Eliminasi Gauss Universitas Trunojoyo Madura
March 2, 2026
Pengertian ¶ Dalam aljabar linier, eliminasi gauss merupakan metode untuk menyelesaikan Sistem Persamaan Linier (SPL) dengan cara merubah SPL menjadi matriks. Tujuan dari eliminasi gauss adalah bentuk eselon baris. Terdapat teknik-teknik tersendiri untuk mencapai bentuk eselon baris.
Representasi Sistem Persamaan Linier dalam Matriks ¶ Sebagai contoh, terdapat sistem persamaan linier sebagai berikut.
a 1 x 1 + b 2 y 1 + c 3 z 1 = d 1 a 1 x 2 + b 2 y 2 + c 3 z 2 = d 2 a 1 x 3 + b 2 y 3 + c 3 z 3 = d 3 \begin{aligned}
a_{1}x_{1} + b_{2}y_{1} + c_{3}z_{1} &= d_1 \\
a_{1}x_{2} + b_{2}y_{2} + c_{3}z_{2} &= d_2 \\
a_{1}x_{3} + b_{2}y_{3} + c_{3}z_{3} &= d_3
\end{aligned} a 1 x 1 + b 2 y 1 + c 3 z 1 a 1 x 2 + b 2 y 2 + c 3 z 2 a 1 x 3 + b 2 y 3 + c 3 z 3 = d 1 = d 2 = d 3 Jika persamaan diatas diubah ke bentuk matriks teraugmentasi (diperbesar), maka hasilnya adalah sebagai berikut.
[ a 1 x 1 b 2 y 1 c 3 z 1 d 1 a 1 x 2 b 2 y 2 c 3 z 2 d 2 a 1 x 3 b 2 y 3 c 3 z 3 d 3 ] \left[\begin{array}{ccc|c}
a_{1}x_{1} & b_{2}y_{1} & c_{3}z_{1} & d_1 \\
a_{1}x_{2} & b_{2}y_{2} & c_{3}z_{2} & d_2 \\
a_{1}x_{3} & b_{2}y_{3} & c_{3}z_{3} & d_3
\end{array}\right] ⎣ ⎡ a 1 x 1 a 1 x 2 a 1 x 3 b 2 y 1 b 2 y 2 b 2 y 3 c 3 z 1 c 3 z 2 c 3 z 3 d 1 d 2 d 3 ⎦ ⎤ Tujuan Eliminasi Gauss ¶ Tujuan utama dari metode eliminasi gauss adalah mengubah matriks menjadi bentuk eselon baris . Eselon baris memiliki beberapa ciri-ciri, antara lain:
Setelah matriks telah mencapai bentuk eselon baris , maka barulah solusi dapat diperoleh dengan metode substitusi balik.
Operasi Baris Elementer ¶ Terdapat tiga operasi dasar yang disebut operasi baris elementer untuk merubah matrix menjadi bentuk eselon baris, antara lain:
Menukar dua baris
Mengalikan satu baris dengan konstanta yang bukan nol
Menambahkan kelipatan suatu baris ke baris lain
Contoh Penyelesaian SPLTV dengan Eliminasi Gauss ¶ Soal ¶ Tentukan himpunan penyelesaian dari persamaan dibawah ini:
x + y + z = 6 x + 2 y − z = 2 2 x − y + z = 3 \begin{align*}
x + y + z &= 6 \\
x + 2y - z &= 2 \\
2x - y + z &= 3
\end{align*} x + y + z x + 2 y − z 2 x − y + z = 6 = 2 = 3 Langkah Penyelesaian ¶ Buat matrix teraugmentasi
[ 1 1 1 6 1 2 − 1 2 2 − 1 1 3 ] \left[\begin{array}{ccc|c}
1 & 1 & 1 & 6 \\
1 & 2 & -1 & 2 \\
2 & -1 & 1 & 3
\end{array}\right] ⎣ ⎡ 1 1 2 1 2 − 1 1 − 1 1 6 2 3 ⎦ ⎤ Kita anggap:
b1 \text{b1} b1
b2 \text{b2} b2
b3 \text{b3} b3
Baris 2 = − 1 × b1 + b2 -1 \times \text{b1} + \text{b2} − 1 × b1 + b2
[ 1 1 1 6 0 1 − 2 − 4 2 − 1 1 3 ] \left[\begin{array}{ccc|c}
1 & 1 & 1 & 6 \\
0 & 1 & -2 & -4 \\
2 & -1 & 1 & 3
\end{array}\right] ⎣ ⎡ 1 0 2 1 1 − 1 1 − 2 1 6 − 4 3 ⎦ ⎤ Baris 3 = − 2 × b1 + b3 -2 \times \text{b1} + \text{b3} − 2 × b1 + b3
[ 1 1 1 6 0 1 − 2 − 4 0 − 3 − 1 − 9 ] \left[\begin{array}{ccc|c}
1 & 1 & 1 & 6 \\
0 & 1 & -2 & -4 \\
0 & -3 & -1 & -9
\end{array}\right] ⎣ ⎡ 1 0 0 1 1 − 3 1 − 2 − 1 6 − 4 − 9 ⎦ ⎤ Baris 3 = 3 × b1 + b3 3 \times \text{b1} + \text{b3} 3 × b1 + b3
[ 1 1 1 6 0 1 − 2 − 4 0 0 − 7 − 21 ] \left[\begin{array}{ccc|c}
1 & 1 & 1 & 6 \\
0 & 1 & -2 & -4 \\
0 & 0 & -7 & -21
\end{array}\right] ⎣ ⎡ 1 0 0 1 1 0 1 − 2 − 7 6 − 4 − 21 ⎦ ⎤ Baris 3 = b3 × ( − 1 7 ) \text{b3} \times (-\frac{1}{7}) b3 × ( − 7 1 )
[ 1 1 1 6 0 1 − 2 − 4 0 0 1 3 ] \left[\begin{array}{ccc|c}
1 & 1 & 1 & 6 \\
0 & 1 & -2 & -4 \\
0 & 0 & 1 & 3
\end{array}\right] ⎣ ⎡ 1 0 0 1 1 0 1 − 2 1 6 − 4 3 ⎦ ⎤ Subsitusi Balik
Dari OBE sebelumnya, didapatkan:
x + y + z = 6 y − 2 z = − 4 z = 3 \begin{align*}
x + y + z &= 6 \\
y - 2z &= -4 \\
z &= 3
\end{align*} x + y + z y − 2 z z = 6 = − 4 = 3 Variabel z z z
z = 3 \begin{align*}
z &= 3
\end{align*} z = 3 Variabel y y y
y − 2 z = − 4 y − 2 ( 3 ) = − 4 y − 6 = − 4 y = − 4 + 6 y = 2 \begin{align*}
y - 2z &= -4 \\
y - 2(3) &= -4 \\
y - 6 &= -4 \\
y &= -4 + 6 \\
y &= 2
\end{align*} y − 2 z y − 2 ( 3 ) y − 6 y y = − 4 = − 4 = − 4 = − 4 + 6 = 2 Variabel z z z
x + y + z = 6 x + 2 + 3 = 6 x + 5 = 6 x = 6 − 5 x = 1 \begin{align*}
x + y + z &= 6 \\
x + 2 + 3 &= 6 \\
x + 5 &= 6 \\
x &= 6 - 5 \\
x &= 1
\end{align*} x + y + z x + 2 + 3 x + 5 x x = 6 = 6 = 6 = 6 − 5 = 1 Himpunan Penyelesaian
Solusi:
x = 1 y = 2 z = 3 \begin{align*}
x &= 1 \\
y &= 2 \\
z &= 3
\end{align*} x y z = 1 = 2 = 3 Maka, himpunan penyelesaiannya adalah { 1 , 2 , 3 } \{1, 2, 3\} { 1 , 2 , 3 }
Contoh Penyelesaian SPL Lima Variabel dengan Eliminasi Gauss ¶ Soal ¶ Tentukan himpunan penyelesaian dari sistem persamaan linear berikut:
a + b + c + d + e = 5 a + 2 b + 2 c + 2 d + 2 e = 9 a + 2 b + 3 c + 3 d + 3 e = 12 a + 2 b + 3 c + 4 d + 4 e = 14 a + 2 b + 3 c + 4 d + 5 e = 15 \begin{align*}
a + b + c + d + e &= 5 \\
a + 2b + 2c + 2d + 2e &= 9 \\
a + 2b + 3c + 3d + 3e &= 12 \\
a + 2b + 3c + 4d + 4e &= 14 \\
a + 2b + 3c + 4d + 5e &= 15
\end{align*} a + b + c + d + e a + 2 b + 2 c + 2 d + 2 e a + 2 b + 3 c + 3 d + 3 e a + 2 b + 3 c + 4 d + 4 e a + 2 b + 3 c + 4 d + 5 e = 5 = 9 = 12 = 14 = 15 Langkah Penyelesaian ¶ Buat matriks teraugmentasi
[ 1 1 1 1 1 5 1 2 2 2 2 9 1 2 3 3 3 12 1 2 3 4 4 14 1 2 3 4 5 15 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
1 & 2 & 2 & 2 & 2 & 9 \\
1 & 2 & 3 & 3 & 3 & 12 \\
1 & 2 & 3 & 4 & 4 & 14 \\
1 & 2 & 3 & 4 & 5 & 15
\end{array}\right] ⎣ ⎡ 1 1 1 1 1 1 2 2 2 2 1 2 3 3 3 1 2 3 4 4 1 2 3 4 5 5 9 12 14 15 ⎦ ⎤ Kita anggap:
b1 \text{b1} b1
b1 \text{b1} b1
b1 \text{b1} b1
b1 \text{b1} b1
b1 \text{b1} b1
Baris 2 (Nolkan a 21 \text{a}_{21} a 21
Operasi: − 1 × b1 + b2 -1 \times \text{b1} + \text{b2} − 1 × b1 + b2
[ 1 1 1 1 1 5 0 1 1 1 1 4 1 2 3 3 3 12 1 2 3 4 4 14 1 2 3 4 5 15 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
0 & 1 & 1 & 1 & 1 & 4 \\
1 & 2 & 3 & 3 & 3 & 12 \\
1 & 2 & 3 & 4 & 4 & 14 \\
1 & 2 & 3 & 4 & 5 & 15
\end{array}\right] ⎣ ⎡ 1 0 1 1 1 1 1 2 2 2 1 1 3 3 3 1 1 3 4 4 1 1 3 4 5 5 4 12 14 15 ⎦ ⎤ Baris 3 (Nolkan a 31 \text{a}_{31} a 31 a 32 \text{a}_{32} a 32
Operasi 1: − 1 × b1 + b3 -1 \times \text{b1} + \text{b3} − 1 × b1 + b3
[ 1 1 1 1 1 5 0 1 1 1 1 4 0 1 2 2 2 7 1 2 3 4 4 14 1 2 3 4 5 15 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
0 & 1 & 1 & 1 & 1 & 4 \\
0 & 1 & 2 & 2 & 2 & 7 \\
1 & 2 & 3 & 4 & 4 & 14 \\
1 & 2 & 3 & 4 & 5 & 15
\end{array}\right] ⎣ ⎡ 1 0 0 1 1 1 1 1 2 2 1 1 2 3 3 1 1 2 4 4 1 1 2 4 5 5 4 7 14 15 ⎦ ⎤ Operasi 2: − 1 × b2 + b3 -1 \times \text{b2} + \text{b3} − 1 × b2 + b3
[ 1 1 1 1 1 5 0 1 1 1 1 4 0 0 1 1 1 3 1 2 3 4 4 14 1 2 3 4 5 15 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
0 & 1 & 1 & 1 & 1 & 4 \\
0 & 0 & 1 & 1 & 1 & 3 \\
1 & 2 & 3 & 4 & 4 & 14 \\
1 & 2 & 3 & 4 & 5 & 15
\end{array}\right] ⎣ ⎡ 1 0 0 1 1 1 1 0 2 2 1 1 1 3 3 1 1 1 4 4 1 1 1 4 5 5 4 3 14 15 ⎦ ⎤ Baris 4 (Nolkan a 41 \text{a}_{41} a 41 a 42 \text{a}_{42} a 42 a 43 \text{a}_{43} a 43
Operasi 1: − 1 × b1 + b4 -1 \times \text{b1} + \text{b4} − 1 × b1 + b4
[ 1 1 1 1 1 5 0 1 1 1 1 4 0 0 1 1 1 3 0 1 2 3 3 9 1 2 3 4 5 15 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
0 & 1 & 1 & 1 & 1 & 4 \\
0 & 0 & 1 & 1 & 1 & 3 \\
0 & 1 & 2 & 3 & 3 & 9 \\
1 & 2 & 3 & 4 & 5 & 15
\end{array}\right] ⎣ ⎡ 1 0 0 0 1 1 1 0 1 2 1 1 1 2 3 1 1 1 3 4 1 1 1 3 5 5 4 3 9 15 ⎦ ⎤ Operasi 2: − 1 × b2 + b4 -1 \times \text{b2} + \text{b4} − 1 × b2 + b4
[ 1 1 1 1 1 5 0 1 1 1 1 4 0 0 1 1 1 3 0 0 1 2 2 5 1 2 3 4 5 15 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
0 & 1 & 1 & 1 & 1 & 4 \\
0 & 0 & 1 & 1 & 1 & 3 \\
0 & 0 & 1 & 2 & 2 & 5 \\
1 & 2 & 3 & 4 & 5 & 15
\end{array}\right] ⎣ ⎡ 1 0 0 0 1 1 1 0 0 2 1 1 1 1 3 1 1 1 2 4 1 1 1 2 5 5 4 3 5 15 ⎦ ⎤ Operasi 3: − 1 × b3 + b4 -1 \times \text{b3} + \text{b4} − 1 × b3 + b4
[ 1 1 1 1 1 5 0 1 1 1 1 4 0 0 1 1 1 3 0 0 0 1 1 2 1 2 3 4 5 15 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
0 & 1 & 1 & 1 & 1 & 4 \\
0 & 0 & 1 & 1 & 1 & 3 \\
0 & 0 & 0 & 1 & 1 & 2 \\
1 & 2 & 3 & 4 & 5 & 15
\end{array}\right] ⎣ ⎡ 1 0 0 0 1 1 1 0 0 2 1 1 1 0 3 1 1 1 1 4 1 1 1 1 5 5 4 3 2 15 ⎦ ⎤ Baris 5 (Nolkan a 51 \text{a}_{51} a 51 a 52 \text{a}_{52} a 52 a 54 \text{a}_{54} a 54
Operasi 1: − 1 × b1 + b5 -1 \times \text{b1} + \text{b5} − 1 × b1 + b5
[ 1 1 1 1 1 5 0 1 1 1 1 4 0 0 1 1 1 3 0 0 0 1 1 2 0 1 2 3 4 10 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
0 & 1 & 1 & 1 & 1 & 4 \\
0 & 0 & 1 & 1 & 1 & 3 \\
0 & 0 & 0 & 1 & 1 & 2 \\
0 & 1 & 2 & 3 & 4 & 10
\end{array}\right] ⎣ ⎡ 1 0 0 0 0 1 1 0 0 1 1 1 1 0 2 1 1 1 1 3 1 1 1 1 4 5 4 3 2 10 ⎦ ⎤ Operasi 2: − 1 × b2 + b5 -1 \times \text{b2} + \text{b5} − 1 × b2 + b5
[ 1 1 1 1 1 5 0 1 1 1 1 4 0 0 1 1 1 3 0 0 0 1 1 2 0 0 1 2 3 6 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
0 & 1 & 1 & 1 & 1 & 4 \\
0 & 0 & 1 & 1 & 1 & 3 \\
0 & 0 & 0 & 1 & 1 & 2 \\
0 & 0 & 1 & 2 & 3 & 6
\end{array}\right] ⎣ ⎡ 1 0 0 0 0 1 1 0 0 0 1 1 1 0 1 1 1 1 1 2 1 1 1 1 3 5 4 3 2 6 ⎦ ⎤ Operasi 3: − 1 × b3 + b5 -1 \times \text{b3} + \text{b5} − 1 × b3 + b5
[ 1 1 1 1 1 5 0 1 1 1 1 4 0 0 1 1 1 3 0 0 0 1 1 2 0 0 0 1 2 3 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
0 & 1 & 1 & 1 & 1 & 4 \\
0 & 0 & 1 & 1 & 1 & 3 \\
0 & 0 & 0 & 1 & 1 & 2 \\
0 & 0 & 0 & 1 & 2 & 3
\end{array}\right] ⎣ ⎡ 1 0 0 0 0 1 1 0 0 0 1 1 1 0 0 1 1 1 1 1 1 1 1 1 2 5 4 3 2 3 ⎦ ⎤ Operasi 4: − 1 × b4 + b5 -1 \times \text{b4} + \text{b5} − 1 × b4 + b5
[ 1 1 1 1 1 5 0 1 1 1 1 4 0 0 1 1 1 3 0 0 0 1 1 2 0 0 0 0 1 1 ] \left[\begin{array}{ccccc|c}
1 & 1 & 1 & 1 & 1 & 5 \\
0 & 1 & 1 & 1 & 1 & 4 \\
0 & 0 & 1 & 1 & 1 & 3 \\
0 & 0 & 0 & 1 & 1 & 2 \\
0 & 0 & 0 & 0 & 1 & 1
\end{array}\right] ⎣ ⎡ 1 0 0 0 0 1 1 0 0 0 1 1 1 0 0 1 1 1 1 0 1 1 1 1 1 5 4 3 2 1 ⎦ ⎤ Substitusi Balik
Dari OBE sebelumnya, didapatkan:
a + b + c + d + e = 5 b + c + d + e = 4 c + d + e = 3 d + e = 2 e = 1 \begin{align*}
a + b + c + d + e &= 5 \\
b + c + d + e &= 4 \\
c + d + e &= 3 \\
d + e &= 2 \\
e &= 1
\end{align*} a + b + c + d + e b + c + d + e c + d + e d + e e = 5 = 4 = 3 = 2 = 1 Variabel e e e
e = 1 \begin{align*}
e &= 1
\end{align*} e = 1 Variabel d d d
d + e = 2 d + 1 = 2 d = 2 − 1 d = 1 \begin{align*}
d + e &= 2 \\
d + 1 &= 2 \\
d &= 2 - 1 \\
d &= 1
\end{align*} d + e d + 1 d d = 2 = 2 = 2 − 1 = 1 Variabel c c c
c + d + e = 3 c + 1 + 1 = 3 c + 2 = 3 c = 3 − 2 c = 1 \begin{align*}
c + d + e &= 3 \\
c + 1 + 1 &= 3 \\
c + 2 &= 3 \\
c &= 3 - 2 \\
c &= 1
\end{align*} c + d + e c + 1 + 1 c + 2 c c = 3 = 3 = 3 = 3 − 2 = 1 Variabel b b b
b + c + d + e = 4 b + 1 + 1 + 1 = 4 b + 3 = 4 b = 4 − 3 b = 1 \begin{align*}
b + c + d + e &= 4 \\
b + 1 + 1 + 1 &= 4 \\
b + 3 &= 4 \\
b &= 4 - 3 \\
b &= 1
\end{align*} b + c + d + e b + 1 + 1 + 1 b + 3 b b = 4 = 4 = 4 = 4 − 3 = 1 Variabel a a a
a + b + c + d + e = 5 a + 1 + 1 + 1 + 1 = 5 a + 4 = 5 a = 5 − 4 a = 1 \begin{align*}
a + b + c + d + e &= 5 \\
a + 1 + 1 + 1 + 1 &= 5 \\
a + 4 &= 5 \\
a &= 5 - 4 \\
a &= 1 \\
\end{align*} a + b + c + d + e a + 1 + 1 + 1 + 1 a + 4 a a = 5 = 5 = 5 = 5 − 4 = 1 Himpunan Penyelesaian
Solusi:
a = 1 b = 1 c = 1 d = 1 e = 1 \begin{align*}
a &= 1 \\
b &= 1 \\
c &= 1 \\
d &= 1 \\
e &= 1
\end{align*} a b c d e = 1 = 1 = 1 = 1 = 1 Maka, himpunan penyelesaiannya adalah 1 , 1 , 1 , 1 , 1 {1, 1, 1, 1, 1} 1 , 1 , 1 , 1 , 1