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Math Help - Solve system of 7 equations with 3 unknowns

  1. #1
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    Solve system of 7 equations with 3 unknowns

    I want to solve this in some form of Ax=b , but getting this into a system of linear equations in matrix form is perplexing me....

    thanks for any insight

    equations:

    x*cos((2*pi*50*z)/2000)) - y*sin((2*pi*50*z)/2000))=.008
    x*cos((2*pi*60*z)/2000)) - y*sin((2*pi*60*z)/2000))=.009
    x*cos((2*pi*70*z)/2000)) - y*sin((2*pi*70*z)/2000))=.006
    x*cos((2*pi*80*z)/2000)) - y*sin((2*pi*80*z)/2000))=.005
    x*cos((2*pi*90*z)/2000)) - y*sin((2*pi*90*z)/2000))=.0009
    x*cos((2*pi*100*z)/2000)) - y*sin((2*pi*100*z)/2000))=.0008
    x*cos((2*pi*110*z)/2000)) - y*sin((2*pi*110*z)/2000))=.0001
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  2. #2
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    Quote Originally Posted by jcf2506 View Post
    I want to solve this in some form of Ax=b , but getting this into a system of linear equations in matrix form is perplexing me....

    thanks for any insight

    equations:

    x*cos((2*pi*50*z)/2000)) - y*sin((2*pi*50*z)/2000))=.008
    x*cos((2*pi*60*z)/2000)) - y*sin((2*pi*60*z)/2000))=.009
    x*cos((2*pi*70*z)/2000)) - y*sin((2*pi*70*z)/2000))=.006
    x*cos((2*pi*80*z)/2000)) - y*sin((2*pi*80*z)/2000))=.005
    x*cos((2*pi*90*z)/2000)) - y*sin((2*pi*90*z)/2000))=.0009
    x*cos((2*pi*100*z)/2000)) - y*sin((2*pi*100*z)/2000))=.0008
    x*cos((2*pi*110*z)/2000)) - y*sin((2*pi*110*z)/2000))=.0001

    If you have an equation of the form

    A\mathbf{x} = \mathbf{b}

    where A is NOT a square matrix, you can create a square matrix by premulitplying both sides by A^T.


    So A^TA\mathbf{x} = A^T\mathbf{b}.


    Now that you have a square matrix on the LHS ( A^TA), you can premultiply both sides by its inverse (provided an inverse exists).


    (A^TA)^{-1}A^TA\mathbf{x} = (A^TA)^{-1}A^T\mathbf{b}

    I\mathbf{x} = (A^TA)^{-1}A^T\mathbf{b}

    \mathbf{x} = (A^TA)^{-1}A^T\mathbf{b}.


    Does this help?
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  3. #3
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    clarification on reply to: solve system of 7 eqns

    First of all, thank you so much for the help, your answer did clarify things immensely, but unfortunately part of my problem is setting up my Ax=b

    what does A and x look like? I know b is a vector of all the answers and I know x will be some combination of my variables in vector form [x y z] , but I am not sure what A looks like to begin the operations you mentioned on the first response of A^TA etc

    thanks again, all the best
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  4. #4
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    Quote Originally Posted by jcf2506 View Post
    I want to solve this in some form of Ax=b , but getting this into a system of linear equations in matrix form is perplexing me....

    thanks for any insight

    equations:

    x*cos((2*pi*50*z)/2000)) - y*sin((2*pi*50*z)/2000))=.008
    x*cos((2*pi*60*z)/2000)) - y*sin((2*pi*60*z)/2000))=.009
    x*cos((2*pi*70*z)/2000)) - y*sin((2*pi*70*z)/2000))=.006
    x*cos((2*pi*80*z)/2000)) - y*sin((2*pi*80*z)/2000))=.005
    x*cos((2*pi*90*z)/2000)) - y*sin((2*pi*90*z)/2000))=.0009
    x*cos((2*pi*100*z)/2000)) - y*sin((2*pi*100*z)/2000))=.0008
    x*cos((2*pi*110*z)/2000)) - y*sin((2*pi*110*z)/2000))=.0001
    Try

    \left[\begin{matrix}\cos{\frac{2\pi\cdot 50z}{2000}} & -\sin{\frac{2\pi \cdot 50z}{2000}} \\ \cos{\frac{2\pi \cdot 60z}{2000}} & -\sin{\frac{2\pi \cdot 60z}{2000}} \\ \cos{\frac{2\pi \cdot 70z}{2000}} & -\sin{\frac{2\pi \cdot 70z}{2000}} \\ \cos{\frac{2\pi \cdot 80z}{2000}} & -\sin{\frac{2\pi \cdot 80z}{2000}}\end{matrix}\right] etc... as A, (it wouldn't let me fit it all on the page)

    \left[\begin{matrix}x \\ y \end{matrix}\right] as \mathbf{x}

    and \left[\begin{matrix}0.008 \\ 0.009 \\ 0.006 \\ 0.005 \\ 0.0009 \\ 0.0008 \\ 0.0001 \end{matrix}\right] as \mathbf{b}.
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