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Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
ISBN: 9781285463230

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BuyFindarrow_forward

Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
ISBN: 9781285463230
Textbook Problem

In Exercise 11 14 , Solve the systems of equations in 7 .

[ 3 ] [ x ] + [ 2 ] [ y ] = [ 1 ] [ 5 ] [ x ] + [ 6 ] [ y ] = [ 5 ]

To determine

The solution of the system of equation in 7

[3][x]+[2][y]=[1][5][x]+[6][y]=[5]

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={[0],[1],[2],[3],[4],[5],[6]} of congruence classes

Consider the system of linear equations

[3][x]+[2][y]=[1].......(1)[5][x]+[6][y]=[5]

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

[y] can be eliminate by multiplying the top equation by [3].

The top equation becomes [9][x]+[6][y]=[3].

It is same as,

[2][x]+[6][y]=[3], since [9]=[2] in 7.

Therefore, system becomes

[2][x]+[6][y]=[3][5][x]+[6][y]=[5]

Now subtract top equation from the bottom one,

[5][x][2][x]=[5][3]

[52][x]=[53]

This simplifies to

[3][x]=[2].

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

Using Euclidean Algorithm to find the multiplicative inverse of [3] 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)[3][2]=[1]

Hence,

[3]1=[2]=[5], as [2]=[72]=[5] in 7

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