   Chapter 2.6, Problem 13E

Chapter
Section
Textbook Problem

In Exercise 11 − 14 , Solve the systems of equations in ℤ 7 . [ 3 ] [ x ] + [ 2 ] [ y ] = [ 1 ] [ 5 ] [ x ] + [ 6 ] [ y ] = [ 5 ]

To determine

The solution of the system of equation in 7

[x]+[y]=[x]+[y]=

Explanation

Formula used:

1) Multiplicative inverses in n:

An element [a] of n has a multiplicative inverse in n, if and only if a and n are relatively prime.

2) [x]=[a]1[b] is the unique solution to the equation [a][x]=[b] in n.

Explanation:

The set 7={,,,,,,} of congruence classes

Consider the system of linear equations

[x]+[y]=.......(1)[x]+[y]=

Need to eliminate one of the element [x] and [y].

[y] can be eliminate by multiplying the top equation by .

The top equation becomes [x]+[y]=.

It is same as,

[x]+[y]=, since = in 7.

Therefore, system becomes

[x]+[y]=[x]+[y]=

Now subtract top equation from the bottom one,

[x][x]=

[x]=

This simplifies to

[x]=.

Since 3 and 7 are relatively prime,  has a multiplicative inverse in 7.

Using Euclidean Algorithm to find the multiplicative inverse of  in 7.

7=(3)(2)+1

3=(1)(3)+0

Substituting the remainders in turn,

1=7(3)(2)

=(7)(1)+3(2)

Thus,

(3)(2)1(mod7)=

Hence,

1==, as == in 7

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