LINEAR ALGEBRA: How to prove system has one unique solution

  • Thread starter phyzz
  • Start date
  • #1
35
0

Homework Statement



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]

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!
 

Answers and Replies

  • #2
22,129
3,297
Why don't you calculate the determinant?
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement



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]

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!

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

RGV
 

Related Threads on LINEAR ALGEBRA: How to prove system has one unique solution

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
171
Replies
7
Views
2K
Replies
9
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
5
Views
796
Replies
5
Views
2K
Replies
3
Views
965
Top