Let S = (0, 1, 2, 3} and the relations A, B, and C on S are as follows: %3D A = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1),(1, 3), (2, 2), (3, 0), (3, 1), (3, 3)}, B = {(0, 0),(1, 1), (0, 2), (0, 3), (2, 3)}, C= {(0, 0), (1, 1), (2, 2)).

Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter6: Linear Transformations
Section6.4: Transistion Matrices And Similarity
Problem 17E
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Let S = (0, 1, 2, 3} and the relations A, B, and C on S are as follows:
%3D
A = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1),(1, 3), (2, 2), (3, 0), (3, 1), (3, 3)},
B = {(0, 0),(1, 1), (0, 2), (0, 3), (2, 3)},
C = {(0, 0), (1, 1), (2, 2)).
C has . properties
symmetric, but neither reflexive nor transitive +
B has . properties
Seç.
B
The equivalance relation is: symmetric, but neither reflexive nor transitive
reflexive, transitive but not symmetric
reflexive, symmetric but not transitive
reflexive, symmetric and transitive
Sorunun tüm bölümlerine ce transitive, symmetric but not reflexive
A has . properties
A
reflexive, but neither symmetric nor transitive
transitive, but neither symmetric nor reflexive
Transcribed Image Text:Let S = (0, 1, 2, 3} and the relations A, B, and C on S are as follows: %3D A = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1),(1, 3), (2, 2), (3, 0), (3, 1), (3, 3)}, B = {(0, 0),(1, 1), (0, 2), (0, 3), (2, 3)}, C = {(0, 0), (1, 1), (2, 2)). C has . properties symmetric, but neither reflexive nor transitive + B has . properties Seç. B The equivalance relation is: symmetric, but neither reflexive nor transitive reflexive, transitive but not symmetric reflexive, symmetric but not transitive reflexive, symmetric and transitive Sorunun tüm bölümlerine ce transitive, symmetric but not reflexive A has . properties A reflexive, but neither symmetric nor transitive transitive, but neither symmetric nor reflexive
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