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Abstract Algebra: M(R) 2x2, units, where does the determinant come from?

  1. Mar 17, 2007 #1
    1. The problem statement, all variables and given/known data

    Prove that
    (a b
    c d)

    is a unit in the ring M(R) if and only if ad-bc !=0. In this case, verify that its inverse is
    (d/t -b/t
    -c/t a/t)

    where t= ad-bc.

    2. Relevant equations

    An element a in a ring R with identity is called a unit if there exists u in R
    s.t. au = 1 = ua (where 1 is the identity of the ring). In thise case
    the element u is called the multiplicative inverse of a and is denoted a^-1.

    3. The attempt at a solution

    I took an arbitrary element from M(R) 2x2, say

    (e f
    g h)

    and I know that if I multiply my unit times that, then I should get
    the identity matrix.

    (a b x (e f ) = (1 0
    c d) . g h ) . 0 1)

    which gives me four equations

    ae + bg = 1 ; ce + dg = 0
    af + bh = 0 ; cf + dh = 1

    This is where I became lost. I started substituting things, and it became
    a mess. I guess this question is really about showing where the
    determinant comes from. Please help, thank you.
    Last edited: Mar 17, 2007
  2. jcsd
  3. Mar 18, 2007 #2
    My attempted solution so far:

    Prove that (a b | c d) is a unit in the ring M(R) iff ad-bc != 0.


    A = (a b | c d) is a unit.

    hence there is an element U in M(R) s.t. AU = 1 = UA where 1 = (1 0 | 0 1).

    This element U can be found through Gaussian Elimination.

    We start with the augmented matrix:

    (a b | c d ) | (1 0 | 0 1)

    and use elementary row operations ( alot of paper) to get

    (1 0 | 0 1) | (d/t -b/t | -c/t a/t) where t = ad - bc

    Hence U = 1/t(d -b |-c a)

    and thus there is a requirement that ad-bc != 0 so that we dont divide
    by zero.

    Hence ad-bc != 0.

    That is my attempt in the -> direction.
    My question is, I feel VERY uncomfortable instantiating gaussian elimination
    in the middle of my abstract algebra book on ring theory. Im not very sure why the augmented [A|I] -> [I|A^{-1}] trick works, and thus im not sure if its "legal" for me to present, especially since I dont know why it works, it just does. Please help
  4. Mar 18, 2007 #3

    matt grime

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    What is R? You use it in two different ways. Is is supposed to mean the Reals or an arbitrary ring? I presume it is the reals, since it is certainly not true that A is invertible in M_2(R) for an arbitrary ring R.

    A is a unit if and only if there is a B such that AB=I. Now take dets.

    You are at no point asked to find the inverse - just show that one exists if and only if det(A)=/=0, and to verify that it is the one given to you.
    Last edited: Mar 18, 2007
  5. Mar 18, 2007 #4

    matt grime

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    It's not a trick. Multiplying matrices is just doing row/column operations. You're just recording the row/column operations that convert A to the identity.
  6. Mar 18, 2007 #5
    First, thank you for your reply. I appreciate it because
    you brought up a point that I had tried to understand but
    wasn't totally conscious of, but knew there was something bad
    lurking in the background.

    That is, if A = (a b | c d) is a unit, there is SOME element B
    s.t. AB = I = BA. I went to #math on irc EFNET and someone said
    to find A^{-1} via Gaussian Elimination. this resulted in me deriving
    where the determinant for a 2x2 comes from. BUt that it is a determinant
    or labled determinant is incidental to the question, as that the question
    or section or book that it is in hasn't up to this point care about determinants or its properties or matrix math like G.E. Which is why I have suspicion that I didn't need to go through G.E.

    What I dont understand is what you mean by "Now take determinants"
    since firstly I dont think im allowed to bring in out-of-book knowledge ..
    and secondly I dont know determinant theory enough to justify my usage of it, or to properly wield determinants, especially in an abstract algebra setting, not a linear algebra setting.

    R in M(R) is the matrix on the field of reals (sorry).

    So, I am asking for your pointers on how to proceed without having to find the inverse. It is presumed to exist in the -> direction, now i need to show that det(a) != 0. But how to i start talking about det(A) (= ad-bc)
    without deriving it naturally from Gaussian elimination? I mean, I can't pull ad-bc
    out of a magic hat.

    ahh, that is a great explanation. thats straight talk. my lin alg book was showing a sequence of compunded elementary row operation matrices and it got alittle confusing. okay i see now what the augmented matrices are about.
    Last edited: Mar 18, 2007
  7. Mar 18, 2007 #6

    matt grime

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    If you're not going to use dets, then just do it via gaussian elemination techniques since that is what you're allowed to use. At some point you will have to divide by ad-bc, and that is valid if and only if ad-bc=/=0
  8. Mar 18, 2007 #7


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    First off, you just verify A=(a b|c d) times the given matrix is the unit matrix and vice versa, you did that right? Now you want to show if ad-bc=0 then there is no inverse. To do this you could use gaussian elimination but a cheaper way is just to find a nonzero vector x that satisfies Ax=0. Can you find one? That would mean A has no inverse.
  9. Mar 19, 2007 #8
    Earlier, in reply #2 when I posted my attempted solution, isn't the process
    of finding [tex]A^{-1}[/tex] through the gaussian elimination going to net me both directions. I.e. once I find that [tex]A^{-1} = \left(\begin{array}{cc}d/(ad-bc)&b/(ad-bc)\\-c/(ad-bc)&a/(ad-bc)\end{array}\right)[/tex], that nets me that if ad-bc!=0 then [tex]A[/tex] is a unit?

    I also did find a nonzero vector x that satisfies [tex]Ax=0[/tex].
    that is [tex]A=\left(\begin{array}{cc}b+d/a+c&b+d/a+c\\-1&-1\end{array}\right)[/tex].

    ALthough it is a computationally cheaper way to show that A has no inverse, I end up having to still go the -> direction, but the Gauss. Elim. in the -> already gets me the <- direction?
    Since I can say, if [tex]ad-bc \neq 0[/tex] then since we are given [tex]A=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)[/tex], and we know that through G.E. that [tex]A^{-1} = \left(\begin{array}{cc}d/(ad-bc)&b/(ad-bc)\\-c/(ad-bc)&a/(ad-bc)\end{array}\right)[/tex], since [tex]ad-bc \neq 0[/tex] then we know [tex]A^{-1}[/tex] is valid since we can divide by [tex]ad-bc[/tex].
  10. Mar 19, 2007 #9


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    I'm a little confused by your 'vector'. I was thinking of x=(d,-c) solving Ax=0. If both c and d happen to be zero there's another obvious choice. But yes, if you've already done gaussian elimination you should be able to argue that that gives you both ways.
  11. Mar 19, 2007 #10


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    Are you aware that det(AB)=det(A)det(B)? It's probably in your textbook in the section you're working in. I thought Ring Theory came after doing Linear Algebra. :confused:

    If not, then go with the other method matt_grime pointed out.
  12. Mar 19, 2007 #11
    Oh haha.. My mistake.. Im just not thinking, x is a 2x1 column vector.

    So you did [tex]\left(\begin{array}{cc}a&b\\c&d\end{array}\right)
    \left(\begin{array}{c}e\\f\end{array}\right) = \left(\begin{array}{c}0\\0\end{array}\right) [/tex]

    and then solve for e and f, and by inspection that [tex]e=d[/tex]and [tex]f = -c[/tex]?
    Last edited: Mar 19, 2007
  13. Mar 19, 2007 #12


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    No. I just did:
    \left(\begin{array}{c}d\\-c\end{array}\right) = \left(\begin{array}{c}0\\0\end{array}\right) [/tex]

    if ad-bc=0.
  14. Mar 19, 2007 #13


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    In other words, yes, I just did what you said. Sorry, didn't read the last line correctly.
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