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Thread: System of 6 linear equations, Cramer's rule, need help with notation.

  1. #1
    Jan 2013
    United Kingdom


    I don't have a particular problem to solve, but I need help with understanding the notation I have found in an academic paper.

    I have a system of linear equations which each take the form:

    pi xi r1 + pi yi r2 - pi u + qi xi r4 + qi yi r5 - qi v + xi x + yi y - w/2 + mi = 0 (for i=1-6)

    Where mi=((Li)^2- (xi)^2- (yi)^2- (pi)^2- (qi)^2 )/2

    w= x^2+ y^2+ z^2

    Where: pi, qi, xi, yi, Li and mi are all known quantities.

    The paper outlines a method to solve for r1, r2, r4, r5, v, u, w, x, y, z

    The paper enters the equations into a 6 x10 matrix and called this matrix 'M'

    The following equation then holds true:

    M * t = 0

    Where t is a 10 x 1 matrix

    t = [r1; r2; u; r4; r5; v; x; y; z; 1]

    (so all variables in the 'M' matrix are known and we are solving for the variables in the 't' matrix)

    The paper then goes on to say that we can solve the system of equations symbolically using Cramer algorithm, regarding r1,r2,r4,r5,u,v as linear unknowns, and obtain the following expressions of those variables with respect to x,y,z.

    a0r1 + a11x + a12y + a13w + a14 = 0
    a0r2 + a21x + a22y + a23w + a24 = 0
    a0u + a31x + a32y + a33w + a34 = 0
    a0r4 + a41x + a42y + a43w + a44 = 0
    a0r5 + a51x + a52y + a53w + a54 = 0
    a0v + a61x + a62y + a63w + a64 = 0


    a0 = det(c1,c2,c3,c4,c5,c6)


    aij = det(c1,...,ci-1,cj+6,ci+1,...,c6)

    Now this is where I am getting stuck.

    I am unsure of exactly what the last line of notation above means.

    I understand that a0 is equal to the determinant of the matrix 'M' when it is reduced to M(6x6).

    But what exactly does the next line mean? I dont know how to calculate the coefficients aij.

    If was looking for the coefficient a11 for instance would this constitute the following matrix:

    a11 = det(c1,..., c0,c7,c2,...,c6) ??????

    There is no column C0 is there? And I'm unsure of what columns of the matrix go into the two gaps.

    Could somebody please help, I'm sure this must be relatively easy for somebody who works with this notation all the time, but I dont understand.

    Thank you

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  2. #2
    MHF Contributor

    Mar 2011

    Re: System of 6 linear equations, Cramer's rule, need help with notation.

    ok, let me see if i can unravel this for you. our system of equations is:

    Mt = 0 (your equations seem to be missing something (i don't see a "z term")).

    as it stands, this system is "under-determined" we have 10 variables to solve for, and only 6 equations.

    so what we are going to do is create a bunch of 6x6 matrices from M, and calculate their determinants. the first matrix, A_0 will just be the first 6 columns of M. this represents our "baseline" (pretending x,y and z aren't a part of our system).

    to calculate a_{ij} = det(M_{ij})$ you need to know how to form the 6x6 matrix M_{ij}. basically this is a choice of "which 6 columns (of M) to use".

    according to your description, for a_{11} we would have:

    a_{11} = \det(c_7,c_2,c_3,c_4,c_5,c_6) (we use column 7 of M in place of column 1 of M...the "i" tells us which column of A_0 to omit, and the "j" tells us which one we're going to replace it with: the j+6-th column of M).

    as another example, we have:

    a_{23} = \det(c_1,c_9,c_3,c_4,c_5,c_6) (we use column 9 of M in place of column 2 of M).

    we have 6 possible columns of A_0 to replace (one of the first 6 columns of M), and 4 possible columns to use instead (columns 7 through 10 of the original M), giving us 24 determinants to calculate.

    this is going to give a system of 6 equations in 4 unknowns ("over-determined") which (if consistent, and it should be, if the original system is) is now certainly solvable.
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