Aljabar Linier Matriks Nilai Eigen dan V

Aljabar Linier Matriks

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Nilai Eigen dan Vektor Eigen

Adri Priadana – ilkomadri.com Halaman 1

Nilai Eigen dan Vektor Eigen Berordo 3x3

Tentukan Nilai Eigen dan Vektor Eigen untuk matriks A =

0 −1 −3

2 3 3

−2 1 1

!

Jawab

Nilai Eigen

| A –λI | = 0

0 −1 −3

2 3 3

−2 1 1

– 0λ 0λ 00

0 0 λ

= 0 → −λ2 3−− λ1 −33

−2 1 1− λ

= 0

– λ ((3 – λ)(1 – λ) – 3.1) + (1) (2(1 – λ) – (3 * – 2)) + (–3) (2*1 – (3 – λ)(-2)) = 0 – λ ((3 – λ)(1 – λ) – 3) + (2(1 – λ) – (–6)) + (–3) (2 – (– 6 + 2 λ)) = 0 –λ (λ2– 4λ + 3 – 3) + (2 – 2λ + 6) + (–3) (2 + 6 – 2 λ) = 0 –λ (λ2– 4λ) + (– 2λ + 8) + (– 6 – 18 + 6λ) = 0

–λ (λ2– 4λ) + (– 2λ + 8) + (– 24 + 6λ) = 0 – λ3 _{+ 4λ}2 _{– 2λ + 8 – 24 + 6λ = 0}

– λ3 _{+ 4λ}2 _{+ 4λ – 16 = 0}

Metode Horner

– 1 4 4 – 16

– 2 2 – 12 16 – 1 6 – 8 0

(λ + 2) (–λ2 + 6λ– 8) = 0

(λ + 2) (–λ + 4) (λ– 2) = 0

λ + 2 = 0 → λ = – 2

– λ + 4 = 0 → λ = 4

λ– 2 = 0 → λ = 2

Aljabar Linier Matriks

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Nilai Eigen dan Vektor Eigen

Adri Priadana – ilkomadri.com Halaman 2

Vektor Eigen

Maka diperoleh vektor eigen: x =