Jawab dengan caranya secara jelas
Tentukan Adjoin Matriks dan Det.Matriks
1.![A = \left[\begin{array}{ccc}-2&0&1\\1&2&1\\0&6&8\end{array}\right]](https://tex.z-dn.net/?f=+A+%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%260%261%5C%5C1%262%261%5C%5C0%266%268%5Cend%7Barray%7D%5Cright%5D+ "A = \left[\begin{array}{ccc}-2&0&1\\1&2&1\\0&6&8\end{array}\right]")
2.![B = \left[\begin{array}{ccc}1&2&2\\1&3&1\\1&3&2\end{array}\right]](https://tex.z-dn.net/?f=B+%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%262%5C%5C1%263%261%5C%5C1%263%262%5Cend%7Barray%7D%5Cright%5D "B = \left[\begin{array}{ccc}1&2&2\\1&3&1\\1&3&2\end{array}\right]")
Tentukan Adjoin Matriks dan Det.Matriks
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det. A = ((- 2 .2 .8 + 0 .1 .0 + 1 .1 .6) - (0 .1 .8 + - 2 .1 .- 6 + 1 . 2 .0)
= (- 32 + 0 + 6) - ( 0 + 12 + 0)
= - 22 - 12
= - 34
det.B = (1 .3 .2 + 2 .1 .1 + 2 .1 .3) - (2 .1 .2 + 1 . 1 .3 + 2 .3 .1)
= (6 + 2 + 6) - (4 + 3 + 6)
= 14 - 13
= 1