Tugas Aljabar Linear

LATIHAN 7.15. Carilah nilai eigen dari matriks berikut:C. Jawab :A = Langkah 1 : untuk mencari nilai eigen..I A = - = = Det(.I A ) = =(()())+(0)(0)(-19)+(-1)(6)(-5)) ((-1)()(-19)+()(0)(-5)+(0)(6)()) =( () ()) + (0) + (30) ) ( () (19) (0) (0) )= () ()) - () (19) + 30= ( 2+4+4) (+4) - 19 - 38 +30= 3 + 4 2 + 2 + 16 + 4 + 16 -19 8= 3 + 8 2 + + 8=( 2 + 1) () = 0 Karena merupakan akar imaginer, maka diabaikan Sehingga, nilai eigen dari matriks A adalah

Langkah 2 : Untuk mencari vektor eigen. Untuk Ax = xAx = -8x = -8

-2 X1 + X3 = -8 X1 6 X1 + X3 = 0-6 X1 2 X2 = -8 X2 -6 X1 + 6 X2 = 0 19 X1 + 5 X2 - 4 X3 = -8 X3 19 X1 + 5 X2 + 4 X3 = 0Missal : X1 = tX1 = X2 , X2 = tX3 = -8X1 + 2X1 = -8t + 2tX3 = -6tJadi, = = t Vektor eigen dari matriks diatas adalah

LATIHAN 8.43. Anggap T : adalah operartor linear yang didefinisikan olehT (a0 + a1x + a2x2) = a0 + a1(x-1) + a2(x-1)2a) Cari matriks untuk T berkenaan dengan basis standar B= {1,x,x2} untuk P2 .b) Tunjukan bahwa matriks [T]B yang diperoleh pada bagian (a) memenuhi Rumus (5a) untuk setiap vector X= a0 + a1x + a2x2 pada P2JAWAB :a) B= {1,x,x2}T (a0 + a1x + a2x2) = a0 + a1(x-1) + a2(x-1)2

T(1) = 1T(x) = x-1T(x2) = (x-1)2 = x2-2x+1

[T(1)]B = [T(x)]B = [T(x2)]B =

[T]B =

b) X = a0 + a1x + a2x2 , B={1,x,x2}[X(1)]B = a0[X(x)]B = a1[X(x2)]B = a2

[X]B =

[T]B [X]B = [T(X)]B

=> [T]B [X]B= [T]B [X]B = T = B = { V1, V2, V3 }, B = {(1, 0, 1), (0, 1, 1), (1, 1, 0)}V1= , V2= , V3= T (V1) = T = = = V1 + V2 + V3T (V2) = T = = = = V1 + V2 + V3T(V3)= T= = = V1 + V2 + V3Diperoleh : T (V1) = 11 v1 + 21 v2 + 31 v3 11 + 21 + 31

11 + 31 = 1(1)21+ 31 = -1(2)11 + 21 = 0 (3)

11 + 31 = 1 (1)11 + 21 = 0 (3) 31 - 21 = 1(4)

31 - 21 = 1(4)31 + 21 = -1(2) 21 = 221 = -1

*) 11 + 21 = 0 (3)*) 11 + 31 = 1 (1)11 - 1 = 0 1 + 31 = 111 = 131 = 0

Jadi, B =

T (V2) = = = 21 + 2223

21 + 23 =-1(5)22 + 23 = 1(6)21 + 22 = -1(7)

21 + 23 =-1(5)21 + 22 = -1(7) 23 - 22= 0 (8) 23 = 22(9)

23 - 22 = 0 (8)23 22 = 1(6) -2 22 = -1

22 = 23 = 22(9) 23 =

*) 21 + 22 = -1 (7) 21 + = -1 21 = Jadi, B =

T (V3) = = 31323331 + 33 = 0 (10)32 + 33 = 0 (11)31 + 32 = 1 (12)

31 + 33 =0(10)31 + 32 =1(12) 33 - 32 = -1(13)

33 - 32 = -1(13)33 + 32 = 0(11)-232 = -132 =

*) + 33 = 0 (11)*) 32+ 31 = 1 (12) +33 = 0 + 31 = 133 = - 31 =

Jadi, B =

matirks representasinya adalah

b.) [T]B [X]B = [T(X)]B [T]B [X]B =

= - + = + + = + + = =

= -1 + + = + + = + + = = - = - = - = = = -

Jadi, [T]B [X]B = = =

[T(X)]B

T(X) =T = T = a + b + c

a + c = (i)b + c = (ii)a + b = (iii)

a + c = (i)b + c = (ii) a b = (iv)

a b = (iv)a + b = (iii) +2a= 3x1-2x2-x3

a=

a + b = x1 x3(iii) b = x1 x3 () b = - - + + b =

b + c = (ii)c = - ( x1 + x2 x3 )c = - + - +

Jadi, [T(X)]B = =