1.
Find
g.c.d, l.c.m and x,y such that d= ax+by, d=(a,b) where
i)
a= 322
and b=770
Solution:
770 = 2´322+126
322 = 2´126+70
126 = 1´70+56
70 = 1´56+14
56 = 4´14+0
Hence g.c.d =d(770,322)=14
Again
dl=ab
14´l= 770´322
\ l = 17710.
Now for x and y
126= 770-2´322
70= 322-2´126
= 322-2´(770-2´322)
= 1´322-2´770+4´322
= 5´322-2´770
56= 126-1´70
= 770-2´322-1´(5´322-2´770)
= 770-2´322-5´322+2´770
= 3´770-7´322
14= 70-1´56
= 5´322-2´770-1´(3´770-7´322)
= 5´322-2´770-3´770+7´322
=-5´770+12´322
\ x = -5 and y= 12.
ii)
a=
11+7i and b= 3+7i in Z(i) where
Solution:
11+7i= (1-i)´(3+7i)+(1+3i)………..(1)
3+7i = (2-i)´(1+3i)+2i-2…………(2)
1+3i = (1-i)´(2i-2)+(i-1)………….(3)
2i-2 = 2´(i-1)+0
Hence g.c.d = d(11+7i,3+7i)
= i-1
And we know
For x and y
1+3i= (11+7i)- (1-i)´(3+7i) [
from(1)]
from(1)]
(2i-2)(1-i)=
(1+3i)-(i-1) [ from(3)]
from(3)]
Multiply (1) by 2-i
Then (11+7i)(2-i)= {(1-i)´(3+7i)+(1+3i)}(2-i)
Or, (11+7i)(2-i)= (1-i)(2-i)´(3+7i)+(1+3i)(2-i)
Or, (11+7i)(2-i)= (2-i-2i-1)(3+7i)+(1+3i)(2-i)
Or, (11+7i)(2-i)= (1-3i)(3+7i)+(1+3i)(2-i)
Or, (1+3i)(2-i)= (11+7i)(2-i)- (1-3i)(3+7i)………….(4)
Multiplying (2) by (1-i)
(3+7i)(1-i) = (2-i)(1-i)´(1+3i)+(2i-2)(1-i)
= (1-i)´(1+3i)(2-i)+ (1+3i)-(i-1)
(i-1) = (1-i){(11+7i)(2-i)-
(1-3i)(3+7i)}+ (11+7i)- (1-i)´(3+7i)-(1-i)(3+7i) [ from(1),(3)&(4)]
from(1),(3)&(4)]
=[(1-i) (2-i)+1](11+7i)-[(1-i)(1-3i)+(1-i)+(1-i)] (3+7i)
= (2-i-2i-1+1) (11+7i) -[1-3i-i-3+2-2i] (3+7i)
= (2-3i) (11+7i)+6i (3+7i)
Hence x = 2-3i
Y = 6i
1.
Find
the multiplicative inverse of
where a = 17.
where a = 17.
Solution:
Here
a=17, n= 27
Now 27= 1´17+10
17=
1´10+7
10=
1´7+3
7= 2´3+1
Now 10=27-1´17
7=
17-1´10
=
17-1´(27-1´17)
=
17-1´27+1´17
= 2´17-1´27
=
27-1´17-1´(
2´17-1´27)
=
27-1´17-2´17+1´27
= 2´27-3´17
1= 7-2´3
= 2´17-1´27-2´( 2´27-3´17)
= 2´17-1´27-4´27+6´17
= 8´17-5´27
Hence,
Multiplicative inverse of
1.
Compute the remainder when 350 is
divided by 37.
Solution:
We know
34 º 7mod37
38 º 12mod37
316º 33mod37
332º 16mod37
So 38.332 = 340º 7mod37
Again,
35º 21mod37
310º 34mod37
So 350=340.310º 16mod 37
Hence the
remainder is 16.
|
Rough part
34=81,
37)81(2
74
7
72=49, 37)49(1
37
12
122=144, 37)144(3
111
33
332=1089,
37)1089(29
74
349
333
16
|
12´16=192
37)192(5
185
7
35=243
37)243(6
222
21
212=441
37)441(11
37
71
37
34
Lastly 34´7=238
37)238(6
222
16
is the
ans.
|
2.
Does
the set of all matrices of the form G=
, 
R under matrices multiplication form a group? Justify your answer.
, R under matrices multiplication form a group? Justify your answer.
Solution:
i)
Closure property
Let A=
and B=
for all 
R
and B= for all R
Now AB= 
= 
= 
ii)
Associative
property
Let A= ,B= and C=
,B= and C=
Then A(BC)= 
= 
= 
Again
(AB)C= 
= 
= 
Hence A(BC)= (AB)C
iii)
Exixtence of
identity
For all 
and A=there
exist 
i.e I= 
=
such that
and A=there
exist i.e I= = such that
AI=
A
A
iv)
Existence of
inverse
For all A= there exist A-I= 
= 
=
=
there exist A-I= = ==
Such that A.A-I=I
Hence it forms a group under matrix multiplication.
3.
Prove
that the set G=Q-{1} is a group under binary operation * defined by a*b =
a+b-ab for all a,b
G.
G.
Solution:
i)
Closure law :
For all a,bÎG , a*b=a+b-abÎG
ii)
Associative
property
For all a,b,cÎG,
a*(b*c)=a*(b+c-bc)=a+(b+c-bc)-a(b+c-bc)=a+b+c-bc-ab-ac+abc
(a*b)*c= (a+b-ab)*c= (a+b-ab)+c-(a+b-ab).c= a+b-ab+c-ac-bc+abc
So a*(b*c)= (a*b)*c
iii)
Existence of
Identity
For all aÎG $ eÎG such that a*e=a
or, a+e-ae=a
or, e(1-a)=0
either e=0 or a=1 but a
1
1
so
e=0 ÎG.
iv)
Existence of
inverse
For all aÎG $ bÎG such that
a*b=e=0
a+b-ab=0
b(1-a)=-a
\b=
ÎG
ÎG
Hence it forms a group under * operation.
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