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Math Help - solve system

  1. #1
    Junior Member
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    solve system

    Hi again, I think I have the correct answer to part a of this question but I I am not sure about part b, so would someone be kind enough to check my working please.

    \begin{bmatrix}6 & -3 \\2 & 6 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} = \begin{bmatrix}3 \\1 \end{bmatrix}

    a) solve in F_p for p=5,11,17
    b) find the number of solutions for p=7


    Working:
    a) det \begin{bmatrix}6 & -3 \\2 & 6 \end{bmatrix} =42=2 \cdot 3 \cdot 7
    so it is not invertible for p=2,3,7
    so I get
    \begin{bmatrix}x_1 \\x_2 \end{bmatrix} = \frac{1}{42} \begin{bmatrix}6 & 3 \\-2 & 6 \end{bmatrix}\begin{bmatrix}3 \\1 \end{bmatrix}= \begin{bmatrix}2^{-1} \\0 \end{bmatrix}

    For p=5: 2^{-1}=3 and 0=0 so solution for p=5 is \begin{bmatrix}3 \\0 \end{bmatrix}

    For p=11, solution is \begin{bmatrix}6 \\0 \end{bmatrix}

    For p=17, solution is \begin{bmatrix}9 \\0 \end{bmatrix}


    b) For p=7 the matrix in not invertible so for the two linear equation one is a scalar of the other.

    3 \cdot (2,6)=(6,18)=(6,-3)

    therefore
    2x_1+6x_2=1=6
    x_1+3x_2=3

    Thus, for each choice of x_2 \in F_7, x_1=3-3x_2.
    So there are 7 possible solutions:
    \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \end{bmatrix}, \begin{bmatrix} 4 \\ 2 \end{bmatrix}, \begin{bmatrix} 5 \\ 4 \end{bmatrix}, \begin{bmatrix} 6 \\ 6 \end{bmatrix}
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  2. #2
    MHF Contributor

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    Re: solve system

    1 does not equal 6 mod 7.

    2x_1 + 6x_2 = 1 \implies x_1 + 3x_2 = 4 \implies x_1 = 4 + 4x_2
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  3. #3
    Junior Member
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    Re: solve system

    oh yes, stupid mistake. Thanks Deveno. Got it now.

    you mean x_1=4-3x_2 \implies x_1 = 4 +(7-3)x_2=4+4x_2
    Last edited by Juneu436; October 3rd 2011 at 04:28 PM.
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