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The number of 2x2 matrices over Z3 (the field with three elements) with determinant 1 is
- a)24
- b)60
- c)(a) 24 (b) 60
- d)(a) 24 (b) 60 (c) 20 (<7)30
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The number of 2x2 matrices over Z3 (the field with three elements) wit...
Consider the binary operation on X, a∗b=a+b+4, for a,b∈X. We are asked to determine the properties satisfied by this operation.
This operation satisfies the properties of an abelian group since it is closed under addition, associative, has an identity element (0), and every element has an inverse.
This operation satisfies the properties of an abelian group since it is closed under addition, associative, has an identity element (0), and every element has an inverse.
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The number of 2x2 matrices over Z3 (the field with three elements) wit...
Number of n x n matrices over Zp
So number of 2 x 2 matrices over
So number of 2 x 2 matrices over
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The number of 2x2 matrices over Z3 (the field with three elements) wit...
The answer is (b) 60.
To find the number of 2x2 matrices over Z3 with determinant 1, we can consider the possible values for each entry in the matrix.
Let's denote the matrix as:
| a b |
| c d |
Since we want the determinant to be 1, we have the equation ad - bc = 1.
Now let's consider the possible values for each entry in the matrix:
For a, b, c, and d, each entry can take on one of the three elements in Z3 (0, 1, or 2).
We can start by fixing the value of a. Since a can take on 3 possible values, we have 3 choices for a.
Now let's consider the possible values for b. Since the determinant equation is ad - bc = 1, we can solve for b:
b = (ad - 1)/c
Since c can take on 3 possible values, and a and d are fixed, there are 3 possible values for b.
Similarly, we can solve for c:
c = (ad - 1)/b
Again, since b can take on 3 possible values, there are 3 possible values for c.
Finally, we can solve for d:
d = (bc + 1)/a
Again, since a can take on 3 possible values, there are 3 possible values for d.
Therefore, there are 3 choices for a, 3 choices for b, 3 choices for c, and 3 choices for d, giving us a total of 3^4 = 81 possible matrices.
However, we need to exclude the cases where a, b, c, and d are all 0, since in that case the determinant would be 0.
So the total number of 2x2 matrices over Z3 with determinant 1 is 81 - 1 = 80.
Therefore, the correct answer is (b) 60.
To find the number of 2x2 matrices over Z3 with determinant 1, we can consider the possible values for each entry in the matrix.
Let's denote the matrix as:
| a b |
| c d |
Since we want the determinant to be 1, we have the equation ad - bc = 1.
Now let's consider the possible values for each entry in the matrix:
For a, b, c, and d, each entry can take on one of the three elements in Z3 (0, 1, or 2).
We can start by fixing the value of a. Since a can take on 3 possible values, we have 3 choices for a.
Now let's consider the possible values for b. Since the determinant equation is ad - bc = 1, we can solve for b:
b = (ad - 1)/c
Since c can take on 3 possible values, and a and d are fixed, there are 3 possible values for b.
Similarly, we can solve for c:
c = (ad - 1)/b
Again, since b can take on 3 possible values, there are 3 possible values for c.
Finally, we can solve for d:
d = (bc + 1)/a
Again, since a can take on 3 possible values, there are 3 possible values for d.
Therefore, there are 3 choices for a, 3 choices for b, 3 choices for c, and 3 choices for d, giving us a total of 3^4 = 81 possible matrices.
However, we need to exclude the cases where a, b, c, and d are all 0, since in that case the determinant would be 0.
So the total number of 2x2 matrices over Z3 with determinant 1 is 81 - 1 = 80.
Therefore, the correct answer is (b) 60.
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