Z_p}$ would not be a field and not every non-zero element would have a multiplicative inverse. Which means some of the entries in the first colomn may not have an inverse, so we cannot construct an invertible matrix with this entry, which makes the argument given by Michael Albanese break down. So the number of $2\times 2$ invertible matrices with entries from the fi – JackpotWizard 180 May 02 '23 at 13:41

2 Answers2

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Hints:

  • A square matrix is invertible if and only if its columns are linearly independent.
  • Two non-zero vectors are linearly independent if and only if one is not a scalar multiple of the other.

How many choices are there for the first column? How many choices are there for the second column?

Michael Albanese
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Remember that a matrix is invertible iff the columns are linearly independent. Then, how many choices for linearly independent columns you got?

Say for the first column, how many vectors are there in $\mathbb{Z}_{p}^2$ ? You have no retrictions there other than the vector being different than 0. So you got $p^2 -1$ for the first vector ( there are $p^2$ vectors and you get rid of the 0 vector ).

Now for your second column, once you fix the first one you only have to make sure it is not a multiple of the first one, how many of those are? Well, there are exactly p-1 linear combinations of the first vector because you have exactly p-1 scalars to pmultiply the first one.

So you got $(p^{2}-1)$ choices for the first one and $p^{2}-1-(p-1)=p^{2}-1-p+1=p^{2}-p$. So you have exactly $(p^{2}-1)(p^{2}-p)$ choices for linealy independent vectors. This is the number of invertible matrices in $\mathbb{Z}_{p}$

Angel
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  • Sorry for the very late question, but why does this only work for p is prime? –  Oct 31 '17 at 11:57