Skip to article frontmatterSkip to article content
Site not loading correctly?

This may be due to an incorrect BASE_URL configuration. See the MyST Documentation for reference.

Perhitungan Determinan 3x3

A=(231415726)A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 1 & 5 \\ 7 & 2 & 6 \end{pmatrix}
(1)

Langkah 1: Gunakan Rumus Ekspansi Baris Pertama

det⁡(A)=a11det⁡(A11)−a12det⁡(A12)+a13det⁡(A13)\det(A) = a_{11}\det(A_{11}) - a_{12}\det(A_{12}) + a_{13}\det(A_{13})
(2)
det⁡(A)=2det⁡(A11)−3det⁡(A12)+1det⁡(A13)\det(A) = 2\det(A_{11}) - 3\det(A_{12}) + 1\det(A_{13})
(3)

Langkah 2: Tentukan Minor

Minor A11A_{11} (coret baris 1 kolom 1)

A11=(1526)A_{11} = \begin{pmatrix} 1 & 5 \\ 2 & 6 \end{pmatrix}
(4)

Minor A12A_{12} (coret baris 1 kolom 2)

A12=(4576)A_{12} = \begin{pmatrix} 4 & 5 \\ 7 & 6 \end{pmatrix}
(5)

Minor A13A_{13} (coret baris 1 kolom 3)

A13=(4172)A_{13} = \begin{pmatrix} 4 & 1 \\ 7 & 2 \end{pmatrix}
(6)

Langkah 3: Hitung Determinan 2x2 (SATU PER SATU)

1. det⁡(A11)\det(A_{11})

det⁡(A11)=(1⋅6)−(5⋅2)\det(A_{11}) = (1 \cdot 6) - (5 \cdot 2)
(7)
=6−10= 6 - 10
(8)
=−4= -4
(9)

2. det⁡(A12)\det(A_{12})

det⁡(A12)=(4⋅6)−(5⋅7)\det(A_{12}) = (4 \cdot 6) - (5 \cdot 7)
(10)
=24−35= 24 - 35
(11)
=−11= -11
(12)

3. det⁡(A13)\det(A_{13})

det⁡(A13)=(4⋅2)−(1⋅7)\det(A_{13}) = (4 \cdot 2) - (1 \cdot 7)
(13)
=8−7= 8 - 7
(14)
=1= 1
(15)

Langkah 4: Substitusi ke Rumus

det⁡(A)=2(−4)−3(−11)+1(1)\det(A) = 2(-4) - 3(-11) + 1(1)
(16)
=−8+33+1= -8 + 33 + 1
(17)
=26= 26
(18)

Hasil Akhir

det⁡(A)=26\det(A) = 26
(19)

Catatan Penting

KOMPUTASI ALJABAR LINEAR
Eliminasi Gauss
KOMPUTASI ALJABAR LINEAR
Rumus Determinan (Ekspansi Kofaktor)