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Transformasi Linear dan Sistem Persamaan Linear Pertemuan 5 Matakuliah: MATRIX ALGEBRA FOR STATISTICS Tahun: 2009.

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Presentasi berjudul: "Transformasi Linear dan Sistem Persamaan Linear Pertemuan 5 Matakuliah: MATRIX ALGEBRA FOR STATISTICS Tahun: 2009."— Transcript presentasi:

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2 Transformasi Linear dan Sistem Persamaan Linear Pertemuan 5 Matakuliah: MATRIX ALGEBRA FOR STATISTICS Tahun: 2009

3 Transformasi Linear Jika T:V  W merupakan fungsi dari ruang vektor V ke ruang vektor W, maka T disebut transformasi linear dari V ke W jika semua vektor u dan v dalam V dan semua skalar c a)T(u+v) = T(u) + T(v) b) T(cu) = cT(u) Transformasi T:V  V disebut linear operator pada V Bina Nusantara University 3

4 Contoh: Diketahui vektor v 1 = (1,1,1), v 2 = (1,1,0), dan v 3 = (1,0,0) membentuk basis pada S untuk R 3 Bila T: R 3  R 2 merupakan transformasi linear sehingga berlaku T(v 1 ) = (1,0), T(v 2 ) = (2,-1), T(v 3 ) = (4,3) Tentukan bentuk transformasi T(x 1,x 2,x 3 ) Jawab: Nyatakan x = (x 1,x 2,x 3 ) sebagai kombinasi linear (x 1,x 2,x 3 ) = c 1 (1,1,1)+ c 2 (1,1,0)+c 3 (1,0,0) Bina Nusantara University 4

5 Sehingga c 1 + c 2 +c 3 = x 1 c 1 + c 2 = x 2 c 1 = x 3 diperoleh c 1 = x 3, c 2 = x 2 - x 3, dan c 3 = x 1 – x 2 maka (x 1,x 2,x 3 ) = x 3 (1,1,1)+ (x 2 - x 3 )(1,1,0)+ (x 1 - x 2 )(1,0,0) T(x 1,x 2,x 3 )= x 3 T(v 1 )+(x 2 - x 3 )T(v 2 )+ (x 1 - x 2 )T(v 3 ) = x 3 (1,0)+(x 2 - x 3 )(2,-1)+ (x 1 - x 2 )(4,3) = (4x 1 -2x 2 -x 3, 3x 1 -4x 2 +x 3 ) Misalnya T (1,2,0) = (0, -5) Bina Nusantara University 5

6 A linear transformation In a vector space V we define A transformation t of V is linear For all vectors u, v and all real numbers r t(u+v) = t(u)+t(v) and t(r.u) = r.t(u) The set of all linear transformations of V is L(V). Examples : t : R x R  R x R : (x,y)  (x+y,x) t : R x R  R x R : (x,y)  (0,y) t : R  R : x  6x Bina Nusantara University 6

7 Image of the vector 0. Let t be a linear transformation of V, then t(0) = t(0v) = 0.t(v) = 0 Hence, the image of the vector 0 is 0. Criterion for the linearity of a transformation of V Theorem : Take a transformation t of V. t is in L(V) For all vectors u, v and all real numbers r, s t(r.u + s.v) = r.t(u) + s.t(v) Bina Nusantara University 7

8 Proof : Part 1 : If t is in L(V) then t(r.u + s.v) = t(r.u) + t(s.v) = r.t(u) + s.t(v) Part 2 : If t(r.u + s.v) = r.t(u) + s.t(v) for all r, s then take r = s = 1 t(u+v) = t(u)+t(v) take s = 0 t(r.u) = r.t(u) Bina Nusantara University 8

9 We show this for dimension(V) = 3, but all can easily be generalized. Theorem : If (e 1, e 2, e 3 ) is an ordered basis of V, and if (u 1, u 2, u 3 ) is an ordered random set of three vectors from V. Then, there is just one linear transformation t of V such that t(e 1 ) = u 1 t(e 2 ) = u 2 t(e 3 ) = u 3 Bina Nusantara University 9 Building linear transformations

10 Prove : A random vector v in V can be written as v = k.e 1 +l.e 2 +m.e 3. A random vector w in V can be written as w = k'.e 1 +l'.e 2 +m'.e 3. Then u + v = (k+k')e 1 + (l+l')e 2 + (m+m')e 3. We start from a transformation t of V defined by t(v) = k.u 1 + l.u 2 + m.u 3 Bina Nusantara University 10

11 t is linear because : t(v + w) = t( (k + k')e 1 + (l + l')e 2 + (m + m')e 3 ) = (k + k')u 1 + (l + l')u 2 + (m + m')u 3 = k.u 1 + l.u 2 + m.u 3 + k'.u 1 + l'.u 2 + m'.u 3 = t(v) + t(w) t(r.v) = t(rk.e 1 + rl.e 2 + rm.e 3 ) = rk.u 1 + rl.u 2 + rm.u 3 = r(k.u 1 + l.u 2 + m.u 3 ) = rt(v) Bina Nusantara University 11

12 Example: There is just one linear transformation of R x R such that t(1,0) = (3,2) t(0,1) = (5,4) We calculate the image of (-1,5). t(-1,5) = t( -1(1,0) + 5(0,1) ) = -1(3,2) + 5(5,4) = (22,18) Bina Nusantara University 12

13 Matrices and linear transformations Example : There is just one linear transformation of R x R such that t(1,0) = (3,2) t(0,1) = (5,4) The matrix of the linear transformation with respect to the basis is Bina Nusantara University 13

14 Example In a 2-dimensional space with basis (e 1, e 2 ), a linear transformation T has matrix Now we take a new basis e 1 ' = e 1 + e 2 e 2 ' = e 1 - e 2 Then the transformation matrix C is Bina Nusantara University 14

15 and from this C -1 is The matrix of the linear transformation t with respect to the new basis is Bina Nusantara University 15

16 SISTEM PERSAMAAN LINEAR (SPL) Bentuk umum persamaan garis di R n : Untuk n variabel x 1, x 2, x 3, …x n dan a 1, a 2, a 3,..., a n, b adalah konstanta a 1 x 1 + a 2 x 2 + a 3 x a n x n +b=0 Bina Nusantara University 16

17 Bentuk umum SPL a 11 x 1 + a 12 x 2 + a 13 x a 1n x n = b 1 a 21 x 1 + a 22 x 2 + a 23 x a 2n x n = b 2 a 31 x 1 + a 32 x 2 + a 33 x a 3n x n = b a m1 x 1 + a m2 x a mn x n = b m Bila b 1, b 2, …, b m =0 maka SPL disebut homogen Bina Nusantara University 17

18 Dalam bentuk matriks Bina Nusantara University 18 Atau dalam bentuk matriks lengkap

19 Penyelesaian dari SPL : Eliminasi Gauss atau menggunakan Operasi Baris Elementer Eliminasi Gauss-Jordan Bina Nusantara University 19

20 Contoh: Selesaikan SPL berikut: 2x 1 - x 2 + 3x 3 = 13 -x 1 + 2x 2 + 3x 3 = 16 x 1 + x 2 + 4x 3 = 21 Atau dalam matriks lengkap Bina Nusantara University 20

21 Lakukan operasi baris elementer Bina Nusantara University 21 b 2 +½b 1 & b 3 -½b 1 

22 Substitusi balik maka diperoleh: -4/3x 3 =-16/2 diperoleh x 3 = 4 Substitusi x 3 = 4 ke 3/2 x 2 +9/2 x 3 = 45/2 diperoleh x 2 = 3 Substitusi kan nilai x 2 dan x 3 ke persamaan 2x 1 -x 2 +3 x 3 = 13 x 2 Diperoleh x 1 = 2 Bina Nusantara University 22

23 Masalah Kuadrat Terkecil (Least Square problem) Bentuk umum persamaan garis Y=AX+ε atau ε =Y-AX, maka = - Bina Nusantara University 23 -

24 Menemukan vektor x sedemikian rupa sehingga minimum ε minimum jika ε tegaklurus ruang kolom A atau CS(A) Bina Nusantara University 24

25 Contoh: Tentukan persamaan garis y=ax+b mewakili data berikut: AX=Y atau. = Bina Nusantara University 25 NoXiXi YiYi

26 A T AX=A T Y = Bina Nusantara University 26

27 Diperoleh (A T A) = A T Y = = Persamaan: Y=2,1X +50 Bina Nusantara University 27 Maka X= (A T A) -1. A T Y


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