Tugas 4 Universitas Trunodjoyo Madura
March 9, 2026
Membuat operasi baris elementar 4 variabel dan 4 persamaan (eliminasi gauss) ¶ x 1 = 1 , x 2 = 2 , x 3 = 3 , x 4 = 4 , x 5 = 5 x1 = 1, x2 = 2, x3 = 3, x4 = 4, x5 = 5 x 1 = 1 , x 2 = 2 , x 3 = 3 , x 4 = 4 , x 5 = 5
Sistem Persamaan Linear ¶ x 1 + x 2 + x 3 + x 4 = 10 2 x 1 + 3 x 2 + x 3 + x 4 = 15 x 1 + 2 x 2 + 3 x 3 + x 4 = 18 x 1 + x 2 + x 3 + 2 x 4 = 14 x_1 + x_2 + x_3 + x_4 = 10 \\
2x_1 + 3x_2 + x_3 + x_4 = 15 \\
x_1 + 2x_2 + 3x_3 + x_4 = 18 \\
x_1 + x_2 + x_3 + 2x_4 = 14 x 1 + x 2 + x 3 + x 4 = 10 2 x 1 + 3 x 2 + x 3 + x 4 = 15 x 1 + 2 x 2 + 3 x 3 + x 4 = 18 x 1 + x 2 + x 3 + 2 x 4 = 14 Matrix Augmented: ¶ [ 1 1 1 1 10 2 3 1 1 15 1 2 3 1 18 1 1 1 2 14 ] \left[
\begin{array}{cccc|c}
1 & 1 & 1 & 1 & 10 \\
2 & 3 & 1 & 1 & 15 \\
1 & 2 & 3 & 1 & 18 \\
1 & 1 & 1 & 2 & 14
\end{array}
\right] ⎣ ⎡ 1 2 1 1 1 3 2 1 1 1 3 1 1 1 1 2 10 15 18 14 ⎦ ⎤ Operasi Baris Elementer ¶ Kolom 1(pivot R1) Pakai baris pertama sebagai jangkar untuk membuat kolom 1 dibawahnya menjadi 0. R 2 ← R 2 − 2 R 1
R_2 \leftarrow R_2 - 2R_1 R 2 ← R 2 − 2 R 1
2 − 2 ( 1 ) = 0 2 - 2(1) = 0 2 − 2 ( 1 ) = 0
3 − 2 ( 1 ) = 1 3 - 2(1) = 1 3 − 2 ( 1 ) = 1
1 − 2 ( 1 ) = − 1 1 - 2(1) = -1 1 − 2 ( 1 ) = − 1
1 − 2 ( 1 ) = − 1 1 - 2(1) = -1 1 − 2 ( 1 ) = − 1
15 − 2 ( 10 ) = − 5 15 - 2(10) = -5 15 − 2 ( 10 ) = − 5
Matriks:
[ 1 1 1 1 10 0 1 − 1 − 1 − 5 0 1 2 0 8 0 0 0 1 4 ] \left[
\begin{array}{cccc|c}
1 & 1 & 1 & 1 & 10 \\
0 & 1 & -1 & -1 & -5 \\
0 & 1 & 2 & 0 & 8 \\
0 & 0 & 0 & 1 & 4
\end{array}
\right] ⎣ ⎡ 1 0 0 0 1 1 1 0 1 − 1 2 0 1 − 1 0 1 10 − 5 8 4 ⎦ ⎤ Kolom 2(Pivot R2) Pakai baris kedua yang baru untuk membersihkan kolom 2. R 4 R_4 R 4 R 3 R_3 R 3
Matriks:
[ 1 1 1 1 10 0 1 − 1 − 1 − 5 0 0 3 1 13 0 0 0 1 4 ] \left[
\begin{array}{cccc|c}
1 & 1 & 1 & 1 & 10 \\
0 & 1 & -1 & -1 & -5 \\
0 & 0 & 3 & 1 & 13 \\
0 & 0 & 0 & 1 & 4
\end{array}
\right] ⎣ ⎡ 1 0 0 0 1 1 0 0 1 − 1 3 0 1 − 1 1 1 10 − 5 13 4 ⎦ ⎤ Baris 4 (R 4 R_4 R 4 R 4 R_4 R 4 x 3 x_3 x 3
Subtitusi Balik ¶ Mencari x 4 x_4 x 4
1 x 4 = 4 ⇒ x 4 = 4 1x_4 = 4 \;\Rightarrow\; \mathbf{x_4 = 4} 1 x 4 = 4 ⇒ x 4 = 4 Mencari x 3 x_3 x 3
3 x 3 + x 4 = 13 3 x 3 + 4 = 13 ⇒ 3 x 3 = 9 ⇒ x 3 = 3 3x_3 + x_4 = 13\\
3x_3 + 4 = 13 \;\Rightarrow\; 3x_3 = 9 \;\Rightarrow\; \mathbf{x_3 = 3} 3 x 3 + x 4 = 13 3 x 3 + 4 = 13 ⇒ 3 x 3 = 9 ⇒ x 3 = 3 Mencari x 2 x_2 x 2
x 2 − x 3 − x 4 = − 5 x 2 − 3 − 4 = − 5 x 2 − 7 = − 5 ⇒ x 2 = − 5 + 7 ⇒ x 2 = 2 x_2 - x_3 - x_4 = -5\\
x_2 - 3 - 4 = -5\\
x_2 - 7 = -5 \;\Rightarrow\; x_2 = -5 + 7 \;\Rightarrow\; \mathbf{x_2 = 2} x 2 − x 3 − x 4 = − 5 x 2 − 3 − 4 = − 5 x 2 − 7 = − 5 ⇒ x 2 = − 5 + 7 ⇒ x 2 = 2 Mencari x 1 x_1 x 1
x 1 + x 2 + x 3 + x 4 = 10 x 1 + 2 + 3 + 4 = 10 x 1 + 9 = 10 ⇒ x 1 = 1 x_1 + x_2 + x_3 + x_4 = 10\\
x_1 + 2 + 3 + 4 = 10\\
x_1 + 9 = 10 \;\Rightarrow\; \mathbf{x_1 = 1} x 1 + x 2 + x 3 + x 4 = 10 x 1 + 2 + 3 + 4 = 10 x 1 + 9 = 10 ⇒ x 1 = 1 Hasil:
x 1 = 1 , x 2 = 2 , x 3 = 3 , x 4 = 4 x_1 = 1,\; x_2 = 2,\; x_3 = 3,\; x_4 = 4 x 1 = 1 , x 2 = 2 , x 3 = 3 , x 4 = 4