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Math Help - Linear Algebra : Fields, Matrices

  1. #1
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    Linear Algebra : Fields, Matrices

    I'm in the subject 'Matrices', but there still is a use of the Fields topic.

    I was asked to find a system of equations with exactly 49 different solutions. (no further information included)

    Now, I thought of the following system:

    3x+7y=14
    6x+14y=28

    x,y(belong to)Z7

    Z7 is the following field:
    (0,1,2,3,4,5,6)

    By looking at the system of equations, there are infinite possibilities. If we are limited only to Z7, then it means that there are 7 options for x, and 7 options for y, and together - 49.

    (0,0), (0,1), (0,2), ... (0,6)
    (1,0) . . . (1,6)
    .
    .
    .
    .
    (6,0), . . . . . (6,6)

    and so on...

    Now, this is the subject of matrices, and I'm really not sure that it's right to use such ways to solve this problem.

    Can you please help me?

    Thank you
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  2. #2
    ynj
    ynj is offline
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    Quote Originally Posted by adam63 View Post
    I'm in the subject 'Matrices', but there still is a use of the Fields topic.

    I was asked to find a system of equations with exactly 49 different solutions. (no further information included)

    Now, I thought of the following system:

    3x+7y=14
    6x+14y=28

    x,y(belong to)Z7

    Z7 is the following field:
    (0,1,2,3,4,5,6)

    By looking at the system of equations, there are infinite possibilities. If we are limited only to Z7, then it means that there are 7 options for x, and 7 options for y, and together - 49.

    (0,0), (0,1), (0,2), ... (0,6)
    (1,0) . . . (1,6)
    .
    .
    .
    .
    (6,0), . . . . . (6,6)

    and so on...

    Now, this is the subject of matrices, and I'm really not sure that it's right to use such ways to solve this problem.

    Can you please help me?

    Thank you
    if there are m variables and the rank of these equations(or matrix) is r,then there is 7^(m-r) solutions.....
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  3. #3
    MHF Contributor

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    Quote Originally Posted by adam63 View Post
    I'm in the subject 'Matrices', but there still is a use of the Fields topic.

    I was asked to find a system of equations with exactly 49 different solutions. (no further information included)

    Now, I thought of the following system:

    3x+7y=14
    6x+14y=28

    x,y(belong to)Z7

    Z7 is the following field:
    (0,1,2,3,4,5,6)

    By looking at the system of equations, there are infinite possibilities. If we are limited only to Z7, then it means that there are 7 options for x, and 7 options for y, and together - 49.

    (0,0), (0,1), (0,2), ... (0,6)
    (1,0) . . . (1,6)
    .
    .
    .
    .
    (6,0), . . . . . (6,6)

    and so on...

    Now, this is the subject of matrices, and I'm really not sure that it's right to use such ways to solve this problem.

    Can you please help me?

    Thank you
    I'm not sure what you mean by that. Certainly Z_7\times Z_7 contains exactly 49 members but what makes you think they are all solutions to the equations? You mention (1,0) but if x= 1, y= 0, the equations become 3(1)+ 7(0)= 14 and 6(1)+ 14(0)= 28 which are not true, even in Z_7.

    In fact, in Z_7, 7, 14, and 48 are all congruent to 0, so your equations become just 3x= 0 and 6x= 0 (mod 7). While y can be any number, x must be 0. There are only 7 solutions, not 49.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    How about the system x=x over the field with 49 elements?
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