Nomor 7.docx

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2 1

7. bentuk matrik suatu operator adalah (

1 ). Hitunglah harga-harga eigennya dan tentukan 2

fungsi eigen bersangkutan Misalkan vektor eigen adalah (𝐢𝐢1 ) dan harga eigen a adalah 2

[

2βˆ’π‘Ž 2 1 𝐢1 ] (𝐢 ) = a (𝐢𝐢1 ) β†’ [ 2 2 1 1 2

1 ] (𝐢1 ) 2 βˆ’ π‘Ž 𝐢2

Untuk itu berlaku [

2βˆ’π‘Ž 1

1 ] = 0 β†’ (2 βˆ’ π‘Ž) (2 βˆ’ π‘Ž) – 1 = 0 2βˆ’π‘Ž 4 -2π‘Ž - 2π‘Ž + π‘Ž2 – 1 = 0 4 - 4π‘Ž + π‘Ž2 – 1 = 0 π‘Ž2 - 4π‘Ž + 3 = 0

X12 = = = = = π‘Ž1 =

βˆ’π‘Β±βˆšπ‘ 2 βˆ’4π‘Žπ‘ 2π‘Ž βˆ’(βˆ’4)±√(βˆ’4)2 βˆ’4.1.3 2.1 4±√16βˆ’12 2 4±√4 2 4Β±2 2 4Β±2 2 6

π‘Ž1 = 2 - 3 π‘Ž2 =

4βˆ’2 2 2

π‘Ž2 = 2 = 1 Untuk π‘Ž1 = 3 [

3βˆ’π‘Ž 1

1 ] (𝐢1 ) = 0 2 βˆ’ 3 𝐢2

[

βˆ’1 1 𝐢1 ]( )=0 1 βˆ’1 𝐢2

-C1 + C2 = 0 β†’ ambil satu saja C1 - C2 = 0 Dengan normalisasi C12 + C22 = 1 C12 + (C12) = 1 2 C12 = 1 1

C12 = 2 C12 = 0,5 C1 = √0,5 C1 = 0,707 Karena C1= C2 Maka C1 = 0,707 C2 = 0,707 Jadi untuk π‘Ž1 = 3 πœ“ = 0,707 πœ‘1 + 0,707 πœ‘2 Untuk π‘Ž2 = 1 [

2βˆ’1 1 ] (𝐢1 ) = 0 1 2 βˆ’ 1 𝐢2

[

1 1 𝐢1 ]( )=0 1 1 𝐢2

C1 + C2 = 0 β†’ ambil satu saja C1 + C2 = 0 C1 = - C2 C2 = - C1

Normalisasi C12 + C22 = 1 C12 + (-C2)2 = 1 C12 + C12 = 1 2 C12 = 1 1

C12 = 2 C12 = 0,5 C1 = √0,5 C1 = 0,707 Karena C2 = - C1 C2 = - 0,707 Jadi untuk π‘Ž2 = 1 πœ“ = 0,707 πœ‘1 + 0,707 πœ‘2 = 0,707 πœ‘1 + 0,707 πœ‘2

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