LINEAR ALGEBRA: How to prove system has one unique solution

In summary: BkbyBub3QgY2FsY3VsYXRlIHRoZSBkZXRlcm1pbnQ/In summary, the system ∑ admits one unique solution. In order to find this solution, you must perform Gauss-Jordan elimination by pivoting on the diagonal and zeroing out both above and below it. However, you may encounter fractions, so it is preferable to find a system that has a compatible number of rows and columns with no secondary unknowns. Additionally, it is important to assume something about the values of ai in order for the result to be true.
  • #1
phyzz
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!
 
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  • #2
Why don't you calculate the determinant?
 
  • #3
phyzz said:

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
 

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations and their representations in vector spaces. It involves the use of matrices, determinants, and systems of linear equations to solve problems related to lines, planes, and higher dimensional objects.

2. How do you prove that a system of linear equations has one unique solution?

To prove that a system of linear equations has one unique solution, we can use the Gaussian elimination method or the inverse matrix method. Both methods involve transforming the system of equations into an equivalent form that is easier to solve. If the resulting equations have a unique solution, then the original system also has a unique solution.

3. Can a system of linear equations have more than one solution?

Yes, a system of linear equations can have more than one solution. This typically happens when there are more variables than equations in the system, leading to an infinite number of solutions. In some cases, there may be no solutions at all.

4. What is a determinant and how is it related to systems of linear equations?

A determinant is a value that can be calculated from a square matrix. It is used to determine whether a system of linear equations has a unique solution, no solution, or infinitely many solutions. If the determinant of a matrix is non-zero, then the system has a unique solution.

5. How is linear algebra used in real life?

Linear algebra has many applications in real life, including computer graphics, data analysis, engineering, and physics. It is used to solve problems related to optimization, regression, and pattern recognition. It also plays a crucial role in machine learning algorithms and data mining techniques.

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