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Linear Algebra, 2020a

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60 Chapter One. <strong>Linear</strong> Systems<br />

and to C’s.<br />

c 1,1 x 1 + c 1,2 x 2 + ···+ c 1,n x n = 0<br />

c 2,1 x 1 + c 2,2 x 2 + ···+ c 2,n x n = 0<br />

.<br />

c m,1 x 1 + c m,2 x 2 + ···+ c m,n x n = 0<br />

(∗∗∗)<br />

With B and C different only in column n, suppose that they differ in row i.<br />

Subtract row i of (∗∗∗) from row i of (∗∗) to get the equation (b i,n −c i,n )·x n = 0.<br />

We’ve assumed that b i,n ≠ c i,n and so we get x n = 0. Thus x n is not a free<br />

variable and so in (∗∗) and (∗∗∗) the n-th column contains the leading entry of<br />

some row, since in an echelon form matrix any column that does not contain a<br />

leading entry is associated with a free variable.<br />

But now, with B and C equal on the first n − 1 columns, by the definition of<br />

reduced echeleon form their leading entries in the n-th column are in the same<br />

row. And, both leading entries would have to be 1, and would have to be the<br />

only nonzero entries in that column. Therefore B = C.<br />

QED<br />

We have asked whether any two echelon form versions of a linear system have<br />

the same number of free variables, and if so are they exactly the same variables?<br />

With the prior result we can answer both questions “yes.” There is no linear<br />

system such that, say, we could apply Gauss’s Method one way and get y and z<br />

free but apply it another way and get y and w free.<br />

Before the proof, recall the distinction between free variables and parameters.<br />

This system<br />

x + y = 1<br />

y + z = 2<br />

has one free variable, z, because it is the only variable not leading a row. We<br />

have the habit of parametrizing using the free variable y = 2 − z, x =−1 + z, but<br />

we could also parametrize using another variable, such as z = 2 − y, x = 1 − y.<br />

So the set of parameters is not unique, it is the set of free variables that is<br />

unique.<br />

2.7 Corollary If from a starting linear systems we derive by Gauss’s Method two<br />

different echelon form systems, then the two have the same free variables.<br />

Proof The prior result says that the reduced echelon form is unique. We get<br />

from any echelon form version to the reduced echelon form by eliminating up, so<br />

any echelon form version of a system has the same free variables as the reduced<br />

echelon form version.<br />

QED<br />

We close with a recap. In Gauss’s Method we start with a matrix and then<br />

derive a sequence of other matrices. We defined two matrices to be related if we

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