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Math Help - proof involving determinants

  1. #1
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    proof involving determinants

    Show that if A is an n x n matrix of integers, then A^-1 is a matrix of integers if and only if det(A) = +/- 1. This is actually part b of a problem. Part a) asks to prove that (x_i)det(A) = det(A_ib) where A_ib is the matrix formed by replacing the ith column of A by the column vector b. x_i is a scalar and is the component of the vector x. So since that was the previous part i tried using that for part b). So if det(A) = +/- 1, then x_i = +/- det(A_ib). but i also know that x = (A^-1)b. so the ith entry of x is also the ith row of A^-1 which i denote as a_i dot product with the vector b. So i have +/- det(A_ib) = a_i * b. i don't know what to do from here. b may or may not be a vector of integers so the det(A_ib) may or may not be an integer. and even if it were i don't know how that would show that A^-1 must be a matrix of integers. am i going about this problem in the correct way? please help.
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  2. #2
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    Quote Originally Posted by oblixps View Post
    Show that if A is an n x n matrix of integers, then A^-1 is a matrix of integers if and only if det(A) = +/- 1. This is actually part b of a problem. Part a) asks to prove that (x_i)det(A) = det(A_ib) where A_ib is the matrix formed by replacing the ith column of A by the column vector b.


    This is known as Cramer's Rule



    x_i is a scalar and is the component of the vector x. So since that was the previous part i tried using that for part b). So if det(A) = +/- 1, then x_i = +/- det(A_ib). but i also know that x = (A^-1)b. so the ith entry of x is also the ith row of A^-1 which i denote as a_i dot product with the vector b. So i have +/- det(A_ib) = a_i * b. i don't know what to do from here. b may or may not be a vector of integers so the det(A_ib) may or may not be an integer. and even if it were i don't know how that would show that A^-1 must be a matrix of integers. am i going about this problem in the correct way? please help.

    You should know by now, besides Cramer's Rule, also the basic formula Adj(A)\cdot A=|A|\cdot I , where Adj(A)= the

    classical adjoint of the matrix A (whose entries are \pm the minors of A ), |A|= the determinant of A,

    and I= the unit matrix.

    From this it follows that if A is invertible, then A^{-1}=\frac{1}{|A|}\,Adj(A) , and then it is straightforward that if A is an

    integer matrix and |A|=\pm 1 , then also A^{-1} is an integer matrix.

    The other direction follows from Cramer's rule: in A^{-1}\cdot |A| = Adj(A) , the entry ij of Adj(A) ,which of course

    in an integer, is given by x_{ij}|A| , where x_{ij}= the ij-entry of A^{-1} . Continue from here.

    Tonio
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    i understand the first part but i have trouble continuing from where you left off in the second part.
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  4. #4
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    Quote Originally Posted by oblixps View Post
    i understand the first part but i have trouble continuing from where you left off in the second part.

    My message entered at 8:52 AM, and your last post entered at 9:08...16 minutes difference: where did the

    "sit down, grab a pencil and paper, THINK REALLY HARD and begin doing maths" fit in this lapse of time?!?

    Tonio
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    sorry i am still not figuring it out. i have that x_ij |A| = Adj(A)_ij where x_ij is the i,jth entry of A^-1 and Adj(A)_ij is the i,jth entry of Adj(A). Then i have formula for Cramer's Rule which is x_i det(A) = det(A_ib). I've tried manipulating the two equations i had together but i keep getting dead ends. I still have trouble seeing how if A^-1 and A are integer matrices then det(A) must be +/- 1 follows from Cramer's rule. Can you give me a hint or two? or maybe show the next step or two? thanks
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  6. #6
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    Quote Originally Posted by oblixps View Post
    sorry i am still not figuring it out. i have that x_ij |A| = Adj(A)_ij where x_ij is the i,jth entry of A^-1 and Adj(A)_ij is the i,jth entry of Adj(A). Then i have formula for Cramer's Rule which is x_i det(A) = det(A_ib). I've tried manipulating the two equations i had together but i keep getting dead ends. I still have trouble seeing how if A^-1 and A are integer matrices then det(A) must be +/- 1 follows from Cramer's rule. Can you give me a hint or two? or maybe show the next step or two? thanks

    It's not that: by the product theorem, we have that det(AB)=det(A)\cdot det(B) , so if A,\,A^{-1} are both integer

    matrices, then both determinants are integers, but

    1 = det(I) = det(AA^{-1})=det(A)\cdot det(A^{-1})=det(A)\cdot det(A)^{-1} .

    Now, how many pairs of integers number do you know that their product equals 1...?
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  7. #7
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    oh so both det(A) and det(A^-1) must be integers and the only two integers that multiply to 1 are 1 and 1. also, since det(A^-1) = det(A)^-1, the inverse of an integer must be an integer and the only integer with that property is 1 and -1. Thank you! i understand now. I was just going about the problem the wrong way since i thought i had to use Cramer's rule somewhere in the proof.
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  8. #8
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    Quote Originally Posted by oblixps View Post
    oh so both det(A) and det(A^-1) must be integers and the only two integers that multiply to 1 are 1 and 1. also, since det(A^-1) = det(A)^-1, the inverse of an integer must be an integer and the only integer with that property is 1 and -1. Thank you! i understand now. I was just going about the problem the wrong way since i thought i had to use Cramer's rule somewhere in the proof.

    To be honest, so did I in the beginning, and perhaps this misled you. But it is way easier than that, as you saw above.

    Tonio
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