Definisi :
Misalkan A adalah matriks kuadrat. Fungsi determinan dinyatakan oleh det, dan kita definisikan det(A) sebagai jumlah semua hasil kali elementer beranda dari A. Jumlah det(A) kita namakan determinan A.
Jika kita punya matriks A2×2, = ![\left [ \begin{array}{rr} a_{11} &a_{12}\\ a_{21} &a_{22} \end{array} \right ]](https://s0.wp.com/latex.php?latex=%5Cleft+%5B+%5Cbegin%7Barray%7D%7Brr%7D+a_%7B11%7D+%26a_%7B12%7D%5C%5C+a_%7B21%7D+%26a_%7B22%7D+%5Cend%7Barray%7D+%5Cright+%5D&bg=ffffff&fg=666666&s=0 "\left [ \begin{array}{rr} a_{11} &a_{12}\\ a_{21} &a_{22} \end{array} \right ]")
Kemudian jika kita punya matriks B3×3 = ![\left [ \begin{array}{rrr} a_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33} \end{array} \right ]](https://s0.wp.com/latex.php?latex=%5Cleft+%5B+%5Cbegin%7Barray%7D%7Brrr%7D+a_%7B11%7D%26+a_%7B12%7D%26+a_%7B13%7D%5C%5C+a_%7B21%7D%26+a_%7B22%7D%26+a_%7B23%7D%5C%5C+a_%7B31%7D%26+a_%7B32%7D%26+a_%7B33%7D+%5Cend%7Barray%7D+%5Cright+%5D&bg=ffffff&fg=666666&s=0 "\left [ \begin{array}{rrr} a_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33} \end{array} \right ]")
det(B) = a11a22a33 – a12a23a31 – a13a21a32 – a13a22a31 – a12a21a33 – a11a23a32.
Tapi jika kita memiliki matriks yang berordo 4×4, 5×5, dan seterusnya, bagaimana cara mencari determinannya? Pada tulisan ini saya akan membahas untuk mencari determinan matriks menggunakan Operasi Baris Elementer dengan mereduksi matriks tersebut pada bentuk eselon baris atau membuat Matriks Segitiga Atas atau Matriks Segitiga Bawah.
Contoh matriks segitiga atas pada matriks 4×4 :
![\left [ \begin{array}{rrrr} a_{11}& a_{12}& a_{13}& a_{14}\\ 0& a_{22}& a_{23}& a_{24}\\ 0& 0& a_{33}& a_{34}\\ 0& 0& 0& a_{44}\\ \end{array} \right ]](https://s0.wp.com/latex.php?latex=%5Cleft+%5B+%5Cbegin%7Barray%7D%7Brrrr%7D+a_%7B11%7D%26+a_%7B12%7D%26+a_%7B13%7D%26+a_%7B14%7D%5C%5C+0%26+a_%7B22%7D%26+a_%7B23%7D%26+a_%7B24%7D%5C%5C+0%26+0%26+a_%7B33%7D%26+a_%7B34%7D%5C%5C+0%26+0%26+0%26+a_%7B44%7D%5C%5C+%5Cend%7Barray%7D+%5Cright+%5D&bg=ffffff&fg=666666&s=0 "\left [ \begin{array}{rrrr} a_{11}& a_{12}& a_{13}& a_{14}\\ 0& a_{22}& a_{23}& a_{24}\\ 0& 0& a_{33}& a_{34}\\ 0& 0& 0& a_{44}\\ \end{array} \right ]")
Contoh matriks segitiga bawah pada matriks 4×4 :
![\left [ \begin{array}{rrrr} a_{11}& 0& 0& 0\\ a_{21}& a_{22}& 0& 0\\ a_{31}& a_{32}& a_{33}& 0\\ a_{41}& a_{42}& a_{43}& a_{34}\\ \end{array} \right ]](https://s0.wp.com/latex.php?latex=%5Cleft+%5B+%5Cbegin%7Barray%7D%7Brrrr%7D+a_%7B11%7D%26+0%26+0%26+0%5C%5C+a_%7B21%7D%26+a_%7B22%7D%26+0%26+0%5C%5C+a_%7B31%7D%26+a_%7B32%7D%26+a_%7B33%7D%26+0%5C%5C+a_%7B41%7D%26+a_%7B42%7D%26+a_%7B43%7D%26+a_%7B34%7D%5C%5C+%5Cend%7Barray%7D+%5Cright+%5D&bg=ffffff&fg=666666&s=0 "\left [ \begin{array}{rrrr} a_{11}& 0& 0& 0\\ a_{21}& a_{22}& 0& 0\\ a_{31}& a_{32}& a_{33}& 0\\ a_{41}& a_{42}& a_{43}& a_{34}\\ \end{array} \right ]")
Untuk menghitung determinan matriks dengan cara ini dijamin oleh sebuah teorema berikut.
Teorema :
Jika A adalah matriks segitiga n x n, maka det(A) adalah hasil kali entri-entri diagonal utama yakni det(A) = a11a22 … ann.
Contoh 1 :
Hitung determinan dari matriks A =
![\left [ \begin{array}{rrrrr} 2& 7& -3& 8& 3\\ 0& -3& 7& 5& 1\\ 0& 0& 6& 7& 6\\ 0& 0& 0& 9& 8\\ 0& 0& 0& 0& 4 \end{array} \right ]](https://s0.wp.com/latex.php?latex=%5Cleft+%5B+%5Cbegin%7Barray%7D%7Brrrrr%7D+2%26+7%26+-3%26+8%26+3%5C%5C+0%26+-3%26+7%26+5%26+1%5C%5C+0%26+0%26+6%26+7%26+6%5C%5C+0%26+0%26+0%26+9%26+8%5C%5C+0%26+0%26+0%26+0%26+4+%5Cend%7Barray%7D+%5Cright+%5D&bg=ffffff&fg=666666&s=0 "\left [ \begin{array}{rrrrr} 2& 7& -3& 8& 3\\ 0& -3& 7& 5& 1\\ 0& 0& 6& 7& 6\\ 0& 0& 0& 9& 8\\ 0& 0& 0& 0& 4 \end{array} \right ]")
Penyelesaian :
Karena matriks A merupakan matriks segitiga atas, maka untuk menghitung determinan dapat langsung menggunakan Teorema diatas.
Jadi, det(A) = (2)(-3)(6)(9)(4) = -1296
Contoh 2 :
Hitung determinan dari matriks B =
![\left [ \begin{array}{rrrr} 2& 1& 3& 1\\ 1& 0& 1& 1\\ 0& 2& 1& 0\\ 0& 1& 2& 3 \end{array} \right ]](https://s0.wp.com/latex.php?latex=%5Cleft+%5B+%5Cbegin%7Barray%7D%7Brrrr%7D+2%26+1%26+3%26+1%5C%5C+1%26+0%26+1%26+1%5C%5C+0%26+2%26+1%26+0%5C%5C+0%26+1%26+2%26+3+%5Cend%7Barray%7D+%5Cright+%5D&bg=ffffff&fg=666666&s=0 "\left [ \begin{array}{rrrr} 2& 1& 3& 1\\ 1& 0& 1& 1\\ 0& 2& 1& 0\\ 0& 1& 2& 3 \end{array} \right ]")
Penyelesaian :
Langkah awal untuk mengerjakan ini adalah dengan cara mereduksi matriks baris tersebut menjadi eselon baris dengan menggunakan OBE.
-
baris kedua : 2B2 – B1
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baris ketiga : B3 + 2B2
baris keempat : B4 + B2
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baris keempat : B4 + B3
![\left [ \begin{array}{rrrr} 2& 1& 3& 1\\ 0& -1& -1& 1\\ 0& 2& 1& 0\\ 0& 1& 2& 3 \end{array} \right ]](https://s0.wp.com/latex.php?latex=%5Cleft+%5B+%5Cbegin%7Barray%7D%7Brrrr%7D+2%26+1%26+3%26+1%5C%5C+0%26+-1%26+-1%26+1%5C%5C+0%26+2%26+1%26+0%5C%5C+0%26+1%26+2%26+3+%5Cend%7Barray%7D+%5Cright+%5D&bg=ffffff&fg=666666&s=0 "\left [ \begin{array}{rrrr} 2& 1& 3& 1\\ 0& -1& -1& 1\\ 0& 2& 1& 0\\ 0& 1& 2& 3 \end{array} \right ]")
![\left [ \begin{array}{rrrr} 2& 1& 3& 1\\ 0& -1& -1& 1\\ 0& 0& -1& 2\\ 0& 0& 1& 4 \end{array} \right ]](https://s0.wp.com/latex.php?latex=%5Cleft+%5B+%5Cbegin%7Barray%7D%7Brrrr%7D+2%26+1%26+3%26+1%5C%5C+0%26+-1%26+-1%26+1%5C%5C+0%26+0%26+-1%26+2%5C%5C+0%26+0%26+1%26+4+%5Cend%7Barray%7D+%5Cright+%5D&bg=ffffff&fg=666666&s=0 "\left [ \begin{array}{rrrr} 2& 1& 3& 1\\ 0& -1& -1& 1\\ 0& 0& -1& 2\\ 0& 0& 1& 4 \end{array} \right ]")
![\left [ \begin{array}{rrrr} 2& 1& 3& 1\\ 0& -1& -1& 1\\ 0& 0& -1& 2\\ 0& 0& 0& 6 \end{array} \right ]](https://s0.wp.com/latex.php?latex=%5Cleft+%5B+%5Cbegin%7Barray%7D%7Brrrr%7D+2%26+1%26+3%26+1%5C%5C+0%26+-1%26+-1%26+1%5C%5C+0%26+0%26+-1%26+2%5C%5C+0%26+0%26+0%26+6+%5Cend%7Barray%7D+%5Cright+%5D&bg=ffffff&fg=666666&s=0 "\left [ \begin{array}{rrrr} 2& 1& 3& 1\\ 0& -1& -1& 1\\ 0& 0& -1& 2\\ 0& 0& 0& 6 \end{array} \right ]")
Jadi det(B) = (2)(-1)(-1)(6) = 12
Sumber :
Anton, H., 1992, Aljabar Linier Elementer, Erlangga, Jakarta.
