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Matrix Puzzle - Math Puzzle

Source: www.puzzletweeter.com

Problem:
Let A,B be 2x2 matrices with integer entries.
Suppose the matrices A,A+B,A+2B,A+3B,A+4B are all invertible and their inverses are also integer matrices.

Then show that A+5B is invertible and it's inverse is an integer matrix.

Update (24 June 2014):
Solution: Posted by Yashoteja Prabhu (Ex-RA at Microsoft Research, IITB CSE 2011 Alumnus) and Sanchit Gupta in comments!


Comments

  1. first guess; A is identity and B is zero matrix?

    ReplyDelete
    Replies
    1. Given A and B such that... you are not supposed to find A and B :)

      Delete
  2. If P and inv(P) are both integer matrices, then both det(P) and det(inv(P)) are integers. Since det(inv(P)) = 1/det(P), this means det(P) = +1 or -1.

    Now det(A+pB) can be written as a quadratic in p, say:
    det(A+pB) = a*p^2 + b*p + c
    Then these 5 equations should hold by assumption:
    |c| = 1,
    |a + b + c| = 1
    |4a + 2b + c| = 1
    |9a + 3b + c| = 1
    |16a + 4b + c| = 1

    Theorem
    =======
    Now a (strictly convex) quadratic can take same value only at 2 distinct domain points. If it takes same value at 3 distinct domain points, this means the quadratic is in fact a constant, i.e., a = b = 0,

    Since we have five equations and each equation has only 2 possible values in rhs (+1 or -1), at least 3 equations will have the same value.
    Hence invoking the theorem, a=b=0,
    which implies |25a + 4b + c| = |c| = 1, which further means:
    inv(A+pB) exists and is integral for all values of p including p=5.

    Proof of theorem
    =================

    If there exists 3 distinct points p,q,r such that: ap^2+bp+c = aq^2+bq+c = ar^2+br+c,
    then subtracting pairs of equations:
    (p-q)(a(p+q)+b) = (q-r)(a(q+r)+b) = 0
    since p!=q and q!=r:
    a(p+q)+b = a(q+r)+b = 0
    since, p!=r, this would mean: a = b = 0

    ReplyDelete
    Replies
    1. Good solution. Thanks

      One clarification required:
      I see that |det(A+pB)| = 1 and so inverse exists
      How did you reach to the conclusion that inverse is integral?

      Delete
    2. Yes it does not immediately follow.
      Inverse of [a b;c d] is [d -b;-c a]/(ad-bc)
      If a,b,c and d are integers and determinant is +/- 1 => inverse will be integral too

      Now for p=5, A+pB is integral, and (as said before) determinant is +/- 1, hence its inverse is integral too (so integrality does not hold for general p)

      Delete
    3. How did you deduce that det(A+pB) can be written as a*p^2 + b*p + c?

      Delete
    4. By following:
      det([p q;r s]) = ps-qr

      Delete
    5. One typo in the solution: "which implies |25a + 5b + c| = |c| = 1, which further means:"
      No harm caused though :-)

      Delete
  3. It has a simple solution
    Since A+2B,A+3B,A+4B are invertible
    (A+2B)*C1 = I
    (A+3B)*C2 = I
    (A+4B)*C3 = I
    Add 2 and 3 and subtract 1
    (A+5B)(C2+C3-C1) = I
    =>A+5B is invertible and since Ci are integer matrices => A+5B has integer inverse

    ReplyDelete
    Replies
    1. Thanks. Of course to prove that A+5B is invertible, you have to prove that there exists X such that X*(A+5B) and (A+5B)*X are both integer matrices. Following your argument, the other part of the proof can also be done similarly. Thanks

      Delete
    2. Possibly silly question:
      Adding 2 and 3 i get
      A(C1 + C2 - C3) + B(2*C1 + 3*C2 - 4*C3) = I
      How did you get (A+5B)(C2+C3-C1)?

      Delete

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