Page 373 - 35Linear Algebra

G.2 Systems of Linear Equations 373

Now for the second system we have

   
5 2 9 5 2 9
1
R 2
 0 5 10  5 ∼  0 1 2 
0 3 6 0 3 6
 
5 2 9
R 3 −3R 2  0
∼ 1 2 
0 0 0
 
5 0 5
∼  0 1 2 
R 1 −2R 2
0 0 0
 
1 0 1
1
5 R 1
∼  0 1 2 
0 0 0
We scale the second and third rows appropriately in order to avoid fractions,
then subtract the corresponding rows as before. Finally scale the first row
and hence we have x = 1 and y = 2 as a unique solution.

Hints for Review Question 10

This question looks harder than it actually is:

Row equivalence of matrices is an example of an equivalence
relation. Recall that a relation ∼ on a set of objects U
is an equivalence relation if the following three properties
are satisfied:
• Reflexive: For any x ∈ U, we have x ∼ x.

• Symmetric: For any x, y ∈ U, if x ∼ y then y ∼ x.
• Transitive: For any x, y and z ∈ U, if x ∼ y and y ∼ z
then x ∼ z.
(For a more complete discussion of equivalence relations, see
Webwork Homework 0, Problem 4)
Show that row equivalence of augmented matrices is an equivalence
relation.

Firstly remember that an equivalence relation is just a more general ver-
sion of ‘‘equals’’. Here we defined row equivalence for augmented matrices
whose linear systems have solutions by the property that their solutions are
the same.
So this question is really about the word same. Lets do a silly example:
Lets replace the set of augmented matrices by the set of people who have hair.

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