Diketahui A=[1 1 -2 6 9 -10 -3 -2 7], tentukan invers

Invers matriks A adalah [[43, -3, -8], [-12, 1, -2], [-15, -1, 3]].

Pembahasan

Untuk menentukan invers matriks A, kita perlu menggunakan rumus:
A⁻¹ = (1 / det(A)) * adj(A)

Langkah-langkahnya adalah sebagai berikut:

1.  **Hitung Determinan (det(A))**: 
    Matriks A = [1 1 -2]
                [6 9 -10]
                [-3 -2 7]

    det(A) = 1 * ( (9*7) - (-10*-2) ) - 1 * ( (6*7) - (-10*-3) ) + (-2) * ( (6*-2) - (9*-3) )
    det(A) = 1 * ( 63 - 20 ) - 1 * ( 42 - 30 ) - 2 * ( -12 - (-27) )
    det(A) = 1 * ( 43 ) - 1 * ( 12 ) - 2 * ( -12 + 27 )
    det(A) = 43 - 12 - 2 * ( 15 )
    det(A) = 43 - 12 - 30
    det(A) = 31 - 30
    det(A) = 1

    Karena determinan A tidak sama dengan nol (det(A) = 1), maka invers matriks A ada.

2.  **Hitung Matriks Adjoin (adj(A))**: Matriks adjoin adalah transpose dari matriks kofaktor.
    Pertama, kita cari matriks kofaktor (C).
    C₁₁ = +( (9*7) - (-10*-2) ) = 63 - 20 = 43
    C₁₂ = -( (6*7) - (-10*-3) ) = -( 42 - 30 ) = -12
    C₁₃ = +( (6*-2) - (9*-3) ) = -( -12 + 27 ) = -15
    C₂₁ = -( (1*7) - (-2*-2) ) = -( 7 - 4 ) = -3
    C₂₂ = +( (1*7) - (-2*-3) ) = +( 7 - 6 ) = 1
    C₂₃ = -( (1*-2) - (1*-3) ) = -( -2 + 3 ) = -1
    C₃₁ = +( (1*-10) - (-2*9) ) = -( -10 + 18 ) = -8
    C₃₂ = -( (1*-10) - (-2*6) ) = -( -10 + 12 ) = -2
    C₃₃ = +( (1*9) - (1*6) ) = +( 9 - 6 ) = 3

    Matriks Kofaktor (C) =
    [ 43  -12  -15 ]
    [ -3    1   -1 ]
    [ -8   -2    3 ]

    Matriks Adjoin (adj(A)) = Cᵀ =
    [ 43   -3   -8 ]
    [ -12   1   -2 ]
    [ -15  -1    3 ]

3.  **Hitung Invers Matriks (A⁻¹)**:
    A⁻¹ = (1 / det(A)) * adj(A)
    A⁻¹ = (1 / 1) * adj(A)
    A⁻¹ =
    [ 43   -3   -8 ]
    [ -12   1   -2 ]
    [ -15  -1    3 ]

Jadi, invers matriks A adalah:
[ 43  -3  -8 ]
[ -12  1  -2 ]
[ -15 -1   3 ]