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Math Help - Linear Algebra - Fields

  1. #1
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    Linear Algebra - Fields

    I came across this question in my course.
    Consider a system AX=B of n linear equations in n unknowns where A and B have integer entries. Prove or disprove : If the system has an integer solution, then it has a solution in F(p) for all p.

    Well, I could see that I have to disprove it. Since the determinant could be 0 (mod p). In which case the system needn't have a solution. But how exactly do i present a formal proof for this?

    One more question,
    It says in my book for the system AX = B
    where
    ( 8 3 = A
    2 6)
    and B = (3 -1)t
    (I hope you understand the matrix notations.) , there is a solution in F(7), even though det(A)=42 ~0 (mod 7). How is this ?
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  2. #2
    MHF Contributor

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    Quote Originally Posted by poorna View Post
    I came across this question in my course.
    Consider a system AX=B of n linear equations in n unknowns where A and B have integer entries. Prove or disprove : If the system has an integer solution, then it has a solution in F(p) for all p.

    Well, I could see that I have to disprove it. Since the determinant could be 0 (mod p). In which case the system needn't have a solution. But how exactly do i present a formal proof for this?
    it's actually true for an obvious reason: let A=[a_{ij}]. write AX = B as a system of equations: \sum_{j=1}^n a_{ij}x_j=b_i, \ 1 \leq i \leq n. for any integer a, let \bar{a} be a \mod p.

    then we'll have \sum_{j=1}^n \bar{a_{ij}} \bar{x_j}=\bar{b_i}, \ 1 \leq i \leq n. thus x=[\bar{x_1} \ \bar{x_2} \ \cdots \ \bar{x_n}]^T is a solution modulo p.


    One more question,
    It says in my book for the system AX = B
    where
    ( 8 3 = A
    2 6)
    and B = (3 -1)t
    (I hope you understand the matrix notations.) , there is a solution in F(7), even though det(A)=42 ~0 (mod 7). How is this ?
    \det A = 0 does not imply that the equation AX=B has no solution. so there's nothing strange about your example.
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