# LINEAR ALGEBRA: How to prove system has one unique solution

## Homework Statement

Show that the system ∑ hereunder admits one unique solution

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

## 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:

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

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!

Why don't you calculate the determinant?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Show that the system ∑ hereunder admits one unique solution

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

## 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:

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

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 ai!

RGV