Umum
-
- A =
Tentukan Nilai dari A−1
kemudian hitung kofaktor dari matriks A
- C11 = -3 C12 = 3 C13 = -2
- C21 = 5 C22 = -7 C23 = 6
- C31 = 5 C32 = -5 C33 = 4
menjadi matriks kofaktor
-
cari adjoint dari matriks kofaktor tadi dengan mentranspose matriks kofaktor di atas, sehingga menjadi
dengan metode Sarrus, kita dapat menghitung determinan dari matriks A
Bentuk 
-
- A =
Tentukan Nilai dari A−1
Diawali dengan
![{\displaystyle \left[{\begin{array}{rrr|rrr}1&5&5&1&0&0\\-1&-1&0&0&1&0\\2&4&3&0&0&1\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aadba6d1f7d90224e7cf3ae06c120b6b3cfdce08)
Operasikan Matriks tersebut
![{\displaystyle \left[{\begin{array}{rrr|rrr}1&5&5&1&0&0\\0&4&5&1&1&0\\2&4&3&0&0&1\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/522dccf647ec010fe5092a208762818aba94eba6)
B2 + B1
![{\displaystyle \left[{\begin{array}{rrr|rrr}1&5&5&1&0&0\\0&4&5&1&1&0\\0&-6&-7&-2&0&1\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c7085bca93450bbe2bfdbf88a1cab3b166458e4)
B3 - 2.B1
![{\displaystyle \left[{\begin{array}{rrr|rrr}1&5&5&1&0&0\\0&4&5&1&1&0\\0&0&{\frac {1}{2}}&-{\frac {1}{2}}&{\frac {3}{2}}&1\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72c4ebac1acd495d4124a513dd6d55a2ccf9d309)
B3 + 3/2.B1
![{\displaystyle \left[{\begin{array}{rrr|rrr}1&5&5&1&0&0\\0&1&{\frac {5}{4}}&{\frac {1}{4}}&{\frac {1}{4}}&0\\0&0&1&-1&3&2\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b39592189f20bf67f00b4a724cf7785ba1d0c18f)
1/4.B2, 2.B3
![{\displaystyle \left[{\begin{array}{rrr|rrr}1&5&5&1&0&0\\0&1&0&{\frac {3}{2}}&-{\frac {7}{2}}&-{\frac {5}{2}}\\0&0&1&-1&3&2\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/311f9e35f819ec95b0e6a50add50a45cc11d34ef)
B2 - 5/4.B3
![{\displaystyle \left[{\begin{array}{rrr|rrr}1&5&0&6&-15&-10\\0&1&0&{\frac {3}{2}}&-{\frac {7}{2}}&-{\frac {5}{2}}\\0&0&1&-1&3&2\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96c86cbc3843a97de4a5fc587376c6bf9c59f0bb)
B1 - 5.B3
![{\displaystyle \left[{\begin{array}{rrr|rrr}1&0&0&-{\frac {3}{2}}&{\frac {5}{2}}&{\frac {5}{2}}\\0&1&0&{\frac {3}{2}}&-{\frac {7}{2}}&-{\frac {5}{2}}\\0&0&1&-1&3&2\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83ebe80059650410ff46f96784792f90486aaf52)
B1 - 5.B2
