KS091206 KALKULUS DAN ALJABAR LINEAR Vektor di...
Transcript of KS091206 KALKULUS DAN ALJABAR LINEAR Vektor di...
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KS091206KS091206KS091206KS091206
KALKULUS DAN ALJABAR KALKULUS DAN ALJABAR KALKULUS DAN ALJABAR KALKULUS DAN ALJABAR LINEARLINEARLINEARLINEARKALKULUS DAN ALJABAR KALKULUS DAN ALJABAR KALKULUS DAN ALJABAR KALKULUS DAN ALJABAR LINEARLINEARLINEARLINEAR
Vektor di Ruang ‘N’Vektor di Ruang ‘N’Vektor di Ruang ‘N’Vektor di Ruang ‘N’
TIM KALIN
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TUJUAN INSTRUKSIONAL KHUSUS
Setelah menyelesaikan pertemuan ini
mahasiswa diharapkan :
– Dapat mengetahui definisi dan dapat
menghitung perkalian vektor di ruang n-
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menghitung perkalian vektor di ruang n-
eucledian
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Ruang-n Euclidean
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(Euclidean n-space)
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Review: Bab 3 membahas Ruang-2 dan Ruang-3
Ruang-n : himpunan yang beranggotakan vektor-vektor dengan n komponen
{ … , v = (v1, v2, v3, v4, …, vn), ….. }
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• Atribut: arah dan “panjang” / norma ||v||
• Aritmatikavektor-vektor di Ruang-n:
1. Penambahan vektor
2. Perkalian vektor dengan skalar
3. Perkalian vektor dengan vektor
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Norma sebuah vektor:Norma Euclidean (Euclidean norm) di Ruang-n :
u = (u1, u2, u3, … , un)
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||u|| = √√√√ u12 + u2
2 + u32 + … + un
2
d(u,v) = ||u-v| |= √√√√ (u1-v1)2 + (u2-v2)2 + (u3-v3)2 + … + (un-vn)2
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Example:
If u = (1, 3, -2, 7) and v = (0, 7, 2, 2) then in the Euclidean space R4.
||u|| = = =
2222 )7()2()3()1( +−++ 63 73
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And
d(u,v) = = 2222 )27()22()73()01( −+−−+−+− 58
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Penambahan vektor: di Ruang-n
u = (u1, u2 , u3, …, un); v = (v1, v2 , v3, …, vn)
w = (w1, w2 , w3, …, wn) = u + v
w = (u1, u2 , u3, …, un) + (v1, v2 , v3, …, vn)
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w = (u1 + v1, u2 + v2, u3 + v3, …, un + vn)
w1 = u1 + v1
w2 = u2 + v2
………..
w2 = un + vn
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Negasi suatu vektor:
u = (u1, u2 , u3, …, un)
– u = (– u1, – u2 , – u3, …, – un)
Selisih dua vektor:
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w = u – v = u + (– v)
= (u1 – v1, u2 – v2, u3 – v3, …, un – vn)
Vektor nol: 0 = (01, 02 , 03, …, 0n)
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Perkalian skalar dengan vektor:
w = kv = (kv1, kv2 , kv3 , …, kvn)
(w1, w2 , w3,…, wn) = (kv1, kv2 , kv3 , …, kvn)
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(w1, w2 , w3,…, wn) = (kv1, kv2 , kv3 , …, kvn)
w1= kv1
w2 = kv2
…..…
wn = kvn
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Perkalian titik: (perkalian Euclidean)
u . v = skalar
u . v = u1v1 + u2v2 + u3v3 + … + unvn
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u . v = 0 jika u dan v ortogonal
Catatan:perkalian silang hanya di Ruang-3
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Example:
The Euclidean inner product of the vectors
u = (-1, 3, 5, 7) and v = (5, -4, 7, 0) in R4 is
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u.v = (-1)(5)+(3)(-4)+(5)(7)+(7)(0) = 18
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Aritmatika vektor di Ruang-n:
Teorema 4.1.1.:u, v, w vektor-vektor di Ruang-n
k, l adalah skalar (bilangan real)
• u + v = v + u
• (u + v) + w = u + (v + w)
• u + 0 = 0 + u = u
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• u + 0 = 0 + u = u
• u + (-u) = (-u) + u = 0
• k(lu) = (kl)u
• k(u + v) = ku + kv
• (k + l) u = ku + lu
• 1u = u
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Teorema 4.1.2:
Vektor-vektor u, v, w di Ruang-n; k adalah skalar
• u . v = v . u
• u . (v + w) = u .v + u .w
• k(u . v) = (ku) . v = u . (kv)
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• v .v > 0 jika v ≠ 0
v . v = 0 jika dan hanya jikav = 0
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Example 2 Theorem 4.1.2 allows us to perform computation
with Euclidean inner products in much the same way that we
perform them with ordinary arithmetic products. For Exmple,
(3u + 2v).(4u + v) = (3u).(4u + v) + (2v).(4u + v)
= (3u).(4u) + (3u).v + (2v).(4u) + (2v).(v)
= 12(u.u) + 3(u.v) + 8(v.u) + 2(v.v)
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= 12(u.u) + 3(u.v) + 8(v.u) + 2(v.v)
= 12(u.u) + 11(u.v) + 2(v.v)
The reader should determine which parts of Theorm 4.1.2 were
used in each step
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Teorema 4.1.3 - 4.1.5:
| u . v | ≤ || u || || v ||
|| u || ≥ 0
|| u || = 0 jika dan hanya jikau = 0
|| ku || = | k | || u ||
|| u + v || ≤ || u || + || v ||
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|| u + v || ≤ || u || + || v ||
d(u, v) ≥ 0
d(u, v) = 0 jika dan hanya jikau = v
d(u, v) = d(v, u)
d(u, v) ≤ d(u, w) + d(w, v)
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kv
v u + v v
Pembuktian Bahwa ||u + v || ≤ || u || || v ||
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(a)
||kv|| = ||k||||v||
u
(b)
||u+v|| ||u|| + ||v||
≤
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Teorema 4.1.6 – 4.1.7:
u . v = ¼ || u + v || 2 – ¼ || u – v || 2
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Teorema Pythagoras
|| u + v || 2 = || u || 2 + || v || 2
u
v u + v
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Example : In the Euclidean space R4 the vectors
u = (-2, 3, 1, 4) and v = (1, 2, 0, -1)
are orthogonal, since
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are orthogonal, since
u.v = (-2)(1) + (3)(2) + (1)(0) + (4)(-1) = 0
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Perkalian Titik (dot product) dikerjakan dengan
perkalian matriksu = (u1, u2 , u3, …, un); v = (v1, v2 , v3, …, vn)
u . v = u1v1 + u2v2 + u3v3 + … + unvn
Kalau vektor u dan vektor v masing-masing ditulis dalam notasi matriks baris, maka
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baris, maka
u . v = (u1 u2 u3 … un) v1
v2
v3 u . v = (u) (v)T
vn
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u = (u1, u2 , u3, …, un); v = (v1, v2 , v3, …, vn)
u . v = u1v1 + u2v2 + u3v3 + … + unvn
Kalau vektor u dan vektor v masing-masing ditulis dalam notasi matriks kolom, maka
u . v = (u1 u2 u3 … un) v1 = v . u
v2
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v3 u . v = (u)T (v)
v . u = (v)T (u)
vn u . v = (v)T (u)
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Matriks A (n x n), u dan v masing-masing vektor kolom
Au . v = u . (ATv) ?
(Au) . v = v . (Au) perkalian titik bersifat komutatif (T.4.1.2)
= vT(Au) v vektor kolom
= (vTA)u perkalian matriks asosiatif (T.1.4.1)
= (ATv)T u (MN)T = NTMT & (MT)T = M (T.1.4.9)
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= (ATv)T u (MN)T = NTMT & (MT)T = M (T.1.4.9)
ATv merupakan vektor kolom; maka (ATv)T vektor baris
= u . (ATv) persamaan (7)
Jadi Au . v = u . (ATv) terbukti
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Diketahui matriks A (n x n), u dan v masing-masing vektor kolom
u . Av = ATu . v ?
u . (Av) = (Av) . u perkalian titik bersifat komutatif
= v . (ATu) baru dibuktikan
di slide sebelum ini: Au . v = u . (ATv)
= (A u) . v perkalian titik komutatif
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= (ATu) . v perkalian titik komutatif
Jadi: u . Av = ATu . v terbukti
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Contoh:
• Hitunglah eucledian norm dari vektor berikut :
– (-2,5)
– (1,-2,2)
– (3,4,0,-12)
– (-2,1,1,-3,4)
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• Hitunglah eucledian inner product u.v
– u = (1,-2) , v = (2,1)
– u = (0,-2,1,1), v = (-3,2,4,4)
– u = (2,-2,2), v = (0,4,-2)
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Let’s Try !
1. Carilah semua skalar k sehingga ||kv||=3, di mana v = (-1, 2, 0, 3).
2. Carilah jarak euclidis di antara u dan v bila u=(6,0,1,3,0) dan
v=(-1,4,2,8,3)
3. u=(3,0,1,2), v=(-1,2,7,-3), dan w=(2,0,1,1). Carilah:
a. ||-2u|| + 2||u|| b. c.
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4. u1=(-1,3,2,0), u2=(2,0,4,-1), u3=(7,1,1,4), dan u4=(6,3,1,2). Carilah skalar
c1,c2,c3, dan c4 sehingga c1u1+c2u2+c3u3+c4u4=(0,5,6,-3)
5. Buktikanlah identitas berikut:
||u+v||2 + ||u-v||2 = 2||u||2 + 2||v||2.
Tafsirkanlah hasil ini secara geometris pada R2.
6. Buktikanlah identitas berikut:
u.v = ¼||u+v||2 - ¼||u-v||2
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