A. Operasi Baris Elementer
-
- Baris pada matriks dapat dipertukarkan
\(\begin{pmatrix}
{\color{red}1} & {\color{red}2} & {\color{red}3} \\
{\color{blue}4} & {\color{blue}5} & {\color{blue}6} \\
7 & 8 & 9 \\
\end{pmatrix}
\rightarrow
\begin{pmatrix}
{\color{blue}4} & {\color{blue}5} & {\color{blue}6} \\
{\color{red}1} & {\color{red}2} & {\color{red}3} \\
7 & 8 & 9 \\
\end{pmatrix}\)
-
- Baris pada matriks dapat dikali dengan konstanta bukan nol
\(\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
{\color{red}7} & {\color{red}8} & {\color{red}9} \\
\end{pmatrix}
\rightarrow
\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
{\color{red}14} & {\color{red}16} & {\color{red}18}\\
\end{pmatrix}\)
Baris ketiga dikali 2 \((2 \times R_3)\)
-
- Baris pada matriks dapat dijumlah/dikurangi dengan baris lainnya
\(\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \\
\end{pmatrix}
\rightarrow
\begin{pmatrix}
{\color{red}-7} & {\color{red}-8} & {\color{red}-9} \\
4 & 5 & 6 \\
7 & 8 & 9 \\
\end{pmatrix}\)
Baris pertama dikurangi oleh dua kali baris kedua \((R_1 - 2R_2)\)
B. Efek Operasi Baris Elementer pada Determinan
Operasi baris elementer pada suatu matriks akan mengubah nilai determinan dari matriks tersebut dengan cara:
| Tindakan | Perubahan Pada Nilai Determinan |
| Menukar salah satu baris dengan baris lainnya | Determinan dikali −1 |
| Mengalikan salah satu baris dengan konstanta k | Determinan dikali k |
| Menambah/mengurangi salah satu baris dengan baris lainnya | Tidak ada perubahan |
C. Operasi Baris Elementer Untuk Menentukan Invers Matriks (Eliminasi Gauss-Jordan)
\(\text{Menentukan invers matriks }\begin{pmatrix}0 & -3 & -2 \\1 & -4 & -2 \\-3 & 4 & 1 \\ \end{pmatrix} \text{ dengan metode eliminasi Gauss-Jordan}\)
Akan dilakukan operasi baris elementer untuk mengubah matriks seperti di bawah ini:
\(\begin{pmatrix}
0 & -3 & -2 & | & 1 & 0 & 0 \\
1 & -4 & -2 & | & 0 & 1 & 0 \\
-3 & 4 & 1 & | & 0 & 0 & 1 \\
\end{pmatrix}
\xrightarrow {\text{operasi baris elementer}}
\begin{pmatrix}
1 & 0 & 0 & | & ? & ? & ? \\
0 & 1 & 0 & | & ? & ? & ? \\
0 & 0 & 1 & | & ? & ? & ? \\
\end{pmatrix}\)
Eliminasi Gauss-Jordan
\(\begin{pmatrix}
0 & -3 & -2 & | & 1 & 0 & 0 \\
1 & -4 & -2 & | & 0 & 1 & 0 \\
-3 & 4 & 1 & | & 0 & 0 & 1 \\
\end{pmatrix} \\
\xrightarrow{R_1 \text{ ditukar dengan }R_2}
\begin{pmatrix}
1 & -4 & -2 & | & 0 & 1 & 0 \\
0 & -3 & -2 & | & 1 & 0 & 0 \\
-3 & 4 & 1 & | & 0 & 0 & 1 \\
\end{pmatrix}\)
\(\begin{pmatrix}
1 & -4 & -2 & | & 0 & 1 & 0 \\
0 & -3 & -2 & | & 1 & 0 & 0 \\
-3 & 4 & 1 & | & 0 & 0 & 1 \\
\end{pmatrix} \\
\xrightarrow{R_3 + 3R_1}
\begin{pmatrix}
1 & -4 & -2 & | & 0 & 1 & 0 \\
0 & -3 & -2 & | & 1 & 0 & 0 \\
0 & -8 & -5 & | & 0 & 3 & 1 \\
\end{pmatrix}\)
\(\begin{pmatrix}
1 & -4 & -2 & | & 0 & 1 & 0 \\
0 & -3 & -2 & | & 1 & 0 & 0 \\
0 & -8 & -5 & | & 0 & 3 & 1 \\
\end{pmatrix} \\
\xrightarrow{-\frac{1}{3}R_2}
\begin{pmatrix}
1 & -4 & -2 & | & 0 & 1 & 0 \\
0 & 1 & \frac{2}{3} & | & -\frac{1}{3} & 0 & 0 \\
0 & -8 & -5 & | & 0 & 3 & 1 \\
\end{pmatrix}\)
\(\begin{pmatrix}
1 & -4 & -2 & | & 0 & 1 & 0 \\
0 & 1 & \frac{2}{3} & | & -\frac{1}{3} & 0 & 0 \\
0 & -8 & -5 & | & 0 & 3 & 1 \\
\end{pmatrix} \\
\xrightarrow[R_1 + 4R_2]{R_3 + 8 R_2}
\begin{pmatrix}
1 & 0 & \frac{2}{3} & | & -\frac{4}{3} & 1 & 0 \\
0 & 1 & \frac{2}{3} & | & -\frac{1}{3} & 0 & 0 \\
0 & 0 & \frac{1}{3} & | & -\frac{8}{3} & 3 & 1 \\
\end{pmatrix}\)
\(\begin{pmatrix}
1 & 0 & \frac{2}{3} & | & -\frac{4}{3} & 1 & 0 \\
0 & 1 & \frac{2}{3} & | & -\frac{1}{3} & 0 & 0 \\
0 & 0 & \frac{1}{3} & | & -\frac{8}{3} & 3 & 1 \\
\end{pmatrix} \\
\xrightarrow{3R_3}
\begin{pmatrix}
1 & 0 & \frac{2}{3} & | & -\frac{4}{3} & 1 & 0 \\
0 & 1 & \frac{2}{3} & | & -\frac{1}{3} & 0 & 0 \\
0 & 0 & 1 & | & -8 & 9 & 3 \\
\end{pmatrix}\)
\(\begin{pmatrix}
1 & 0 & \frac{2}{3} & | & -\frac{4}{3} & 1 & 0 \\
0 & 1 & \frac{2}{3} & | & -\frac{1}{3} & 0 & 0 \\
0 & 0 & 1 & | & -8 & 9 & 3 \\
\end{pmatrix} \\
\xrightarrow[R_1 - \frac{2}{3}R_3]{R_2 - \frac{2}{3}R_3}
\begin{pmatrix}
1 & 0 & 0 & | & 4 & -5 & -2 \\
0 & 1 & 0 & | & 5 & -6 & -2 \\
0 & 0 & 1 & | & -8 & 9 & 3 \\
\end{pmatrix}\)
Invers matriks adalah \(\begin{pmatrix} 4 & -5 & -2 \\ 5 & -6 & -2 \\ -8 & 9 & 3 \end{pmatrix}\)
D. Operasi Baris Elementer Untuk Menyelesaikan Sistem Persamaan (Eliminasi Gauss-Jordan)
Selesaikan sistem persamaan di bawah ini:
\begin{equation*}
\begin{split}
7x + 2y + z & = 21\\
3y - z & = 5\\
-3x + 4y -2z & = -1
\end{split}
\end{equation*}
Akan dilakukan operasi baris elementer untuk mengubah matriks seperti di bawah ini:
\(\begin{pmatrix}
7 & 2 & 1 & | & 21 \\
0 & 3 & -1 & | & 5 \\
-3 & 4 & -2 & | & -1 \\
\end{pmatrix}
\xrightarrow {\text{operasi baris elementer}}
\begin{pmatrix}
1 & 0 & 0 & | & ? \\
0 & 1 & 0 & | & ? \\
0 & 0 & 1 & | & ? \\
\end{pmatrix}\)
Operasi baris elementer
\(\begin{pmatrix}
7 & 2 & 1 & | & 21 \\
0 & 3 & -1 & | & 5 \\
-3 & 4 & -2 & | & -1 \\
\end{pmatrix} \\
\xrightarrow{\frac{1}{7} R_1}
\begin{pmatrix}
1 & \frac{2}{7} & \frac{2}{7} & | & 3 \\
0 & 3 & -1 & | & 5 \\
-3 & 4 & -2 & | & -1 \\
\end{pmatrix}\)
\(\begin{pmatrix}
1 & \frac{2}{7} & \frac{1}{7} & | & 3 \\
0 & 3 & -1 & | & 5 \\
-3 & 4 & -2 & | & -1 \\
\end{pmatrix} \\
\xrightarrow{R_3 + 3R_1}
\begin{pmatrix}
1 & \frac{2}{7} & \frac{1}{7} & | & 3 \\
0 & 3 & -1 & | & 5 \\
0 & \frac{34}{7} & -\frac{11}{7} & | & 8 \\
\end{pmatrix}\)
\(\begin{pmatrix}
1 & \frac{2}{7} & \frac{1}{7} & | & 3 \\
0 & 3 & -1 & | & 5 \\
0 & \frac{34}{7} & -\frac{11}{7} & | & 8 \\
\end{pmatrix} \\
\xrightarrow{\frac{1}{3}R_2}
\begin{pmatrix}
1 & \frac{2}{7} & \frac{1}{7} & | & 3 \\
0 & 1 & -\frac{1}{3} & | & \frac{5}{3} \\
0 & \frac{34}{7} & -\frac{11}{7} & | & 8 \\
\end{pmatrix}\)
\(\begin{pmatrix}
1 & \frac{2}{7} & \frac{1}{7} & | & 3 \\
0 & 1 & -\frac{1}{3} & | & \frac{5}{3} \\
0 & \frac{34}{7} & -\frac{11}{7} & | & 8 \\
\end{pmatrix} \\
\xrightarrow[R_3 - \frac{34}{7}R_2]{R_1 - \frac{2}{7} R_2}
\begin{pmatrix}
1 & 0 & \frac{5}{21} & | & \frac{53}{21} \\
0 & 1 & -\frac{1}{3} & | & \frac{5}{3} \\
0 & 0 & \frac{1}{21} & | & -\frac{2}{21} \\
\end{pmatrix}\)
\(\begin{pmatrix}
1 & 0 & \frac{5}{21} & | & \frac{53}{21} \\
0 & 1 & -\frac{1}{3} & | & \frac{5}{3} \\
0 & 0 & \frac{1}{21} & | & -\frac{2}{21} \\
\end{pmatrix} \\
\xrightarrow{21R_3}
\begin{pmatrix}
1 & 0 & \frac{5}{21} & | & \frac{53}{21} \\
0 & 1 & -\frac{1}{3} & | & \frac{5}{3} \\
0 & 0 & 1 & | & -2 \\
\end{pmatrix}\)
\(\begin{pmatrix}
1 & 0 & \frac{5}{21} & | & \frac{53}{21} \\
0 & 1 & -\frac{1}{3} & | & \frac{5}{3} \\
0 & 0 & 1 & | & -2 \\
\end{pmatrix} \\
\xrightarrow[R_2 + \frac{1}{3}R_3]{R_1 - \frac{5}{21}R_3}
\begin{pmatrix}
1 & 0 & 0 & | & 3 \\
0 & 1 & 0 & | & 1 \\
0 & 0 & 1 & | & -2 \\
\end{pmatrix}\)
\(x = 3, y = 1, z = -2\)