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

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 T(u+v) = T(u) + T(v) T(cu) = cT(u) Transformasi T:V  V disebut linear operator pada V Bina Nusantara University

Contoh: Diketahui vektor v1 = (1,1,1), v2 = (1,1,0), dan v3 = (1,0,0) membentuk basis pada S untuk R3 Bila T: R3  R2 merupakan transformasi linear sehingga berlaku T(v1) = (1,0), T(v2) = (2,-1), T(v3) = (4,3) Tentukan bentuk transformasi T(x1,x2,x3) Jawab: Nyatakan x = (x1,x2,x3) sebagai kombinasi linear (x1,x2,x3) = c1(1,1,1)+ c2(1,1,0)+c3(1,0,0) Bina Nusantara University

diperoleh c1= x3, c2 = x2 - x3, dan c3 = x1 – x2 maka Sehingga c1+ c2+c3 = x1 c1+ c2 = x2 c1 = x3 diperoleh c1= x3, c2 = x2 - x3, dan c3 = x1 – x2 maka (x1,x2,x3) = x3(1,1,1)+ (x2- x3)(1,1,0)+ (x1- x2)(1,0,0) T(x1,x2,x3)= x3T(v1)+(x2- x3)T(v2)+ (x1- x2)T(v3) = x3(1,0)+(x2- x3)(2,-1)+ (x1- x2)(4,3) = (4x1-2x2-x3, 3x1-4x2+x3) Misalnya T (1,2,0) = (0, -5) Bina Nusantara University

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

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

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

Building linear transformations We show this for dimension(V) = 3, but all can easily be generalized. Theorem : If (e1, e2, e3) is an ordered basis of V, and if (u1, u2, u3) is an ordered random set of three vectors from V. Then, there is just one linear transformation t of V such that t(e1) = u1 t(e2) = u2 t(e3) = u3 Bina Nusantara University

A random vector w in V can be written as w = k'.e1+l'.e2+m'.e3. Prove : A random vector v in V can be written as v = k.e1+l.e2+m.e3. A random vector w in V can be written as w = k'.e1+l'.e2+m'.e3. Then u + v = (k+k')e1 + (l+l')e2 + (m+m')e3. We start from a transformation t of V defined by t(v) = k.u1 + l.u2 + m.u3 Bina Nusantara University

= t(v) + t(w) t(r.v) = t(rk.e1 + rl.e2 + rm.e3) t is linear because : t(v + w) = t( (k + k')e1 + (l + l')e2 + (m + m')e3 ) = (k + k')u1 + (l + l')u2 + (m + m')u3 = k.u1 + l.u2 + m.u3 + k'.u1 + l'.u2 + m'.u3 = t(v) + t(w) t(r.v) = t(rk.e1 + rl.e2 + rm.e3) = rk.u1 + rl.u2 + rm.u3 = r(k.u1 + l.u2 + m.u3) = rt(v) Bina Nusantara University

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

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

Example In a 2-dimensional space with basis (e1, e2), a linear transformation T has matrix Now we take a new basis e1' = e1 + e2 e2' = e1 - e2 Then the transformation matrix C is Bina Nusantara University

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

SISTEM PERSAMAAN LINEAR (SPL) Bentuk umum persamaan garis di Rn: Untuk n variabel x1, x2, x3, …xn dan a1, a2, a3, . . . , an, b adalah konstanta a1x1+ a2x2+ a3x3+ . . . + anxn+b=0 Bina Nusantara University

Bentuk umum SPL a11x1+ a12x2+ a13x3+ . . . + a1nxn = b1 . . . . . . . . . . . . . . . . . . am1x1+ am2x2+ . . . + amnxn = bm Bila b1, b2, …, bm =0 maka SPL disebut homogen Bina Nusantara University

Dalam bentuk matriks Atau dalam bentuk matriks lengkap Bina Nusantara University

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

Contoh: Selesaikan SPL berikut: 2x1 - x2 + 3x3 = 13 Atau dalam matriks lengkap Bina Nusantara University

Lakukan operasi baris elementer b2+½b1 & b3-½b1  Bina Nusantara University

Substitusi balik maka diperoleh: -4/3x3 =-16/2 diperoleh x3 = 4 Substitusi x3 = 4 ke 3/2 x2 +9/2 x3 = 45/2 diperoleh x2 = 3 Substitusi kan nilai x2 dan x3 ke persamaan 2x1-x2+3 x3 = 13 x2 Diperoleh x1 = 2 Bina Nusantara University

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

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

Contoh: Tentukan persamaan garis y=ax+b mewakili data berikut: AX=Y atau . = No Xi Yi 1 2 3 4 5 10 15 20 25 30 70 80 95 105 110 Bina Nusantara University

ATAX=ATY = Bina Nusantara University

Diperoleh (ATA) = ATY Maka X= (ATA)-1 . ATY Persamaan: Y=2,1X +50 = = = = Persamaan: Y=2,1X +50 Maka X= (ATA)-1 . ATY Bina Nusantara University