Matakuliah
Tahun
: MATRIX ALGEBRA FOR STATISTICS
: 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
a) T(u+v) = T(u) + T(v)
b) T(cu) = cT(u)
Transformasi T:V  V disebut linear operator pada V
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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)
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Sehingga
c1+ c2+c3 = x1
c 1+ c 2
= 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)
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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
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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)
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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)
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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
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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
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• 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)
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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)
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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
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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
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and from this C-1 is
The matrix of the linear transformation t with respect to the
new basis is
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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
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Bentuk umum SPL
a11x1+ a12x2+ a13x3+ . . . + a1nxn = b1
a21x1+ a22x2+ a23x3+ . . . + a2nxn = b2
a31x1+ a32x2+ a33x3+ . . . + a3nxn = b3
.
.
.
.
.
.
am1x1+ am2x2+
.
.
.
.
.
.
...
.
.
.
.
.
.
+ amnxn = bm
Bila b1, b2, …, bm =0 maka SPL disebut homogen
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Dalam bentuk matriks
Atau dalam bentuk matriks lengkap
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Penyelesaian dari SPL :
Eliminasi Gauss atau menggunakan
Operasi Baris Elementer
• Eliminasi Gauss-Jordan
•
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Contoh:
Selesaikan SPL berikut:
2x1 - x2 + 3x3 = 13
-x1+ 2x2 + 3x3 = 16
x1+ x2 + 4x3 = 21
Atau dalam matriks lengkap
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Lakukan operasi baris elementer
b2+½b1 & b3-½b1 
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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
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Masalah Kuadrat Terkecil (Least Square problem)
-
Bentuk umum persamaan garis Y=AX+ε
atau ε =Y-AX , maka
=
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Menemukan vektor x sedemikian rupa
sehingga
minimum
ε minimum jika ε tegaklurus ruang kolom A
atau CS(A)
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Contoh:
Tentukan persamaan garis y=ax+b mewakili data berikut:
No
1
2
3
4
5
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Xi
10
15
20
25
30
Yi
70
80
95
105
110
AX=Y atau
.
=
25
ATAX=ATY
=
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Diperoleh (ATA) = ATY
Maka X= (ATA)-1 . ATY
=
=
Persamaan: Y=2,1X +50
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