Tugas i Aljabar Linier d

7/23/2019 Tugas i Aljabar Linier d

http://slidepdf.com/reader/full/tugas-i-aljabar-linier-d 1/4

TUGAS I ALJABAR LINIER

OLEH:

NAMA : HARTINA HUSAIN

NIM : H12113005

KELAS : D

PRODI STATISTIKA

JURUSAN MATEMATIKA

FAKULTAS MATEMATIKA DAN ILMU PENGETAHUAN ALAM

UNIVERSITAS HASANUDDIN

2014



Soal

1. M!al"a# P3 a$ala% %&'(#a# 'ol#o& )*+$*+a,a- 3 a#/ -*+-(-('

-*+%a$a' o'*+a! -a&)a% $a# o'*+a! '*+"ala# !ala+. P*+"!a

a'a"a% %&'(#a# 'ol#o& )*+)*#-(" a0a2a22a33 &*+('a"a#

!() +(a#/ $a+ P3.

Jaa) :M!al"a# a0a2a22a33 &*+('a"a# !()%&'(#a# $a+

!*)(a% P3 &a"a $#a&a"a# !()+(a#/6subspace 7 P3.Sa+a- a0a2a22a33 $!*)(- !()+(a#/ $a+ P3 a$ala% :

1. W ≠ 8 92. W ⊆ P3

W vu ∈, W vu ∈+

3. J"a &a"a

W uk ∈ W u ∈

4. J"a $a# k ∈ Rl &a"a

Bukti :1. )("a# %&'(#a# "o!o#/ "a+*#a

∀ a0

, a2

, a3∈W dimana a

0, a

2, a

3 ∈ R . o#-o%#a 2332;3.

2. J*la! )a%a W ⊆ P3

3. A&)l !*&)a+a#/ (<= ∈W

M!al"a# u= b0+b2 x+b2 x 2+b3 x 3 $a# v = c0+c2 x+c2 x 2+c3 x 3

u+v = (b0+b2 x+b2 x 2+b3 x 3 ) + (c0+c2 x+c2 x 2+c3 x 3 ) = b0+c0 + b2 x+ c2 x + b2 x 2+ c2 x 2+ b3 x 3+ c3 x 3 = (b0+c0 ) + (b2+ c2 )x + (b2+ c2 )x 2 + (b3+ c3 )x 3

Da'a- $l%a- )a%a (= ∈

W u ∈

4. A&)l !*&)a+a#/ k ∈ Rl :

M!al"a# ( )0)2)22)33

k.u = k (b0+b2 x+b2 x 2+b3 x 3 )W uk ∈

= kb0+kb2 x+kb2 x 2+kb3 x 3

I# &*#(#,(""a# )a%aS*%#//a $a'a- $!&'(l"a# )a%a W= a0+a2 x+a2 x

2+a3 x

3

merupakan subruang dari P3.



2. M!al"a# $"*-a%( &a-+"! 22:D&a#a :

A [3 2

0 1] > [ 0 2

−2 4 ] [ 1 1

−2 5 ] D

[ 4 5

−2 10]A'a"a% &a-+"! D $a'a- $-(l! !*)a/a "o&)#a! l#*+ $a+ A<><

$a# ? Jaa) :

Ma-+"! D $"a-a"a# "o&)#a! l#*+ $a+ A<> $a# ,"a &*&*#(% :

D22 "1.A22"2.>22"3.22

D&a#a "1<"2<$a# "3 a$ala% "o#!-a#-a. A"a# $'*+"!a a'a"a% a$a

"1<"2<$a# "3 !*%#//a "*!a&aa# -*+!*)(- -*+'*#(%.

D22 "1 . [3 2

0 1] "2 . [ 0 2

−2 4 ] "3 . [ 1 1

−2 5 ]

[ 4 5

−2 10] [3 k 1 2k 1

0 k 1 ] [

0 2 k 2

−2 k 2

4 k 2] [

k 3 k 3

−2 k 3

5 k 3]

[ 4 5

−2 10] [ 3k 1+k

3 2k

1+2k

2+k

3

−2k 2−2k

3 k

1+4k

2+5k

3]

S*%#//a $ '*+ol*% '*+!a&aa#@'*+!a&aa# :3k

1+k

3=4

2 k 1+2 k

2+k

3=5

−2 k 2−2 k

3=−2

k 1+4 k

2+5 k

3=10

A-a( $a'a- ,(/a $-(l! !*'*+-:



[3

2

0

0 1

2 1

−2 −2

1 4 5 ]

[k

1

k 2k

3]

[ 4

5

−2

10 ]D*#/a# O>E $a'a- "-a '*+ol*% :

[ 3 0 1 4

2 2

0 −2

1 4

1 5

−2 −2

5 10] B

1( 1

3 )[ 1 0 1/3 4 /3

2 2

0 −2

1 4

1 5

−2 −2

5 10]

B21(−2 )

B41

(−1 )[ 1 0 1/3 4 /3

0 2

0 −2

0 4

1/3 7 /3

−2 −2

14 /3 26 /3] B2(

1

2) [

1 0 1/3 4 /3

0 1

0 −2

0 4

1/6 7/6

−2 −2

14 /3 26 /3 ]B32(2)

B42(−4)[

1 0 1/3 4/3

0 1

0 0

0 0

1/6 7 /6

−5/3 1/3

4 4] B

3(−3

5) [

1 0 1/3 4/3

0 1

0 0

0 0

1/6 7 /6

1 −1/5

4 4]

B43(−4)[

1 0 1/3 4 /3

0 1

0 0

0 0

1/6 7 /6

1 −1/5

0 24/5]

Ka+*#a >a+! -*+a"%+ 'a$a &a-+"! # &*#(#,(""a# )a%a S!-*&

'*+!a&aa# l#*a+ 6SPL7 -*+!*)(- a$ala% -$a" "o#!!-*# 6-$a"

&*&'(#a !ol(! 7. Ja$< -$a" a$a #la "1< "2< $a# "3 a#/

&*&*#(% !*%#//a tidak dapat din!atakan sebagaik"mbinasi #inear dari A$B dan %.

Tugas i Aljabar Linier d

Documents

Transcript of Tugas i Aljabar Linier d