LINEAR ALGEBRA: How to prove system has one unique solution

[phyzz]

Homework Statement

Show that the system ∑ hereunder admits one unique solution

<br /> ∑ =<br /> \left[\begin{array}{cc}<br /> 1 &amp; a_{1} &amp; a_{1}{}^{2} &amp; a_{1}{}^{3} &amp; | &amp; b_{1}\\<br /> 1 &amp; a_{2} &amp; a_{2}{}^{2} &amp; a_{2}{}^{3} &amp; | &amp; b_{2}\\<br /> 1 &amp; a_{3} &amp; a_{3}{}^{2} &amp; a_{3}{}^{3} &amp; | &amp; b_{3}\\<br /> 1 &amp; a_{4} &amp; a_{4}{}^{2} &amp; a_{4}{}^{3} &amp; | &amp; b_{4}<br /> \end{array}\right]<br />

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:

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

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!

 

[micromass]

Why don't you calculate the determinant?

 

[Ray Vickson]

Homework Statement

Show that the system ∑ hereunder admits one unique solution

<br /> ∑ =<br /> \left[\begin{array}{cc}<br /> 1 &amp; a_{1} &amp; a_{1}{}^{2} &amp; a_{1}{}^{3} &amp; | &amp; b_{1}\\<br /> 1 &amp; a_{2} &amp; a_{2}{}^{2} &amp; a_{2}{}^{3} &amp; | &amp; b_{2}\\<br /> 1 &amp; a_{3} &amp; a_{3}{}^{2} &amp; a_{3}{}^{3} &amp; | &amp; b_{3}\\<br /> 1 &amp; a_{4} &amp; a_{4}{}^{2} &amp; a_{4}{}^{3} &amp; | &amp; b_{4}<br /> \end{array}\right]<br />

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:

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

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!

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

RGV