MODUL MATA KULIAH ALJABAR LINEAR
Disusun Oleh:
Candra Mecca Sufyana
1
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Daftar Isi Daftar isi...................... ................................... ......................... ......................... ......................... .......................... ........................... ................... ...... 3 Daftar Gambar.................... ................................. ........................... ........................... ......................... ......................... ...................... ......... 5 ................................... ........................... ......................... ......................... ........................... ......................... ............ 6 Daftar Tabel...................... Deskripsi Mata Kuliah………….……………………………….………….. 7 Tujuan Mata Kuliah Umum...….……………………………….………….. 7 Tujuan Mata Kuliah Khusus.….……………………………….………….. 7 BAB I
PENDAHULUAN 1.1 1.2
BAB II
Diagram alur pokok bahasan............................. bahasan................................................ ................... 9 Materi pokok aljabar linear........................................... linear................................................... ........ 9
SISTEM PERSAMAAN LINEAR 2.1 Sistem Persamaan Linear Linear Dua Variabel ………………......... ………………......... 13 2.1.1 Metode grafik………………………………………….. 13 2.1.2 Metode eliminasi………………………………………. 15 2.1.3 Metode subsitusi……………………………………….. 16 2.1.4 Metode Campuran………………………………………17 2.2 Metode Eliminasi Gauss......................…........ Gauss......................…............................... ....................... .20 2.3 Sistem Persamaan Linear Homogen……….………............. Homogen……….………............... 23
BAB III
MATRIKS
3.1 Definisi Definis i matriks… matriks … ……………………….………………... 3.2 Jenis-jenis Jenis-j enis matriks………………………………..………… matriks ………………………………..………… 3.3 Operasi Matriks…..…………………………….……….…… 3.4 Matriks Elementer..…………………………….……….…… 3.5 Invers Matriks…..…………………………….……….……..
BAB IV
27 28 29 30 31
DETERMINAN MATRIKS
4.1 Definisi determinan determinan matriks……………….………… matriks……………….………………... ……... 34 4.2 Sifat-sifat determinan matriks……………………..…… matriks……………………..………… …… 35 4.3 Minor dan Kofaktor…………………………….……….…… 35 4.4 Reduksi Baris..…………………………….……….………… 37 4.5 Aturan Cramer…..…………………………….……….…….. 38
BAB V
VEKTOR 5.1 Persamaan dua buah vektor 5.1.1 Penjumlahan vektor...………………………………….. 41 5.1.2 Pengurangan vektor……………………………………. 43
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5.1.2 Hasil kali silang…………………………………………. 47 5.3 Persamaan Bidang…………………..…….………. Bidang…………………..…….………......... ................ .......... 48 5.4 Jarak titik terhadap bidang....………..… bidang....………..…….………......... ….………..................48 .........48
BAB VI
RUANG VEKTOR
6.1 Ruang vektor umum…...……………………………………... 51 6.2 Ruang Lingkup Vektor 6.1.1 Vektor di ruang 2D......…………………………………..52 6.1.2 Vektor di ruang 3D …………..…………………………. 53 6.3 Sub Ruang…………………………..…………………………54 6.4 Vektor Bebas Linear dan Tak Linear.……… Linear.…………………………54 …………………54 6.5 Kombinasi Linear Linear …………………..…………………………55 6.6 Dimensi dan Basis…………………..…………………………56 6.7 Row Space, Coloumn Space , Null Space ……………………57 6.8 Rank dan Nullity..…………………..…………………………58 Nulli ty..…………………..…………………………58 6.9 Ruang Hasil Kali Dalam..…………..…………………………60 Dalam..…………. .…………………………60 6.10Basis Ortonormal.…………………..…………………………61
BAB VII TRANSFORMASI LINEAR 7.1 7.2 7.3 7.4
Transformasi Transform asi Linear…...……………………………………... Linear…... ……………………………………... 64 Jenis-jenis transformasi transformasi linear bidang............................ bidang....................................... ........... 65 Kernel dan Jangkauan..…………….. Jangkauan..……………..…………………………67 …………………………67 Nilai dan Vektor Eigen......................……… Eigen......................……………………… …………………68 …68
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Daftar Gambar Gambar 1.1 2.1 2.2 2.3 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 6.1 6.2 6.3
Diagram alur pokok bahasan Koordinat kartesius SPL 2 variabel Kemungkinan Solusi Sistem Persamaan Linear
Bagan SPL Homogen Empat buah vector yang dikatakan sama Perpindahan A dan B Penjumlahan 2 buah vektor Penjumlahan 4 buah vector Sifat Asosiatif Penjumlahan Vektor Pengurangan Vektor Komponen-komponen vector Vektor bergantung kuadran Vektor Satuan Metode komponen vector Perkalian dua vector Ruang vector 2D Vektor a=OA=5i+3j Ruang vector 3D
Halaman 9 13 19 24 41 42 42 42 43 43 44 44 45 45 46 52 53 54
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Daftar Tabel Tabel 3.1 4.1
Keterangan Contoh tabel sebagai kumpulan array Permutasi
Halaman 27 35
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A. Deskripsi Mata Kuliah Secara garis besar, mata kuliah ini akan membicarakan tentang pengertian matriks, operasi dasar matriks, dan jenis-jenis matriks, determinan, operasi baris elementer (OBE) dan operasi kolom elementer (OKE), matriks ekivalen, matriks invers dan sifat-sifatnya, sistem persamaan linear, beberapa aplikasi matriks, ruang vektor, basis dan dimensi, transformasi linear, ruang inner product , eigen vektor dan eigenvalues ei genvalues.
B. Tujuan Kompetensi Umum Mata kuliah ini mempersiapkan mahasiswa untuk dapat menyelesaikan masalah yang terkait dengan aljabar matriks dan konsep ruang vektor. Disamping itu, mata kuliah ini membekali mahasiswa dengan logical reasoning dan abstraksi matematika. Oleh karena itu, keterlibatan aktif dari siswa memegang peranan penting.
C. Tujuan Kompetensi Khusus Sesuai dengan tujuan pemelajaran Matematika, perkulaihan Aljabar Linier mempunyai dua tujuan utama yang saling terkait yaitu mengasah kemampuan bernalar dan problem solving. Secara rinci, tujuan tersebut dijabarkan dalam sasaran pemelajaran berikut ini: Sasaran pembelajaran terminal 1. Apabila diberi suatu sistem persamaan linier, mahasiswa mampu memilih strategi yang paling efektif untuk menentukan penyelesaiannya atau menetukan penyelesaian kuadrat terkecil (LSS). 2. Mahasiswa mampu mengidentifikasi matriks-matriks persegi yang dapat didiagonalkan secara orthogonal, dan dapat membuat prosedur untuk mendiagonalkannya. 3. Mahasiswa mampu melakukan melakukan operasi-operasi vektor vektor pada bidang dan dan ruang (ruang 2 3 vektor Euclid R dan R ) baik secara aljabar maupun geometris. 4. Mahasiswa mampu mengidentifikasi apakah suatu fungsi merupakan transformasi linier, mampu menentukan matriks transformasi linier, dan dapat menginterpretasikan menginterpretasikan sifat-sifat transformasi linier pada bidang dan ruang.
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BAB I PENDAHULUAN
A. Tujuan Kompetensi Khusus
• Mahasiswa mengetahui dan • •
memahami definisi, tujuan, dan maksud dari pembelajaran materi aljabar linear. Mahasiswa mengetahui mengetahui diagram alur pokok bahasan aljabar linear Mahasiswa mengetahui kaitan keilmuan aljabar linear dengan mata kuliah lain dan manfaatnya dalam kasus informatika
B. Uraian Materi Ketika mahasiswa mengambil mata kuliah aljabar linear, yang paling seru adalah tidak menanyakan menanyakan materi apa saja yang akan dipelajari dipelajari di mata kuliah aljabar linear tersebut, tersebut, karena secara otomastis menyangka ga akan jauh berbeda dengan matematika. Sudah menjadi rahasia umum bahwa matematika merupakan momok menakutkan bagi sebagaian besar pelajar. Diantaranya karena matematika atau aljabar secara umum menggunakan dummy variabel berbentuk huruf (misalnya x, y, z, p, q, m, n) yang umumnya sulit untuk dapat dipahami. Pola pikir “menyatakan sulit sebelum berperang” tersebut juga yang justru menjadi masalah utamanya, tanpa melihat terlebih dahulu manfaat, tantangan, dan keilmuanya. Aljabar Linear Elementer yang termasuk salah satu cabang ilmu Matematika, memiliki karakteristik tersendiri dalam isi maupun proses pembelajarannya. Aljabar Linear Elementer (ALE) ini identik dengan penggunaan logika dalam pemecahan masalah dan pencarian solusi. Pembelajaran ALE yang harus melalui pra syarat penguasaan Logika Matematika & Himpunan tentunya tidak lepas dari aplikasi penggunaannya dalam kehidupan sehari-hari, baik secara konkret maupun abstrak. Aljabar linear ini mempelajari tentang matriks, vektor, ruang vektor, transformasi linear dan sistem persamaan linear. Aljabar linear mempunyai penerapan pada berbagai bidang ilmu alam dan ilmu sosial serta teknologi khususnya teknologi informasi dan komunikasi (infokom) yang saat ini sedang berkembang pesat. Pada dasarnya aljabar linier adalah metode yang sangat kuat ketika berhadapan dengan beberapa variabel, dan ada manfaat yang luar biasa untuk menggunakan ini sebagai landasan teoritis ketika merancang algoritma.
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1.1 Diagram alur pokok bahasan bahasan
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5.
6. 7. 8. 9.
Determinan, meliputi determinan dengan ekspansi kofaktor, Sifat-sifat determinan terhadap Operasi Baris Elementer, Matrik Adjoin, Matrik Invers dengan Matrik Adjoin, Aturan Cramer Ruang Vektor, meliputi Ruang n Euclides, Definisi Ruang Vektor, Sub Ruang, Bebas Linier, Membangun, Basis, dan Dimensi Ruang Hasil Kali Dalam, meliputi Definisi, Panjang dan Sudut di Ruang Hasil Kali Dalam, Ortonormalisasi Basis Nilai dan Vektor Eigen, meliputi Persamaan Karakteristik, Diagonalisasi, dan Diagonalisasi secara Ortogonal Transformasi Linier, meliputi Definisi, Kernel, Rank, Koordinat sebagai bentuk Transformasi dari Ruang vektor sebarang ke Rn, Matrik Transformasi
C. Rangkuman Aljabar linear adalah bidang studi matematika yang mempelajari sistem persamaan linear dan solusinya, vektor, serta transformasi linear. Matriks dan operasinya juga merupakan hal yang berkaitan erat dengan bidang aljabar linear.
D. Tugas Dibawah ini adalah berbagai komentar tentang bagaimana aljabar linear digunakan dalam algoritma. Berikan komentar dan analisis anda terhadap komentar-komentar dan pertanyaan tersebut
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E. Evaluasi 1. Tuliskan bagaimana sejarah keilmuan aljabar? 2. Sebutkan apa saja manfaat dan keuntungan mempelajari aljabar linear? 3. Sebutkan materi apa saja yang dipelajari di aljabar linear dan relevansinya terhadap manfaat?
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BAB II SISTEM PERSAMAAN LINEAR
A. Tujuan Kompetensi Khusus Apabila diberikan sistem persamaan linier (SPL) konsisten berukuran kecil, mahasiswa mampu menetukan konsistensinya; dan menyelesaikan dengan metode eliminasi-substitusi, geometris, dan metode Gauss-Jordan dengan tepat.
B. Uraian Materi Sistem persamaan linear adalah dua persamaan linear atau lebih yang disajikan bersamaan dan mempunyai satu jawaban persekutuan. Pasangan sistem persamaan yang dibentuk dapat berupa linear dan linear, linear dan kuadrat, atau kuadrat-kuadrat. Dalam kehidupan sehari-hari sering kita temui persoalan-persoalan yang dapat diselesaikan dengan memakai model matematika yang berbentuk sistem persamaan linear. Model matematika adalah cara mengubah bentuk penulisan dari bahasa sehari-hari menjadi bahasa matematika. Misalnya: Anto membeli alat tulis untuk keperluan sekolah yaitu 3 buah pulpen dan 2 buah pensil dengan harga Rp. 10.500,00. Pada toko yang sama Budi membeli 2 buah pulpen dan 3 buah pensil dengan dengan harga Rp. Rp. 9.500,00 .
Harga masing-masing 1 buah pulpen dan 1 buah pensil dapat anda ketahui dengan memakai model matematika yang berbentuk sistem persamaan Linear. Dari contoh diatas terdapat Dua Variabel yaitu pulpen dan pensil yang akan diperoleh penyelesaian tunggal dari nilai x dan y. Jadi penyelesian Sistem Persamaan linear adalah nilai x dan y yang memenuhi kedua persamaan linear yang dimaksud. Penulisannya ditulis dalam bentuk Himpunan Penyelesaian Penyelesaian (HP) : {(x,y)} Ada tiga kemungkinan untuk menentukan himpunan penyelesaian, penyelesaian, yaitu : • Sistem persamaan linear akan memiliki penyelesaian jika dipenuhi syarat : (a/p) ≠ (b/q). • Sistem persamaan linear tidak akan memiliki penyelesaian jika penyelesaian jika dipenuhi syarat : (a/p) = (b/q) ≠(c/r).
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2.1.
Sistem Persamaan Linear Dua Variabel
2.1.1
Metode Grafik Penyelesaian dengan metode metode grafik grafik secara secara umum adala adala menggambar kedua persamaan garis pada satu koordinat Cartesius. Adapun l ngkah-langkah secara lengkap adalah s bagai berikut: • Buatla Buatlah h tabe tabel pasangan terurut (x,y) dengan mencari titik potong dengan masing-masing su sumbu X dan Su Sumbu Y dari se setiap iap pe persa aan garis. • Perpotongan sumbu X diperoleh pada saat nilai y = 0 dan perpotongan dengan sumbu Y di eroleh pada saat nilai x = 0. Jadi perpotongan dengan sumbu koordinat adalah : Perpotongan den gan Sumbu Sumbu X : (a,0) (a,0) dan dan Perpoton Perpotongan gan denga denga Sumbu Y : ( 0,b) Karena ada dua persa persamaa maan n garis garis maka maka anda anda harus harus memb membu u t dua tabel dan akan diperoleh empat titik (a,0), (0,b) dan (c,0), (0,d). Lukislah masing masi masing ng persa rsamaa maan pa pada satu satu koord oordin inaat Ca Cart sius !
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Jawab : Persamaan (1) : x + 2y = 60000 Perpotongan den gan Sumbu X (y = 0) 3x + 2y = 6 000 3x = 6 000 x = 2 000, Diperoleh titik (20000,0) Perpotongan dengan Sumbu Y (x = 0) 3x + 2y = 60 00 2y = 30 00, Diperoleh Diperoleh titik ( 0,30000) 0,30000) Jadi perpotongan dengan sumbu koordinat adalah adalah : (20000,0) (20000,0) ( 0,30000),
3x + 2 y = 60000
Persamaan (2) : 5x + y = 65000 Perpotongan den gan Sumbu X (y = 0) 5x + y = 65 00 5x + y = 65 00
X
0
20000
Y
30000
0
(0,30000)
(20000,0)
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Dari pasangan titik (20000,0), ( 0,30000), dan (13000,0), ( 0,65000) maka akan diperoleh dua garis pada satu sumbu koordinat. Dari kedua garis tersebut nampak bahwa ada perpotongan antara keduanya sehingga terdapat satu penyelesaian sistem persamaan linear yaitu titik (10000,15000) (10000,15000) harga tiap kg apel Rp. 10000 dan anggur Rp.15000 2.1.2
Metode Eliminasi Metode Eliminasi Eliminasi adalah cara penyelesaian sistem persaman linear dengan menghilangkan salah satu variabel untuk mencari nilai variabel yang lain. Adapun langkah-langkah secara lengkap adalah sebagai berikut : Untuk mengeliminasi suatu variabel samakan nilai kedua koefisien variabel yang akan dihilangkan. Pada langkah ini anda mengalikan kedua koefisien dengan bilangan tertentu sedemikian sehingga nilai koefisiennya menjadi sama Misalkan pada bentuk umum, anda akan menghilangkan variabel x, maka anda harus mengalikan koefisien variabel x pada kedua persamaan dengan p untuk persaman pertama dan mengalikan dengan a untuk persamaan kedua ax +by = c X p → apx + bpy = cp px + qy = r X a → apx + aqy = ar – (bp-aq) y = cp – ar y = (cp-ar)/(bp-aq) Pada langkah di atas anda akan mendapatkan sebuah persamaan Linear, dan anda akan dapat menghitung nilai variabel pertama yaitu y dengan mudah. Setelah anda menemukan nilai variabel y sekarang akan menghitung nilai variabel x, maka anda harus menghilangkan variabel y, dengan mengalikan koefisien variabel y pada kedua persamaan dengan q untuk persaman pertama dan mengalikan dengan b untuk persamaan kedua ax +by = c X q → aqx + bqy = cq
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10x + 3y = 100 ……………….. (2) Untuk mengeliminasi mengeliminasi variable x samakan nilai kedua koefisien koefisien variable x. Pada langkah ini anda mengalikan kedua koefisien dengan bilangan tertentu sedemikian sehingga nilai koefisiennya menjadi sama. 50x + 40y = 1000 | X 1 | 50x + 40y = 1000 10x + 3y = 100 | X 5 | 50x + 15y = 500 25y = 500 y = 500/25 y = 20 Karena variabel yang akan dieliminasi mempunyai koefisien tanda sama maka untuk menghilangkan variabel x, kedua persamaan harus dikurangkan. Untuk mengeliminasi mengeliminasi variabel y samakan nilai kedua koefisien koefisien variable y. Pada langkah ini anda mengalikan kedua koefisien dengan bilangan tertentu sedemikian sehingga nilai koefisiennya menjadi sama. 50x + 40y = 1000 X 3 >> 150x + 120y = 3000 10x + 3y = 100 X 40 >> 400x + 120y = 20000 -250x + 0y = -17000 -17000 x = -17000/-250 x = 38 Karena variabel yang akan dieliminasi mempunyai koefisien tanda sama maka untuk menghilangkan variabel y, kedua persamaan harus dikurangkan. Dari perhitungan di atas anda memperoleh nilai variabel x = 38 dan nilai variabel y = 20. Jadi Himpunan Penyelesaian : {(38,20)} Hal ini berarti bahwa banyaknya pesawat yang dibutuhkan untuk mengangkut semua tentara dan semua perlengkapan dalam satu kali pemberangkatan pasukan adalah 38 pesawat Hercules dan 20 pesawat Helikopter.
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nilai variabel y tersebut pada persamaan (3). Dari keterangan di atas maka anda dapat menemukan pasangan (x,y) yang merupakan penyelesaian dari sistem persamaan linear tersebut. Contoh: Anda membeli alat tulis untuk keperluan sekolah yaitu 3 buah pulpen dan 2 buah pensil dengan harga Rp. 10.500,00. Pada toko yang sama teman anda membeli 2 buah pulpen dan 3 buah pensil dengan harga Rp. 9.500,00. Bagaimana menghitung harga tiap 1 buah pulpen dan pensil ? Coba anda diskusikan ! Misalkan x = 1 y = 1 Anda membeli 3 buah pulpen dan 2 buah pensil dengan harga Rp. 10.500,00 3 buah pulpen + 2 buah pensil = Rp. 10.500,00 3x + 2y = 10500 ………………. (1) Teman anda membeli 2 buah pulpen dan 3 buah pensil dengan harga Rp. 9.500,00 2 buah pulpen + 3 buah pensil = Rp. 9.500,00 2x + 3y = 9500 …………………. (2) Untuk mengganti (subsitusi) variabel x dengan variabel y, ubahlah salah satu persamaan menjadi persamaan x dalam y. Kemudian gantikan hasil tersebut pada persamaan yang yang lain. Pada langkah langkah ini anda mengubah persamaan pertama (1) menjadi persamaan x dalam y, 3x + 2y = 10500 3x = -2y + 10500 x = -(2/3)y + 10500/3 x = -(2/3)y -(2/3) y + 3500 ……………… (3) Dari persamaan (2) dan (3) 2x + 3y = 9500 2{-(2/3)y + 3500} + 3y = 9500 -(4/3)y + 7000 + 3y = 9500
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Kemudian nilai variabel y ini disubsitusikan ke dalam salah satu persamaan sehingga diperoleh nilai variabel yang lain. px + qy = r px + q{(cp-ar)/(bp-aq)} = r Disini anda akan memperoleh nilai variabel x. Jadi anda akan mendapatkan pasangan (x,y) dengan dua metode yaitu eliminasi dan substitusi. Metode yang digunakan terlebih dahulu sangat tergantung pada soal yang disajikan, akan tetapi biasanya digunakan terlebih dahulu metode eliminasi baru kemudian metode substitusi. Contoh: Tentukan penyelesaian penyelesaian dari : 2/x + 3/y = 5 dan 3/x – 4/y = 16 Jawab : 2/x + 3/y = 5 ………. (1) 3/x – 4/y = 16 ………. (2) Gunakan Metode Campuran !! Metode Eliminasi kemudian Substitusi !! Dengan metode campuran : Langkah pertama dengan metode eliminasi : 2/x + 3/y = 5 X 3 >> 6/x + 9/y = 15
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dimana: aij untuk i=1 s/d m dan j=1 s/d n adalah koefisien atau persamaan persamaan simultan xi untuk i=1 s/d n adalah variabel bebas pada persamaan simultan Solusi Sistem Persamaan Linier adalah solusi setiap persamaan linier yang terdapat dalam Sistem Persamaan Linier tersebut. Penyelesaian persamaan linier adalah penentuan nilai xi untuk semua i=1 s/d n yang memenuhi semua persamaan yang diberikan. Permasalahan persamaan linier simultan merupakan permasalahan yang banyak muncul ketika berhubungan dengan permasalahan multi-variabel dimana setiap persamaan merupakan bentuk persamaan linier atau dengan kata lain setiap variabel berpangkat paling besar satu. Persamaan linier simultan di atas dapat dinyatakan sebagai bentuk matrik yaitu :
Matrik A dinamakan dengan Matrik Koefisien dari persamaan linier simultan, atau ada
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2.1 Me Metode Elimi asi Gauss Meto Metode de Elim Elimin inaasi Gauss merupakan metode yang dikem bangkan dari metode eliminasi, yaitu menghilangkan atau mengurangi ju jumlah v riabel sehingga dapat diperoleh nilai ari ari sua suatu tu varia variabl blee beba bebas. s. Untu Untuk k meng menggu guna nakan metode eliminasi Gaus Gausss ini, ini, terl terleb eb h dahulu bentuk matrik diubah menjadi aug ented matrik sebagai berikut : a11 a12 ... a1n b1
a21 a22 ... a2 n ... ... ... ...
b2
...
b
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Sebagai contoh reduksi baris, kita kit a selesaikan kumpulan persamaan berikut, 2 x – z = 2 6 x + 5 y + 3 z = 7 2 x – y = 4 Persamaannya 2 x – z = 2 6 x + 5 y + 3 z = 7 2 x – y = 4 (a) Bersamaan pertama tetap; persamaan dua dikurangi 3 kali persamaan pertama diperoleh persamaan kedua baru; persamaan ketiga di – kurangi persamaan pertama di dapatkan per – samaan ketiga yang baru 2 x - z = 2 5 y + 6 z = 1
hubungannya hubungannya dengan matriks
(a) Baris kedua dikurangi 3 kali ba - ris pertama dan baris ketga dikura- baris pertama
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Contoh. Kasus empat variabel. Selesaikan susunan persamaan berikut:
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2.2 Sistem Persamaan Linear Homogen Sistem Persamaan Linier Homogen adalah Sistem Persamaan Linier yang semua suku konstannya nol, sehingga bentuk umum SPL homogen, sebagai berikut:
a 22 x 2
a31 x1
+ + +
...
...
a11 x1 a 21 x1
...
a 23 x3
a32 x 2
+ + +
a33 x3
+ ... + + ... + + ... +
...
...
...
... ... ...
a12
x 2
...
a13
x3
...
a3n x n
= 0 = 0 = 0
...
... ...
a1n
x n
a 2 n x n
...
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C. Rangkuman 1. Sistem persamaan linear adalah dua persamaan linear atau lebih yang disajikan bersamaan dan mempunyai satu jawaban atau solusi persekutuan. 2. Solusi Persamaan Linier yang dimaksud adalah sehimpunan bilangan terurut yang jika disubtitusikan kedalam kedalam Persamaan Persamaan Linier tersebut, menjadi menjadi valid. 3. Model matematika adalah cara mengubah bentuk penulisan dari bahasa sehari-hari menjadi bahasa matematika. 4. Metode Penyelesaian untuk kasus dua variabel
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D. Tugas Coba Coba Jela Jelask skan an dala dala
sebu sebuah ah tuli tulisa san n untu untuk k melu meluki kisk skan an keti ketiga ga dia dia ram di bawah ini!
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E. Evaluasi 1.
2.
3.
Tentukan Himpunan penyelesaian penyelesaian dari sistem persamaan berikut : a. 2x + y = 8 dan x - y = 1 b. 2x - 5y = 15 dan 3x + 4y = 11 c. x + 3y = 1 dan 2x - y = 9 Sebuah bilangan bilangan terdiri dari dua angka, penjumlahan tiga angka angka puluhan dan angka angka satuannya adalah 27, dan selisihnya angka puluhan dann satuannya satuannya adalah adalah 5. Bilangan itu adalah .... Suatu perusahaan perusahaan garmen memproduksi memproduksi dua jenis pakaian A dan B. Jumlah yang
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BAB III MATRIKS
A. Tujuan Kompete si Khusus Apabila diberikan m triks-matriks, ma mahasiswa ma mampu me mela ukan operasioperasi aritmetika dengan tepat, dan dan mam mampu pu mene menent ntuk ukan an inve invers rsee mat matri riks ks per per segi secara efektif.
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E. Evaluasi
1. 2. 3. 4.
5.
Diketahui a = ( 1,2 ) , b = ( –2,–3 ) dan c = ( 1,3 ) Apakah c merupakan kombinasi kombinasi linier dari a dan b ? Tunjukkan bahwa v =(3,9,-4,-2) merupakan kombinasi linier u1= (1,-2,0,3), u2 = (2,3,0,-1) dan u3= (2,-1,2,1) Apakah u = ( 1,2,3 ) , v = ( 2,4,6 ) dan w = ( 3,4,7 ) membangun R3 Diketahui u = ( 1,2 ) , v = ( 2,2 ) , w = ( 1,3 ) a. Apakah u , v dan w membangun R2 ? b. Apakah u , v dan w bebas linier ? Misal v1=(1,2,1), v2=(2,9,0), dan v3=(3,3,4).Tunjukkan bahwa S=(v1,v2,v3) adalah basis untuk R3
himpunan
2 1 0 − 3 1 4 a. Vektor baris A adalah… b. Vektor kolom A adalah…
F. Pustaka
Anton, Howard. Dasar-dasar Aljabar Linear Jilid 2 Edisi 7 . 2000. Penerbit Interaksara. Jakarta Kartika Firdausy – UAD blog.uad.ac.id/kartikaf blog.uad.ac.id/kartikaf
BAB VII TRANSFORMASI LINEAR A. Tujuan Kompetensi Khusus Mahasiswa mampu mengidentifikasi apakah suatu fungsi merupakan transformasi linier, mampu menentukan matriks transformasi linier, dan dapat menginterpretasikan sifat-sifat transformasi linier pada bidang dan ruang, dan dapat menjelaskan ruang eigen sebagai
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W adalah Jika F:V W adalah sebuah fungsi dari ruang vektor V ke ke dalam ruang vektor W , maka F disebut transformasi linier, jika : (i). F(u+v F(u+v)) = F(u F(u) + F(v F(v), untuk semua vektor u dan v di V (ii). F(k u) = kF(u kF(u) untuk semua vektor u di dalam V dan dan semua skalar k Contoh 2 3 R adalah sebuah fungsi yang didefinisikan oleh : Misal F:R R x, x+y, x-y) F(v F(v) = ( x, x1 + x2 , y1 + y2) v = ( x Jika u=( x1 , y1) dan v=( x2 , y2) maka u + v = Sehingga , F(u + v) x1 + x2 , [ x x1 + x2]+[ y1 + y2], [ x x1 + x2]-[ y1 + y2]) = ( x x1 , x1 + y1 , x1 - y1) + ( x x2 , x2 + y2 , x2 – y2) = ( x = F(u) + F(v) x1, k y1) sehingga Demikian juga jika k adalah sebuah skalar, k u = (k F(k u) = (k x1 , k x1 + k y1 , k x1 - k y1) = k( x x1 , x1 + y1 , x1 - y1) = k F(u) Jadi F adalah sebuah transformasi linier n
m
7.1 Transformasi Linier dari R R n R dan misalkan A adalah sebuah matrik m Misalkan e1, e2, . . . , e n adalah basis baku untuk R x n yang mempunyai T( e1), T(e2), . . . , T(e n) sebagai vektor – vektor kolomnya. 2
2
Misal jika T:R R diberikan oleh :
x1 x1 + 2 x 2 T = x 2 x1 − x 2 Maka
1 T(e1) = T = 0 1 A = 1
Jadi
2
− 1
1 1
0 2 dan T(e2) = T = 1 − 1
adalah matrik baku untuk T di atas.
3
Misalkan T:R → W adalah transformasi matriks, dan misalkan
1 0 0 3 1 4 T 0 = , T 1 = , T 0 = 0 0 0 1 1 − 7 x 1 Carilah T y dan T 3
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Jawab: (x,y,z) = k1(1,0,0)+k2(0,1,0)+k3(0,0,1) k1(1,0,0)+k2(0,1,0)+k3(0,0,1) Didapat: x = k1 ; y=k2 ; z=k3 T(x,y,z)=k1(3,0)+k2(1,1)+k3(4,-7) T(x,y,z)= x(3,0)+y(1,1)+z(4,-7)=(3x+y x(3,0)+y(1,1)+z(4,-7)=(3x+y+4z,y-7z) +4z,y-7z) T(1,3,8)=(3.1+3+4.8,3-7.8) T(1,3,8)=(3.1+3+4.8,3-7.8) = (38,-53) 7.2
Jenis – jenis Transformasi Linier bidang
1.
Rotasi (Perputaran)
cos θ − sin θ sin θ cos θ
Matrik baku untuk T adalah :
2. Refleksi Refleksi terhadap sebuah garis l adalah transformasi yang memetakan masing – masing titik pada bidang ke dalam bayangan cerminnya terhadap terhadap l Matrik baku untuk : x − x − 1 0 a. refleksi terhadap sumbu y ( yang mengubah menjadi ) adalah : y y 0 1
b. c.
3.
x x refleksi terhadap sumbu x ( yang mengubah menjadi ) adalah : y − y x y refleksi refleks i terhadap garis y = x ( yang mengubah menjadi ) adalah : y x
1 0 0 − 1 0 1 1 0
Ekspansi dan kompresi
Jika koordinat x dari masing – masing titik pada bidang dikalikan dengan konstanta k yang positif dimana k > 1, maka efeknya adalah memperluas gambar bidang dalam arah x. Jika 0 < k < 1 maka efeknya adalah mengkompresi gambar bidang dalam arah x. Disebut dengan ekspansi (kompresi) dalam arah x dengan faktor k k 0 Matrik baku untuk transformasi ini adalah : 0 1 Demikian juga , jika koordinat y dari masing – masing titik pada bidang dikalikan dengan konstanta k yang positif dimana k > 1, maka efeknya adalah memperluas gambar bidang dalam arah y. Jika 0 < k < 1 maka efeknya adalah mengkompresi gambar bidang dalam arah y. Disebut dengan ekspansi (kompresi) dalam arah y dengan faktor k 1 0 Matrik baku untuk transformasi ini adalah : 0 k 4. Geseran Sebuah geseran dalam arah x dengan faktor k adalah transformasi yang menggerakkan masing – masing titik (x,y) sejajar dengan sumbu x sebanyak ky menuju kedudukan yang baru (x + ky, y)
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Sebuah geseran dalam arah y dengan faktor k adalah transformasi yang menggerakkan masing – masing titik (x,y) sejajar dengan sumbu y sebanyak kx menuju kedudukan yang baru (x , y + kx) 1 0 Matrik baku untuk transformasi ini adalah : k 1 n
m
Jika dilakukan banyak sekali transformasi matrik dari R ke R secara berturutan, maka hasil yang sama dapat dicapai dengan transformasi matrik tunggal. Jika transformasi - transformasi matrik T1(x) = A1x, T2(x) = A2x, , .... , Tn(x) = Anx, n m Dari R ke R dilakukan berurutan, maka hasil yang sama dapat dicapai dengan transformasi matrik tunggal T( x) = Ax, dimana A = Ak . . . A 2 A1 Contoh 2 2 Carilah transformasi matrik dari R ke R yang mula – mula menggeser dengan faktor sebesar 2 dalam arah x dan kemudian merefleksikannya terhadap y = x 2 2 a. Carilah transformasi matrik dari R ke R yang mula – mula merefleksikannya terhadap y = x dan kemudian menggeser dengan faktor sebesar 2 dalam arah x b. Matrik baku untuk refleksi yang diikuti dengan geseran Jawab : Jawab : 1 2 a). Matrik baku untuk geseran adalah A 1 = 0 1
0 1 Dan untuk refleksi terhadap y = x adalah A 2 = 1 0 Jadi matrik baku untuk geseran yang diikuti dengan refleksi adalah 0 1 1 2 0 1 A2. A1 = = 1 0 0 1 1 2 b).
Matrik baku untuk refleksi yang diikuti dengan geseran adalah 1 2 0 1 2 1 A1. A2 = 1 0 = 1 0 0 1
Dari contoh di atas, perhatikan bahwa A 2. A1 ≠ A1. A2 Jika T:R2 R2 adalah perkalian oleh sebuah matrik A yang punya invers, dan misalkan T memetakan titik (x,y) ke titik (x’, y’), maka x' x x -1 x ' = A Dan = A y ' y y y '
Contoh Carilah persamaan bayangan sebuah garis y = 2x + 1 yang dipetakan oleh matrik A = 3 1
2 1
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x y =
−1
3 1 x' 1 − 1 x' = 2 1 y' − 2 3 y '
Sehingga x = x’ – y’ y = -2x’ + 3y’ Substitusikan ke y = 2x + 1 maka dihasilkan : -2x’ + 3y’ = 2(x’ – y’) + 1 -2x’ + 3y’ = 2x’ – 2y’ + 1 5y’ = 4x’ + 1 y’ = 4 5 x’ + 15 7.3 Kernel ( inti ) dan Jangkauan Diketahui transformasi linear T : V W dengan fungsi T( u ) , u € V, Kernel dari T (disingkat Ker(T) ) adalah himpunan u sedemikian hingga T( u ) = 0 atau { u | T( u ) = 0 }. Ker (T) juga disebut ruang nol dari T. Himpunan dari b sedemikian hingga T( u ) = b disebut Jangkauan dari T atau disingkat R(T) R.(T) disebut juga dengan bayangan u oleh T( u ). Contoh 8.2.1 Tentukan basis dan dimensi dari Ker(T) dan R(T) dari transformasi linear 1 − 1 2 3 2 3 T : R R dengan T( u ) = A u , dengan u € R dan A = − − 2 2 4
a. Kernel Ker(T) adalah ruang nol dari T( u ) = A u = 0 . Jadi Ker(T) merupakan ruang solusi dari SPL Au = 0 . Dengan melakukan eliminasi Gauss– Jordan didapatkan solusi SPL adalah s − 2t 1 − 2 1 − 2 u
= + = 1 0 , Jadi basis Kert(T) K ert(T) = ( ) s t Kert T 1 , 0 t 0 1 0 1
=
s
dan dim Ker(T) = 2 b. Jangkauan R(T) merupakan himpunan dari b dengan A u = b . Kalau kita perhatikan maka R(T) merupakan ruang kolom dari A. Dari eliminasi Gauss – Jordan pada A didapatkan A ~ …~ 1 − 1 2 A = 0 0 0
1 dan dim R(T) = 1 − 2
Jadi basis R(T) merupakan basis ruang kolom A yaitu : A =
7.4 Nilai Eigen dan Vektor Eigen n x di dalam R dinamakan vektor eigen dari eigen dari Jika A adalah matrik n x n, maka vektor tak nol x di x adalah kelipatan skalar dari x A jika A x adalah dari x,, yaitu,
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1 Vektor x Vektor x = = adalah vektor eigen dari A = 2
3 0 8 − 1
Yang bersesuaian dengan nilai λ = 3 karena 3 0 1 3 1 Ax = = = 3 2 8 − 1 2 6 Untuk mencari nilai eigen matrik A yang berukuran n x n maka kita menuliskannya x = λ x sebagai x sebagai A x = x = λI x x ( λI – A) x = x = 0 kembali A x = ( 3 2 Carilah nilai – nilai eigen dari dari A = − 1 0 Karena
1 0 3 2 λ − 3 − 2 λI – A = λ = 1 λ 0 1 − 1 0 Det(λI – A) = (λ-3) λ - (-2) = 0 2 = λ - 3λ + 2 = 0 λ1 = 2, λ2 = 1 Jadi nilai – nilai eigen dari A adalah λ1 = 2 dan λ2 = 1 C. Rangkuman 1.
2.
Jika F:V W adalah adalah sebuah fungsi dari ruang vektor V ke ke dalam ruang vektor W , maka F disebut transformasi linier, jika : (i).F(u+v F(u+v)) = F(u F(u) + F(v F(v), untuk semua vektor u dan v di V (ii).F(k u) = kF(u kF(u) untuk semua vektor u di dalam V dan dan semua skalar k Jenis – jenis Transformasi Linier bidang a. Rotasi (Perputaran) cos θ − sin θ Matrik baku untuk T adalah : sin θ cos θ b.
c.
Refleksi Refleksi terhadap sebuah garis l adalah transformasi yang memetakan masing – masing titik pada bidang ke dalam bayangan cerminnya terhadap l Ekspansi dan kompresi k 0 Matrik baku untuk transformasi transformasi ini adalah :
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• Cari tahu apa saja yang menjadi fungsi dari transformasi linear ini! • Cari tahu apa saja yang menjadi fungsi dari vektor Eigen! E. Evaluasi 1.
3
Periksa apakah T : R R
2
a T b = (abc) + ( a + b) x + (a + c) x 2 c Merupakan transformasi linear 2.
V dibangun oleh vektor – vektor yang orthonormal a 1, a 2, …, a n . Periksa apakah proyeksi orthogonal z terhadap V merupakan transformasi linear !
3.
Diketahui transformasi nol T: R3 R2 , dengan T( x ) = A x . Tentukan basis dan dimensi dari ker (T) dan R(T) !
4.
Misalkan T:R → W adalah transformasi matriks, dan misalkan
3
1 0 1 2 3 − 2 T 1 = , T 0 = , T 0 = 0 0 1 1 1 − 7 x Carilah T y dan z
− 1 T 9 7
F. Pustaka Anton, Howard. Dasar-dasar Aljabar Linear Jilid 2 Edisi 7. 2000. Penerbit Interaksara. Jakarta Buku Ajar Aljabar Linear Oleh Yuliant Sibaroni 2002 Aljabar Linier Elementer. Elementer. Mahmud ’Imrona. 2002 PAUL CALTER, 1979, Theory and Problems of Technical Technical Mathematics Mathematics , Schaum’s outline, Mc GRAW.HILL BOOK COMPANY Slide: AgusSoft, dll.