Skip to content
Hub Nexus

No author yetClaim it

Version 1Updated 1 min read

Support

After developing the mechanics of Gauss' method,
we observed that it can be done in more than one way.
One example is that we sometimes have to swap rows and there can be more than
one row to choose from.
Another example is that from this matrix

( \begin{matrix}2 & 2 \\ 4 & 3\end{matrix} )

Gauss' method could derive any of these echelon form matrices.

( \begin{matrix}2 & 2 \\ 0 & − 1\end{matrix} ) \ ( \begin{matrix}1 & 1 \\ 0 & − 1\end{matrix} ) \ ( \begin{matrix}2 & 0 \\ 0 & − 1\end{matrix} )

The first results from − 2 {ρ}_{1} + {ρ}_{2}.
The second comes from following ( 1 / 2 ) {ρ}_{1} with − 4 {ρ}_{1} + {ρ}_{2}.
The third comes
from − 2 {ρ}_{1} + {ρ}_{2} followed by {}2 {ρ}_{2} + {ρ}_{1}
(after the first pivot the matrix is already in
echelon form so the second one is extra work
but it is nonetheless a legal row operation).

The fact that the echelon form outcome of Gauss' method
is not unique leaves us with some questions.
Will any two echelon form versions of a system have the same number of
free variables?
Will they in fact have exactly the same variables free?
In this section we will answer both questions "yes".
We will do more than answer the questions.
We will give a way to decide if one linear system
can be derived from another by row operations.
The answers to the two questions will follow from this larger result.

Where this page came from

This page was imported from Wikibooks. From “Linear Algebra” on Wikibooks, by its contributors, under CC BY-SA 4.0. Changed here: set as a page; navigation and edit links left out; each image under its own licence, credited in its caption.

Nobody has written it yet — it is the source material at a new address, which is why search engines are asked to skip it and why no one earns from it. It is up for grabs: take it on, and it is yours to rewrite and to earn from.

Comments

Spinner Logo

Next in Linear Algebra

CC BY-SA 4.0
Gaussian elimination coupled with back-substitution solves linear systems, but it's not the only method possible. Here is an extension of Gauss' method that has some advantages. Example 1.1 To solve $\begin{matrix}x & + & y & − & 2 z & = & − 2 \\ & & y & + & 3 z & = & 7 \\ x & & & − & z & = & − 1\en
CC BY-SA 4.0
We will close this section and this chapter by proving that every matrix is row equivalent to one and only one reduced echelon form matrix. The ideas that appear here will reappear, and be further developed, in the next chapter. The underlying theme here is that one way to understand a mathematical
CC BY-SA 4.0
The linear systems in this chapter are small enough that their solution by hand is easy. But large systems are easiest, and safest, to do on a computer. There are special purpose programs such as LINPACK for this job. Another popular tool is a general purpose computer algebra system, including both
CC BY-SA 4.0
An economy is an immensely complicated network of interdependences. Changes in one part can ripple out to affect other parts. Economists have struggled to be able to describe, and to make predictions about, such a complicated object. Mathematical models using systems of linear equations have emerged
CC BY-SA 4.0
Gauss' method lends itself nicely to computerization. The code below illustrates. It operates on an $n \ × \ n$ matrix a, pivoting with the first row, then with the second row, etc. for (pivot_row = 1; pivot_row <= n - 1; pivot_row++) { for (row_below = pivot_row + 1; row_below <= n; row_below++) {
CC BY-SA 4.0
The diagram below shows some of a car's electrical network. The battery is on the left, drawn as stacked line segments. The wires are drawn as lines, shown straight and with sharp right angles for neatness. Each light is a circle enclosing a loop.