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Math Help - Integer Valued Matrix

  1. #1
    MHF Contributor Drexel28's Avatar
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    Integer Valued Matrix

    Problem( \star\star): Let A,B\in\text{Mat}_2\left(\mathbb{Z}\right) be such that A,A+B,A+2B,A+3B and A+4B are invertible and such that A^{-1},\left(A+B\right)^{-1},\left(A+2B\right)^{-1},\left(A+3B\right)^{-1},\left(A+4B\right)^{-1}\in\text{Mat}_2\left(\mathbb{Z}\right). Prove that A+5B is invertible and \left(A+5B\right)^{-1}\in\text{Mat}_2\left(\mathbb{Z}\right).

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    Last edited by CaptainBlack; December 13th 2010 at 11:03 PM. Reason: typo correction (I hope)
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Problem( \star\star): Let A,B\in\text{Mat}_2\left(\mathbb{Z}\right) be such that A,A+B,A+2B,A+3B and A+5B are invertible and such that A^{-1},\left(A+B\right)^{-1},\left(A+2B\right)^{-1},\left(A+3B\right)^{-1},\left(A+4B\right)^{-1}\in\text{Mat}_2\left(\mathbb{Z}\right). Prove that A+5B is invertible and \left(A+5B\right)^{-1}\in\text{Mat}_2\left(\mathbb{Z}\right).

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    Hint:

    Spoiler:


    First try proving that if A\in\text{Mat}_n\left(\mathbb{Z}\right) then a necessary and sufficient condition for A to be invertible and A\in\text{Mat}_n\left(\mathbb{Z}\right) is that \det A=\pm 1
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Since no one seems interested I'll just give the answer for logical soundness:

    Spoiler:


    To solve this we first make the observation that if A\in\text{Mat}_n\left(\mathbb{Z}\right) then A is invertible and A^{-1}\in\text{Mat}_n\left(\mathbb{Z}\right) if and only if \det A=\pm 1. Indeed, clearly if \det A=\pm 1 the result follows. Conversely, the conclusion follows by noticing that since A^{-1}\in\text{Mat}_n\left(\mathbb{Z}\right) that \det A^{-1}=\frac{1}{\det A}\in\mathbb{Z} from where the conclusion follows.

    So, with this observation let p(x)=\det (A+xB). Then, since A,B\in\text{Mat}_2\left(\mathbb{Z}\right) we see that \deg p\leqslant 2. Note though that by assumption of the problem and the above observation we have that p(x)\in\{-1,1\},\quad x\in\{0,1,2,3,4\}. But, by the Pigeon Hole principle we know that p(x) takes the value either 1 or -1 at least three times on \{0,1,2,3,4\}. But, since \deg p\leqslant 2 this implies that p is constant and p(x)=1 or p(x)=-1. Thus, by our observation we may conclude that A+xB is invertible and \left(A+xB\right)^{-1}\in\text{Mat}_2\left(\mathbb{Z}\right) for every x\in\mathbb{Z}. \blacksquare

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