A=
( [1] [1] [0]
[2] [3] [3]
[3] [1] [-2] )
Determiante 'normal' berechnen.
Det(A) = -6+9+0 - (0+3-4) = 5-(-1) = 6
6 modulo 2 ist 0. ==> Nicht invertierbar für p=2
6 modulo 5 ist 1 ==> Invertierbar für p=5
Also nun noch die Inverse für p=5 berechnen.
Zum Beispiel die folgende Matrix mit geeigneten Zeilenumformungen so umformen, dass die Einheitsmatrix links erscheint. Rechne am besten gleich modulo 5.
A= [modulo 5 gerechnet]
( [1] [1] [0] 1 0 0
[2] [3] [3] 0 1 0
[3] [1] [3] 0 0 1)
-->
( [1] [1] [0] 1 0 0
[0] [1] [3] -2 1 0
[0] [-2] [3] -3 0 1)
--->
( [1] [1] [0] 1 0 0
[0] [1] [3] 3 1 0
[0] [3] [3] 2 0 1)
-->
( [1] [1] [0] 1 0 0
[0] [1] [3] 3 1 0
[0] [0] [-6] -7 -3 1)
--->
( [1] [1] [0] 1 0 0
[0] [1] [3] 3 1 0
[0] [0] [-1] 3 2 1)
-->
( [1] [1] [0] 1 0 0
[0] [1] [3] 3 1 0
[0] [0] [1] -3 -2 -1)
--->
( [1] [1] [0] 1 0 0
[0] [1] [0] 12 7 3
[0] [0] [1] -3 -2 -1)
---->
( [1] [0] [0] -1 -2 0
[0] [1] [0] 2 2 3
[0] [0] [1] 2 3 4)
-->
( [1] [0] [0] 4 3 -3
[0] [1] [0] 2 2 3
[0] [0] [1] 2 3 4)
--->
( [1] [0] [0] 4 3 2
[0] [1] [0] 2 2 3
[0] [0] [1] 2 3 4)
Multipliziere zur Kontrolle
A^{-1} * A
=
(4+6+6, 4+9+2, 9+6
......
......)
= (modulo 5)
( 1, 0, 0
........
.......)
Einheitsmatrix rausgekommen? ==> ok.