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What's wrong with my algebra here? (dealing with determinants)

  1. Oct 19, 2009 #1
    Let A, B and C be 3x3 invertible matrices where det(A)=−4 ,det(B)=−2 and det(C) is some non-zero scalar. Find:

    det[−2(A^2)^T x C^2 x B^−1 (C^−1)^2]

    So (A^2)^T is just A^2 since the transpose's det is the same.
    (C^-1)^2 = C^-2
    C^-2 * C^2 = 1 (so just cancelled it out)
    Inverse of B = 1/-2 = -1/2
    So we have det[-2(A^2) x B^-1

    I then sub in -2(-4^2) x -1/2 = -2(16)/-2

    Which evaluates to 16, however this answer is wrong :(

    Don't know what I did wrong. Can anyone spot a mistake?

    Thanks :D
     
  2. jcsd
  3. Oct 19, 2009 #2

    Fredrik

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    I didn't read the whole thing, but you appear to have used det(XY)=det X det Y incorrectly (or rather not at all), in at least one spot.
     
  4. Oct 19, 2009 #3

    Office_Shredder

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    Your work is very sloppy, because you freely ignore determinant notation while canceling matrices and saying things like AT=A (things which you can't say about matrices, but you can say about their determinants)

    But for all that, as far as I can tell it looks like you did the right thing up until here:
    So we have det[-2(A^2) x B^-1]

    You essentially re-write this as -2det(A2B-1) which isn't true
     
  5. Oct 19, 2009 #4
    Ok, I'll keep trying. Thanks for the help so far :D
     
  6. Oct 19, 2009 #5
    I still can't figure this out. Is this because of the order I'm doing the math? Is my bedmas wrong? >.> What's wrong with my last steps?

    Sorry for the trouble and thanks for the help :D
     
  7. Oct 19, 2009 #6
    Is this what you're trying to find? det[−2(A²)TC²B-1C-2]

    Perhaps writing it out more organized and neat will make it easier to get help.
    Use facts about determinants for your problem:

    det(AB)=det(A)det(B)
    det(A)=det(AT)
    det(A-1)=1/det(A)
    det(cA)=cnA, where n is the size of the matrix

    It looks like you've used these in your solution but it's hard to tell since it's so sloppy and poorly written.

    You've reduced it to det[−2(A²)B-1], so apply the first and last properties
     
  8. Oct 20, 2009 #7

    Fredrik

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    I was actually wrong about the mistake I thought you made. (It's possible you made that mistake somewhere else, but you didn't make it where I thought you did). I read your post too quickly and I think the sloppy notation contributed to my misunderstanding. Still, you can't go wrong if you just apply the rules that VeeEight wrote down for you.
     
  9. Oct 20, 2009 #8
    Oh man thanks a lot, that makes sense now. I forgot to apply the det(cA) = c^n.A rule, since our prof only vaguely mentioned it in one example.

    Anyways, for future reference, is there an easy way to do the actual ascii math symbols that you did?

    i.e. det[−2(A²)TC²B-1C-2]

    That looks like it'd help a lot when I'm describing my problems, so it's not just messy looking. And yeah, sorry for my sloppy notation, I'll try and do better next time. Appreciate all the help :D
     
  10. Oct 20, 2009 #9

    Office_Shredder

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    You can write A2 as A[*sup]2[*/sup] without the *'s. A better choice is to learn Latex so you can write it as [itex]A^2[/itex] (this is superior when you have long expression to write, but using sup tags - and sub tags for subscripts, is a good start)
     
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