Linear Algebra: A Modern Introduction

Published by Cengage Learning
ISBN 10: 1285463242
ISBN 13: 978-1-28546-324-7

Chapter 1 - Vectors - 1.1 The Geometry and Algebra of Vectors - Exercises 1.1 - Page 17: 48

Answer

No solution in $\mathbb{Z}_{5}$.

Work Step by Step

The elements of $\;\mathbb{Z}_{5}\;$ is 0,1,2,3,4 So, if there is a solution for this equation, then x is one of these numbers. $If \;\;x=0\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2.(0)=1 \;\;\; {\color{Red}\rightarrow } \;\;\;\;\;\; 0\neq 1\\ so \;\;0 \;is \;not \;solution\\\\ If \;x=1\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;2.(1)=1\;\; {\color{Red}\rightarrow } \;\;\; 2\neq 1\\ so\; 1\; is\; not\; solution.\\\\ If\;x=2\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;2.(2)=1 \;\;\;\;{\color{Red}\rightarrow }\;\;\;\; 4\neq 1\\ so\; 2\; is\; not\; solution\\\\ If \;x=3\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;2.(3)=1 \;\;\;{\color{Red}\rightarrow } \;\;\;\; 6\neq 1\\ so\; 2\; is\; not\; solution\\\\ If\; x=4\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2.(4)=1 \;\;\;{\color{Red}\rightarrow } \;\;\;8\neq 1\\ so\; 4 \;is\; not\; solution\\\\$ No solution in $\mathbb{Z}_{5}$.
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