1. Tentukan Nilai eigen dari matriks tersebut
2. Tentukan ruang-ruang eigen dari matriks tersebut
1.
[tex]det(A - \lambda I) = 0[/tex]
[tex]det(\left[\begin{array}{ccc}2&-4\\-3&1\\\end{array}\right] - \lambda \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]) = 0[/tex]
[tex]det(\left[\begin{array}{ccc}2&-4\\-3&1\\\end{array}\right] - \left[\begin{array}{ccc} \lambda&0\\0& \lambda\\\end{array}\right]) = 0[/tex]
[tex]det(\left[\begin{array}{ccc}{2-\lambda}&{-4}\\{-3}&{1-\lambda}\\\end{array}\right]) = 0[/tex]
(2 - λ)(1 - λ) - (-3 * -4)= 0
( 2 - 2λ - λ + λ² ) - (12) = 0
2 - 2λ - λ + λ² - 12 = 0
-10 -3λ + λ² = 0
λ² -3λ -10 = 0
(λ + 2)(λ - 5) = 0
Ada dua nilai eigen.
λ = -2
λ = 5
2. Seharusnya basis ruang eigen
Untuk λ = -2
Ax = λx
A[tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex] = -2 [tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex] Masukin matriks A nya, 2 juga dimasukin
[tex]\left[\begin{array}{ccc}2&-4\\-3&1\\\end{array}\right]\left[\begin{array}{ccc}a\\b\\\end{array}\right] = \left[\begin{array}{ccc}-2a\\-2b\\\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}2a+-4b\\-3a+b\\\end{array}\right]= \left[\begin{array}{ccc}-2a\\-2b\\\end{array}\right][/tex] (cari nilai b nya dengan ambil persamaan yang baris kedua)
> -3a + b = -2b
> -3a = -2b - b
> -3a = -3b
> a = b
Lihat ke X lagi, substitusi b = a
x = [tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}a\\a\\\end{array}\right][/tex] (Keluarkan a nya dari matriks)
= [tex]a \left[\begin{array}{ccc}1\\1\\\end{array}\right][/tex]
Untuk λ = 5
Ax = λx
A[tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex] = 5 [tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex] Masukin matriks A nya, 5 juga dimasukin
[tex]\left[\begin{array}{ccc}2&-4\\-3&1\\\end{array}\right]\left[\begin{array}{ccc}a\\b\\\end{array}\right] = \left[\begin{array}{ccc}5a\\5b\\\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}2a+-4b\\-3a+b\\\end{array}\right]= \left[\begin{array}{ccc}5a\\5b\\\end{array}\right][/tex] (cari nilai b nya dengan ambil persamaan yang baris pertama)
> -3a + b = 5b
> -3a = 5b - b
> -3a = 4b
> b = -3/4 a (pindahkan 4nya, huruf b nya jangan ikut pindah)
Lihat ke X lagi, substitusi b = -3/4 a
x = [tex]\left[\begin{array}{ccc}a\\b\\\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}a\\{(-\frac{3}{4}a) }\\\end{array}\right][/tex] (Keluarkan a nya dari matriks)
= [tex]a \left[\begin{array}{ccc}1\\{-\frac{3}{4} }\\\end{array}\right][/tex]
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