Expert: Sherry Wallin - 5/2/2011

Question
QUESTION: the numbers of elements of S5(the symmetric group of 5 letters ) which are their own inverse equals
a. 10
b. 11
c. 25
d. 26
pleas tell me how to solve this and what is the logic to get the answer of this

ANSWER: Geeta~

How many 2-cycles are there in S5?

First think of the elements as being (1,2,3,4,5) for simplicity

Next pair them up, i.e., write out all the transpositions (2-cycles):

(1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5), (1,1)

Note 1:  that in this list I have omitted (2,1) because it is the same as (1,2), therefore if (a,b) is used I don't have (b,a) in the list

Note 2: I used (1,1) for an identity. I could just as easily used (2,2), (3,3),...(5,5) but this is the same as saying (a,a) where a belongs in (1,2,3,4,5)

Count the 'unique' transpositions:(1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5), (1,1) there are 11 of course.

Is this clear? Have I helped you find the answer and the logic behind it?


Math Prof

---------- FOLLOW-UP ----------

QUESTION: sir please help me.
i have one more question.
the number of 2*2 matrices over z3(the field with three elements) with determinant 1 is
a. 24
b. 60
c. 20
d. 30
please also tell the logic behind it.

ANSWER: Geeta~

First you need to know how many 2x2 matrices can be made up of the elements in Z3.
Z3 = {0,1,2} and a 2x2 matrix has 4 elements so for position i,j = 1,1 there are 3C1 = 3 possibilities, likewise for positions i,j = 1,2; 2,1; 2,2. This means there are 3^4 = 81 possible 2x2 matrices that can be made with the elements from Z3.

That should feel pretty good since 81 is larger than any of the numbers you have to choose from :)
What does it take to get a determinant of 1? Well, it takes the main diagonal to be 1 and the 'other' diagonal to be 0:

[ a   b ]   you want ad-bc = 1 thus ad has to equal 1 and bc must be zero.
[ c   d ]

Now you might ask yourself how many of the matrices have that characteristic (above) where ad = 1 and bc = 0. In Z3 1*1 = 1 and 2*2 = 1 and that is it, all the rest will give us either a product of 0 or 2. However you need to realize that if ad-bc = -2 this is the same as 1,  as well as ad-bc = 4 is 1 also.

I'm sure there is a quicker, dirtier method than the grunt them out method but you will want to know when ad = bc+1

Here are the matrices I found and I don't think there are any others:

[0 1]  [0 1]  [0 1]
[2 0]  [2 1]  [2 2]  all these determinants are 0-2 = -2 = 1

[0 2]  [0 2]  [0 2]
[1 0]  [1 1]  [1 2]  all these determinants are 0 - 2 = -2 = 1

[1 0]  [1 0]  [1 0]
[0 1]  [1 1]  [2 1]  all these determinants are 1 - 0 = 1

[1 1]  [1 1]  [1 1]
[0 1]  [1 2]  [2 0]  all these determinants are  equal to 1 and they are respectively ad-bc = 1, 1, -2 = 1

[1 2]  [1 2]  [1 2]
[0 1]  [1 0]  {2 2]     all these determinants are  equal to 1 and they are respectively ad-bc = 1, -2 = 1, 2-4 = -2 =1

[2 0]  [2 0]  [2 0]
[0 2]  [1 2]  [2 2]  all these determinants are  equal to 1 and they are respectively ad-bc = 4-0 = 1, 4-0 = 1, 4-0 = 1

[2 1]  [2 1]  [2 1]
[0 2]  [1 1]  [2 0]   all these determinants are  equal to 1 and they are respectively ad-bc = 4 = 1, 1, 0-2 = 1

[2 2] [2 2]  [2 2]
[0 2] [1 0]  [2 1]   all these determinants are  equal to 1 and they are respectively ad-bc = 4-0 = 1, 0-2 = 1, 0

Count them, there are 24!

Math Prof

---------- FOLLOW-UP ----------

QUESTION: let T:R^3->R^3 be a linear transformation such that T(1,2,3)=(1,2,3)
T(1,5,0)=(2,10,0)
T(-1,2,-1)=(-3,6,-3)
the dimension of the vector space spanned by all the eigenvalues of T is
a. 0
b. 1
c. 2
d. 3

actually i solved this question please tell me that its answer is 3??

just one more little question is:
if A and B are matrices then is following correct??
det(A.B.inverse of A)=detA.detB.det of A inverse.
and if  det B=1 then this implies B is invertible implies det of B inverse=1??

Answer
Geeta~

We can make the communication of this problem simpler by just defining A inverse as A^-1. Also, are you using '.' to mean multiplication? If so * works well universally. If you are using '.' as multiplication then are you saying det(A*B*A^-1)?

Next because you haven't said so, are A and B nxn matrices? det(A*B) = det(A) * det(B) only if they are each square matrices. I will assume these matrices that you are asking about are square and answer your questions from that point of view.

If matrix A is square then A's inverse will also be square (and of the same size) and since we assumed B is also square then it is true that
det(B*A^-1) = det A * det B * det A^-1
Now if det B = 1 then det A* det B * det A^-1 = det A * det A^-1 = det(A * A^-1) = det (e) where e is the identity matrix, ie.,  the same size square matrix with ones down the diagonal and 0's everywhere else and hence it's determinant is 1.

If the det B = 1 then certainly B is invertible. The only thing that stops a square matrix from being invertible is if the det is 0, in other words the det does NOT have to be 1. Can you find a square matrix that has a determinant of 1 and it's inverse does not?

If you don't know a theorem and you have to prove what you want to show then do this:

Let A =  [a  b]  and find it's inverse: [a  b | 1  0]
         [c  d]          [c  d | 0  1]

You will find that if the det A is ad-bc not equal to 0,  then the det A^-1 is (ad-bc)/(ad-bc) = 1
In other words, it does not matter what the det A is as long as it is non zero but it's inverse will always be 1

Math Prof