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LINEAR ALGEBRA: How to prove system has one unique solution

  1. Oct 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that the system ∑ hereunder admits one unique solution

    [tex]
    ∑ =
    \left[\begin{array}{cc}
    1 & a_{1} & a_{1}{}^{2} & a_{1}{}^{3} & | & b_{1}\\
    1 & a_{2} & a_{2}{}^{2} & a_{2}{}^{3} & | & b_{2}\\
    1 & a_{3} & a_{3}{}^{2} & a_{3}{}^{3} & | & b_{3}\\
    1 & a_{4} & a_{4}{}^{2} & a_{4}{}^{3} & | & b_{4}
    \end{array}\right]
    [/tex]

    3. The attempt at a solution

    I know I have to perform Gauss-Jordan elimination fully by pivoting on the diagonal and zeroing out both on top and bottom of the diagonal.
    After just two steps I'm stuck. I came up with this:

    [tex]
    ∑ ∼
    \left[\begin{array}{cc}
    1 & a_{1} & a_{1}{}^{2} & a_{1}{}^{3} & | & b_{1}\\
    0 & a_{2} - a_{1} & a_{2}{}^{2} - a_{1}{}^{2} & a_{2}{}^{3} - a_{1}{}^{3} & | & b_{2} - b_{1}\\
    0 & 0 & - & - & | & -\\
    0 & 0 & - & - & | & -
    \end{array}\right]
    [/tex]

    I don't know how to continue to fill in the _ spaces as I don't know what operation zeroed out the values under the pivot.

    I don't like dividing the pivots through just so they're of value 1 as to avoid working with fractions.

    I know I'm looking for a compatible system (i.e. (n - r) = 0, n being #rows and r being the rank) that also gives 0 secondary unknowns (i.e. (p - r) = 0, p being #columns not including the RHS and r = rank).

    Thank you very much!
     
  2. jcsd
  3. Oct 11, 2012 #2

    micromass

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    Why don't you calculate the determinant?
     
  4. Oct 11, 2012 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    As stated the result is false; you need to assume something about the ai!

    RGV
     
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