(adsbygoogle = window.adsbygoogle || []).push({}); - Jim Hefferon: Theorem 1.5 says that we must get the same solution set no matter how we proceed but if we do Gauss’ method in two ways must we get the same number of free variables in each echelon form system? Must those be the same variables, that is, is solving a problem one way to get y and w free and solving it another way to get y and z free impossible? In the rest of this chapter we will answer these questions. The answer to each is 'yes'.Linear algebra

Theorem 1.5 (Gauss' method). If a linear system is changed to another by one of these operations

(1) an equation is swapped with another

(2) an equation has both sides multiplied by a nonzero constant

(3) an equation is replaced by the sum of itself and a multiple of another

then the two systems have the same set of solutions.

Here is a counter example to the claim that the number of free variables is constant. Take the following coefficient matrix in echelon form:

[tex]\begin{bmatrix}

2 & 3\\

0 & 1

\end{bmatrix}[/tex]

It shows one free variable, by the definition quoted below. But it can be transformed by row operation 3 into to another matrix, also in echelon form, which shows no free variables:

[tex]\begin{bmatrix}

2 & 0\\

0 & 1

\end{bmatrix}.[/tex]

1.10DefinitionIn each row of a system, the first variable with a nonzero coefficient is the row’s leading variable. A system is in echelon form if each leading variable is to the right of the leading variable in the row above it (except for the leading variable in the first row).Should "echelon form" be replaced with "reduced echelon form" in definition 2.2 and the statement I quoted at the beginning of this post? 2.2DefinitionIn an echelon form linear system the variables that are not leading are free.

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# Free variables constant?

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