TD 3 : Matrices et Déterminants Exercice 1 Exercice 2 Exercice 3

A=
12345
23456
34567
45678
, B =
1 1 0 1
3 2 −1 3
a3−2 0
−1 0 −4 3
et C=
1 4 −124
2 0 −3−1 7
−2 3 2 1 4 
.
M=A B
C D rang(M) = rang(A) + rang(D−CA−1B)
M∈ Mn(K)
X1MX1
X1
M M = 0. . .
M1!
M1∈ Mn−1(K)tr(M1) = 0
M
A B M =AB −
BA
M
In+M MX = 0 t(MX)(MX))
A= (In−M)(In+M)−1tAA =In
AK
Ak=In(k6= 0) B=In+A+A2+. . . +Ak−1u, v
KnAet B
Ker(u−id) = Imv, Im(u−id) = Kerv, Kerv ⊕Imv =Kn
trB =k rangA
fR4R3
(I, J, K, L) (i, j, k)
4 5 −7 7
2 1 −1 3
1−121
B= (I, J, 4I+J−3L, −7I+K+ 5L)
B0= 4i+ 2j+k, 5i+j−k, k)
fBet B0

A=
1100
0110
0011
0001
B=
1234
0123
0012
0001
A
B
C0
nC1 n. . . Cp n
C0
n+1 C1 n+1 . . . Cp
n+1
. . .
C0
n+pC1 n+p. . . Cp
n+p
a1 0 0 . . . 0 0 0
1a1 0 . . . 0 0 0
0 1 a1. . . 0 0 0
0 0 1 a . . . 0 0 0
. . .
0 0 . . . 0. . . 1a1
0 0 . . . 0. . . 0 1 a
a1b1. . . . . . b1
b2a2+b2b2. . . b2
bn−1
bn. . . . . . bnan+bn
0 1 2 . . . n −1
1 0 1
2 1 1 2
1
n−1. . . 2 1 0
A, B ∈ Mn(R)
AB =BA det(A2+B2)≥0
M=A B
B A detM =det(A+B)det(A−B)
A comA
com(com(A)A
rangA ≤n−2comA = 0
rangA =n−1rang(comA) = 1
com(comA) = det(A)n−2A
A B com(AB) = (comA)(comB)
A, B, C, D ∈ Mn(K)A AC =CA
M=A B
C D ∈ M2n(K)det(M) = det(AD −CB)
u, v CE
u v uov =vou
det(v) = 0 v
det(idE+v) = 1
det(u+v) = detu
F G
f E detf = (detf)|F(detf )|G