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Subgroup help !

  1. Oct 9, 2006 #1
    Subgroup help plz! Urgent!

    Can anyone help me with the following question? Many thanks! Again I'm working from a book so I'm must trying to apply theory to exercises -__-


    How do I show that the identity lies in this? And also show that we have closure? Do I just choose any values of a, b and c and multiply two matrices? :confused:
  2. jcsd
  3. Oct 9, 2006 #2

    What conditions do you need on a, b, and c in order to make the identity? Are they within the constraints of this set? I think so. Just choose those values of a,b,c.

    In order to show inverses exist for every element of the set, you know something regarding determinants that guarantees inverses.

    Choose two exeplary elements from the matrix, ie a matrix with values a,b,c and another with d,e,f. Multiply them, and see if you get another matrix of the same form.
  4. Oct 9, 2006 #3
    Okay, sorry I can't put this in math format but hopefully you understand what I mean :rolleyes:

    In order for it to fulfill identity, I've put in ac not to be equal to 0 if a and c are not equal to 0..is that right?

    And if say i take the given matrix in the question and another matrix which has elements d, e and f, lie in K, then when multiplied you get the products ad, ae+bf, cf in the new matrix? And ad, ae+bf and cf are not equal to 0 hence we have closure? :confused:
  5. Oct 9, 2006 #4

    matt grime

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    What is the identity matrix? Does it satisfy the rule that tells you which matrices belong to that set?
  6. Oct 9, 2006 #5
    Right I can't type the identity matrix up in its correct format but its [ 1, 0, (first row) 0,1 (second row) ]. So it does satisfy the rule? :confused:

    Is my evaluation on closure incorrect?
  7. Oct 9, 2006 #6

    matt grime

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    Why are you asking a question about the identity. What are the two rules it, the identity, must satisfy to be in the set. Are they both satisfied? Yes.

    You composition one is not correct. You say that since ad, cf and ae+bf are not zero that it is in the group. What the upper right hand entry is is immaterial. All that matters is that the lower left is zero, and the the determinant (which is ac, for the matrix in post 1) is not zero.
  8. Oct 9, 2006 #7


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    The identity is
    [tex]\left(\begin{array}{cc}1 && 0\\0 && 1\end{array}\right)[/tex]
    Your general matrix is
    [tex]\left(\begin{array}{cc}a && b\\0 && c\end{array}\right)[/tex]
    with a and c not 0. What should a, b, c be to give the identity matrix?
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