inversion de matrices

0
0
M∈ Mn(R)N∈ Mn(R)
MN =NM =InN M−1
M
N N0MN =NM =I MN0=N0M=I NMN0=
(NM)N0=N0NMN0=N(MN0) = N N =N0
InIn
A=2 1
3 2 B=2−1 −3 2 
M=
000
213
1−1 2
3I3
A=
a11 0. . . 0
0a22
0
0. . . 0ann
A−1=
a−1
11 0. . . 0
0a−1
22
0
0. . . 0a−1
nn

•M M−1(M−1)−1=M
•M, N ∈ Mn(R)2MN (MN)−1=
N−1M−1
•M Mkk∈N(Mk)−1= (M−1)k
•M, N ∈ Mn(R)MN =I M N
MM−1=M−1M=I
M M−1
(N−1M−1)(MN) = N−1(M−1M)N=N−1IN =N−1N=I(MN)(N−1M−1) = I
N−1M−1MN
Mk(M−1)k= (M−1)kMk=I
M M−1
I
M
MA =MB A =B M −1
A B AB = 0
A
B
M=
−342
−231
2−2 0
M2=
5−4−2
2−1−1
−2 2 2
M2= 2I−M M +M2= 2I1
2M(M+I) = I
M1
2(M+I)M−1=
−1 2 1
−1 2 1
2
1−11
2
(S)
a11x1+a12x2+. . . a1pxp=b1
a21x1+a22x2+. . . a2pxp=b2
an1x1+an2x2+. . . anpxp=bn
A= (aij )16i6n
16j6p
X= (xi)16i6nB= (bi)16i6n(S)
AX =B
A X n i
p
X
j=1
aij xjAX =B⇔ ∀16i6n
p
X
j=1
aij xj=bj(S)

(S)
2x−3y+z= 5
−5x+ 4y−z=−1
x+y−3z=−6
2−3 1
−5 4 −1
1 1 −3
x
y
z
=
5
−1
−6
(S)
A AX =B
X=A−1B(S)
A
0
(Li)16i6n
Li↔LjLi←αLi
Li←Li+αLj
•A
Bk× · · · × B1k
I
•
A
•
A I
Bk0× · · · × B1A
I

A
Li↔Lj
(Li)
(Lj)
1 0 . . . . . . . . . . . . . . . . . . . . . . . . 0
0
1
0. . . . . . . . . 1
1
1
1. . . . . . . . . 0
1
0
0. . . . . . . . . . . . . . . . . . . . . . . . 0 1
Li←αLi(Li)
1 0 . . . . . . . . . . . . 0
0
1
α
1
0
0. . . . . . . . . . . . 0 1
Li←Li+αLj
(Li)
(Lj)
1 0 . . . . . . . . . . . . 0
0
1. . . α
1
0
0. . . . . . . . . . . . 0 1

A I
A=
1 2 −1
2 4 −1
−2−5 3
I=
1 0 0
0 1 0
0 0 1
1 2 −1
0 0 1
0−1 1
1 0 0
−2 1 0
2 0 1
L2←L2−2L1
L3←L3+ 2L1
1 2 −1
0−1 1
0 0 1
1 0 0
2 0 1
−2 1 0
L2↔L3
120
0−1 0
001
−110
4−1 1
−210
L1←L1+L3
L2←L2−L3
100
0−1 0
001
7−1 2
4−1 1
−210
L1←L1+ 2L2
100
010
001
7−1 2
−4 1 −1
−2 1 0
L2← −L2
A−1=
7−1 2
−4 1 −1
−2 1 0
A
A P
P−1AP
A
D=P−1AP n Dn=P−1AnP
A=
10 −6−3
−4 12 2
−4−4 6
P=
3
40−3 2
0−1
2
3
2
3
21 0
P P −1=
2
3
2
3
1
3
−1−11
2
−1
3
1
3
1
6