Cardinality of Set S with 2x2 Invertible Matrices from {0,1,2}?

  • Thread starter Thread starter Oster
  • Start date Start date
  • Tags Tags
    Cardinality Set
Oster
Messages
84
Reaction score
0
S is the set containing all 2x2 invertible matrices such that the entries come from the the set {0,1,2}. What is the cardinality(number of elements) of this set?

I got 50. Is this correct? What is the best way to go about solving this problem?
 
Physics news on Phys.org
One way to solve it might be to use the determinant ab-cd where a, b, c, and d, come from the set {0, 1, 2}. Figure out which allows for invertibility, and then how many ways you can get that value for the determinant.
 
Uh I got 42 this way. What's the answer?
 
What have you tried?

Show some of your work.

Explain some of your thinking.
 
OK! first i tried enumerating the non-degenrate matrices by counting the number of matrices which had one 0 then two 0s then one 1 and no 0s and then two 1s no 0s and last 3 1s and no 0s. I got 50.

Then I looked at all the values the determinant could take (-4 to 4?) and counted the number of matrices which gave the particular value of the determinant. I got 42.

THEN i wrote a program to count the number of degenerate matrices. I got 50. So 50?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top