A. Operasi Baris Elementer

• • Baris pada matriks dapat dipertukarkan

$\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)\to \left(\begin{array}{ccc}4& 5& 6\\ 1& 2& 3\\ 7& 8& 9\end{array}\right)$

 

• • Baris pada matriks dapat dikali dengan konstanta bukan nol

$\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)\to \left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 14& 16& 18\end{array}\right)$

Baris ketiga dikali 2 $(2\times R_3)$

 

• • Baris pada matriks dapat dijumlah/dikurangi dengan baris lainnya

$\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)\to \left(\begin{array}{ccc}-7& -8& -9\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)$

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)

$\text{Menentukan invers matriks }\left(\begin{array}{ccc}0& -3& -2\\ 1& -4& -2\\ -3& 4& 1\end{array}\right)\text{ dengan metode eliminasi Gauss-Jordan}$

 

Akan dilakukan operasi baris elementer untuk mengubah matriks seperti di bawah ini:

 

$\left(\begin{array}{ccccccc}0& -3& -2& |& 1& 0& 0\\ 1& -4& -2& |& 0& 1& 0\\ -3& 4& 1& |& 0& 0& 1\end{array}\right)\stackrel{\text{operasi baris elementer}}{\to }\left(\begin{array}{ccccccc}1& 0& 0& |& ?& ?& ?\\ 0& 1& 0& |& ?& ?& ?\\ 0& 0& 1& |& ?& ?& ?\end{array}\right)$

 

Eliminasi Gauss-Jordan

$$\begin{gathered} \left(\begin{array}{ccccccc}0& -3& -2& |& 1& 0& 0\\ 1& -4& -2& |& 0& 1& 0\\ -3& 4& 1& |& 0& 0& 1\end{array}\right) \\ \stackrel{R_1\text{ ditukar dengan }{R}_2}{\to }\left(\begin{array}{ccccccc}1& -4& -2& |& 0& 1& 0\\ 0& -3& -2& |& 1& 0& 0\\ -3& 4& 1& |& 0& 0& 1\end{array}\right) \end{gathered}$$

 

$$\begin{gathered} \left(\begin{array}{ccccccc}1& -4& -2& |& 0& 1& 0\\ 0& -3& -2& |& 1& 0& 0\\ -3& 4& 1& |& 0& 0& 1\end{array}\right) \\ \stackrel{R_3+3R_1}{\to }\left(\begin{array}{ccccccc}1& -4& -2& |& 0& 1& 0\\ 0& -3& -2& |& 1& 0& 0\\ 0& -8& -5& |& 0& 3& 1\end{array}\right) \end{gathered}$$

 

$$\begin{gathered} \left(\begin{array}{ccccccc}1& -4& -2& |& 0& 1& 0\\ 0& -3& -2& |& 1& 0& 0\\ 0& -8& -5& |& 0& 3& 1\end{array}\right) \\ \stackrel{-\frac{1}{3}{R}_2}{\to }\left(\begin{array}{ccccccc}1& -4& -2& |& 0& 1& 0\\ 0& 1& \frac{2}{3}& |& -\frac{1}{3}& 0& 0\\ 0& -8& -5& |& 0& 3& 1\end{array}\right) \end{gathered}$$

 

$$\begin{gathered} \left(\begin{array}{ccccccc}1& -4& -2& |& 0& 1& 0\\ 0& 1& \frac{2}{3}& |& -\frac{1}{3}& 0& 0\\ 0& -8& -5& |& 0& 3& 1\end{array}\right) \\ \underset{R_1+4R_2}{\overset{R_3+8R_2}{\to }}\left(\begin{array}{ccccccc}1& 0& \frac{2}{3}& |& -\frac{4}{3}& 1& 0\\ 0& 1& \frac{2}{3}& |& -\frac{1}{3}& 0& 0\\ 0& 0& \frac{1}{3}& |& -\frac{8}{3}& 3& 1\end{array}\right) \end{gathered}$$

 

$$\begin{gathered} \left(\begin{array}{ccccccc}1& 0& \frac{2}{3}& |& -\frac{4}{3}& 1& 0\\ 0& 1& \frac{2}{3}& |& -\frac{1}{3}& 0& 0\\ 0& 0& \frac{1}{3}& |& -\frac{8}{3}& 3& 1\end{array}\right) \\ \stackrel{3R_3}{\to }\left(\begin{array}{ccccccc}1& 0& \frac{2}{3}& |& -\frac{4}{3}& 1& 0\\ 0& 1& \frac{2}{3}& |& -\frac{1}{3}& 0& 0\\ 0& 0& 1& |& -8& 9& 3\end{array}\right) \end{gathered}$$

 

$$\begin{gathered} \left(\begin{array}{ccccccc}1& 0& \frac{2}{3}& |& -\frac{4}{3}& 1& 0\\ 0& 1& \frac{2}{3}& |& -\frac{1}{3}& 0& 0\\ 0& 0& 1& |& -8& 9& 3\end{array}\right) \\ \underset{R_1-\frac{2}{3}{R}_3}{\overset{R_2-\frac{2}{3}{R}_3}{\to }}\left(\begin{array}{ccccccc}1& 0& 0& |& 4& -5& -2\\ 0& 1& 0& |& 5& -6& -2\\ 0& 0& 1& |& -8& 9& 3\end{array}\right) \end{gathered}$$

 

Invers matriks adalah $\left(\begin{array}{ccc}4& -5& -2\\ 5& -6& -2\\ -8& 9& 3\end{array}\right)$

D. Operasi Baris Elementer Untuk Menyelesaikan Sistem Persamaan (Eliminasi Gauss-Jordan)

Selesaikan sistem persamaan di bawah ini:

$$\begin{array}{rl}7x+2y+z& =21\\ 3y-z& =5\\ -3x+4y-2z& =-1\end{array}$$

 

Akan dilakukan operasi baris elementer untuk mengubah matriks seperti di bawah ini:

 

$\left(\begin{array}{ccccc}7& 2& 1& |& 21\\ 0& 3& -1& |& 5\\ -3& 4& -2& |& -1\end{array}\right)\stackrel{\text{operasi baris elementer}}{\to }\left(\begin{array}{ccccc}1& 0& 0& |& ?\\ 0& 1& 0& |& ?\\ 0& 0& 1& |& ?\end{array}\right)$

 

Operasi baris elementer

$$\begin{gathered} \left(\begin{array}{ccccc}7& 2& 1& |& 21\\ 0& 3& -1& |& 5\\ -3& 4& -2& |& -1\end{array}\right) \\ \stackrel{\frac{1}{7}{R}_1}{\to }\left(\begin{array}{ccccc}1& \frac{2}{7}& \frac{2}{7}& |& 3\\ 0& 3& -1& |& 5\\ -3& 4& -2& |& -1\end{array}\right) \end{gathered}$$

 

$$\begin{gathered} \left(\begin{array}{ccccc}1& \frac{2}{7}& \frac{1}{7}& |& 3\\ 0& 3& -1& |& 5\\ -3& 4& -2& |& -1\end{array}\right) \\ \stackrel{R_3+3R_1}{\to }\left(\begin{array}{ccccc}1& \frac{2}{7}& \frac{1}{7}& |& 3\\ 0& 3& -1& |& 5\\ 0& \frac{34}{7}& -\frac{11}{7}& |& 8\end{array}\right) \end{gathered}$$

 

$$\begin{gathered} \left(\begin{array}{ccccc}1& \frac{2}{7}& \frac{1}{7}& |& 3\\ 0& 3& -1& |& 5\\ 0& \frac{34}{7}& -\frac{11}{7}& |& 8\end{array}\right) \\ \stackrel{\frac{1}{3}{R}_2}{\to }\left(\begin{array}{ccccc}1& \frac{2}{7}& \frac{1}{7}& |& 3\\ 0& 1& -\frac{1}{3}& |& \frac{5}{3}\\ 0& \frac{34}{7}& -\frac{11}{7}& |& 8\end{array}\right) \end{gathered}$$

 

$$\begin{gathered} \left(\begin{array}{ccccc}1& \frac{2}{7}& \frac{1}{7}& |& 3\\ 0& 1& -\frac{1}{3}& |& \frac{5}{3}\\ 0& \frac{34}{7}& -\frac{11}{7}& |& 8\end{array}\right) \\ \underset{R_3-\frac{34}{7}{R}_2}{\overset{R_1-\frac{2}{7}{R}_2}{\to }}\left(\begin{array}{ccccc}1& 0& \frac{5}{21}& |& \frac{53}{21}\\ 0& 1& -\frac{1}{3}& |& \frac{5}{3}\\ 0& 0& \frac{1}{21}& |& -\frac{2}{21}\end{array}\right) \end{gathered}$$

 

$$\begin{gathered} \left(\begin{array}{ccccc}1& 0& \frac{5}{21}& |& \frac{53}{21}\\ 0& 1& -\frac{1}{3}& |& \frac{5}{3}\\ 0& 0& \frac{1}{21}& |& -\frac{2}{21}\end{array}\right) \\ \stackrel{21R_3}{\to }\left(\begin{array}{ccccc}1& 0& \frac{5}{21}& |& \frac{53}{21}\\ 0& 1& -\frac{1}{3}& |& \frac{5}{3}\\ 0& 0& 1& |& -2\end{array}\right) \end{gathered}$$

 

$$\begin{gathered} \left(\begin{array}{ccccc}1& 0& \frac{5}{21}& |& \frac{53}{21}\\ 0& 1& -\frac{1}{3}& |& \frac{5}{3}\\ 0& 0& 1& |& -2\end{array}\right) \\ \underset{R_2+\frac{1}{3}{R}_3}{\overset{R_1-\frac{5}{21}{R}_3}{\to }}\left(\begin{array}{ccccc}1& 0& 0& |& 3\\ 0& 1& 0& |& 1\\ 0& 0& 1& |& -2\end{array}\right) \end{gathered}$$

 

$x=3,y=1,z=-2$