tp maple : algèbre linéaire - correction
TRANSCRIPT
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(1.1)(1.1)
TP 02Maple et algbre linaire
CPGE - Laayoune MP1-2013-2014
Essaidi Ali Exercice 01 :
Construire les lments de la base canonique de = 10 :restart ; with LinearAlgebra :base d NULL :for k from 1 to 10 do f d i / piecewise i = k , 1 ; base d base , Vector 10, f ;
end do :Base d base ;
Base :=
1
0
0
0
0
0
00
0
0
,
0
1
0
0
0
0
00
0
0
,
0
0
1
0
0
0
00
0
0
,
0
0
0
1
0
0
00
0
0
,
0
0
0
0
1
0
00
0
0
,
0
0
0
0
0
1
00
0
0
,
0
0
0
0
0
0
10
0
0
,
0
0
0
0
0
0
01
0
0
,
0
0
0
0
0
0
00
1
0
,
0
0
0
0
0
0
00
0
1
base d NULL :for k from 1 to 10 do
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(2.2)(2.2)
O O
(2.1)(2.1)
(1.2)(1.2)
O O
f d i / f l oor1
abs i K k C 1;
base d base , Vector 10, f ; end do :Base d base ;
Base :=
10
0
0
0
0
0
0
00
,
01
0
0
0
0
0
0
00
,
00
1
0
0
0
0
0
00
,
00
0
1
0
0
0
0
00
,
00
0
0
1
0
0
0
00
,
00
0
0
0
1
0
0
00
,
00
0
0
0
0
1
0
00
,
00
0
0
0
0
0
1
00
,
00
0
0
0
0
0
0
10
,
00
0
0
0
0
0
0
01
Exercice 02 :Vrifier si les matrices suivantes sont nilpotentes et dterminer, si cest le cas, lindicede nilpotence :
A =
1 0 1 0
0 0 0 0
2 2 2 1
1 3 1 1
B =
1 1 1 2
3 3 1 2
2 2 2 0
2 2 2 0
C =
0 2 0 0
1 0 1 0
0 2 0 0
0 2 2 2
restart ; with LinearAlgebra :A d Matrix 1, 0, 1, 0 , 0, 0, 0, 0 , 2, 2, 2, 1 , 1, 3, 1,
1 ;
A :=
1 0 1 0
0 0 0 0
2 2 2 11 3 1 1
A4 d MatrixPower A , 4 ;
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(2.8)(2.8)
(2.2)(2.2)
(2.7)(2.7)
(2.10)(2.10)
(2.6)(2.6)
(2.5)(2.5)
(2.4)(2.4)
(2.3)(2.3)
O O
(2.9)(2.9)
A4 :=
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
verify A4, Zer oMat r i x 4 , ' Matrix ' ;true
A2 d MatrixPower A , 2 ;
A2 :=
1 2 1 1
0 0 0 0
1 1 1 1
0 1 0 0
verify A2, Zer oMat r i x 4 , ' Matrix ' ; false
A3 d MatrixPower A , 3 ;
A3 :=
0 1 0 0
0 0 0 0
0 1 0 0
0 0 0 0
verify A3, Zer oMat r i x 4 , ' Matrix ' ; false
Bd
Matrix 1, 1, 1, 2 , 3, 3, 1, 2 , 2, 2, 2, 0 , 2, 2,2, 0 ;
B :=
1 1 1 2
3 3 1 2
2 2 2 0
2 2 2 0
B4 d MatrixPower B , 4 ;
B4 :=
0 0 0 0
0 0 0 00 0 0 0
0 0 0 0
verify B4, Zer oMat r i x 4 , ' Matrix ' ;true
B2 d MatrixPower B , 2 ;
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(3.2)(3.2)
(3.1)(3.1)
(3.5)(3.5)
(3.6)(3.6)
O O
(3.4)(3.4)
(3.8)(3.8)
O O
3.3.
O O
(3.3)(3.3)
(3.7)(3.7)
O O
O O
Rsoudre les systmes AX = Y et BX = Y pour les deux cas Y =
1
1
1
et Y =
1
1
0
.
restart ; with LinearAlgebra :
A d Matrix 1, 4, 3 , 2, 1, 2 , 3, 2, 0 ;
A :=
1 4 3
2 1 2
3 2 0
B d Matrix 1, 0, 3 , 5, 2, 2 , 6, 2, 1 ;
B :=
1 0 3
5 2 2
6 2 1
NullSpace A ;
ColumnSpace A ;1
0
0
,
0
1
0
,
0
0
1
NullSpace B ;3
172
1
ColumnSpace B ;1
0
1
,
0
1
1
M d MatrixAdd A , B , 1, ;
M :=
1 C 4 3 K 3
2 K 5 1 C 2 2 K 2
3 K 6 2 C 2
S d Determinant M ;S := 106 C 87
2C 31
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(3.10)(3.10)
O O
(3.9)(3.9)
solve S , ;
= 5387
C4
87 7 , = 53
87 K
487
7
Y d Vector 1, 1, 1 ; LinearSolve A , Y ; LinearSolve B , Y ;
Y :=
1
11
731
531
631
Er r or , ( i n Li near Al gebr a: - LA_Mai n: - Li near Sol ve) i nconsi st entsyst emY d Vector 1, 1, 0 ; LinearSolve A , Y ; LinearSolve B , Y ;
Y :=
1
1
0
231
331
1531
1 C 3 _t2 3
3 C 172
_t2 3
_t23
Exercice 04 :Construire la matrice A = a
ij2 M
mn= ) dans les cas suivants :
1 m =3, n = 4, aij
= 2i3
j 2 m =5, n =3, a
ij= i K j 3 m = n
=3, aij
= iji C j
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O O
(4.7)(4.7)
(4.5)(4.5)
O O
O O
O O
O O
(4.1)(4.1)
(4.4)(4.4)
(4.3)(4.3)
(4.2)(4.2)
(4.6)(4.6)
(4.8)(4.8)
O O
O O
O O
O O
O O
4 m =3, n = 4, aij
=1 i C j est paire
0 sinon 5 m = n =7, a
ij=
1 si i = 1 ou j = n
0 sinon 6 m
= n = 5, aij
=0 si i O j
j K i C 1 sinon
On reprend les matrices prcdentes dans le cas m = n = 5. Calculer le rang, la trace, lenoyau, le dterminant et l'inverse (si la matrice est inversible).
restart ; with LinearAlgebra :
f d i , j / 2i
$3j;
f := i, j / 2i 3 j
A d Matrix 3, 4, f ;
A :=
6 18 54 162
12 36 108 324
24 72 216 648
A d Matrix 5, 5, f ;
A :=
6 18 54 162 486
12 36 108 324 972
24 72 216 648 1944
48 144 432 1296 3888
96 288 864 2592 7776
Rank A ; 1
Trace A ;9330
NullSpace A ;81
0
0
0
1
,
9
0
1
0
0
,
27
0
0
1
0
,
3
1
0
0
0
Determinant A ;0
g d i , j / abs i K j ;g := i, j / i K j
B d Matrix 5, 3, g ;
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(4.12)(4.12)O O
(4.10)(4.10)
(4.16)(4.16)
(4.14)(4.14)
(4.11)(4.11)
O O
(4.13)(4.13)
O O
O O
O O
(4.15)(4.15)
O O
(4.9)(4.9)
O O
B :=
0 1 2
1 0 1
2 1 0
3 2 1
4 3 2
B d Matrix 5, 5, g ;
B :=
0 1 2 3 4
1 0 1 2 3
2 1 0 1 2
3 2 1 0 1
4 3 2 1 0
Rank B ;5
Trace B ;0
NullSpace B ;
Determinant B ;32
MatrixInverse B ;3
8
1
2 0 0
1
812
1 12
0 0
0 12
1 12
0
0 0 12
1 12
18
0 0 12
38
h d i , j /i $ j
i C j;
h := i, j / i ji C j
C d Matrix 3, 3, h ;
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O O
O O
(4.24)(4.24)
O O
O O
(4.17)(4.17)
(4.21)(4.21)
(4.20)(4.20)
O O
(4.22)(4.22)
(4.18)(4.18)
O O
(4.23)(4.23)
(4.19)(4.19)
C :=
12
23
34
23
1 65
3
4
6
5
3
2C d Matrix 5, 5, h ;
C :=
12
23
34
45
56
23
1 65
43
107
34
65
32
127
158
45
43
127 2
209
56
107
158
209
52
Rank C ;5
Trace C ;152
NullSpace C ;
Determinant C ;1
4667544000
MatrixInverse C ;450 2100 4200 3780 1260
2100 11025 23520 22050 7560
4200 23520 156800
350400 17640
3780 22050 50400 992252
17640
1260 7560 17640 17640 317525
k d i , j / piecewise is i C j , even , 1 ;k := i, j / piecewise is i C j, even , 1
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O O (4.26)(4.26)
O O
O O
O O
O O
O O
(4.25)(4.25)
O O
(4.32)(4.32)
(4.31)(4.31)
(4.28)(4.28)
(4.27)(4.27)
O O
(4.29)(4.29)
(4.30)(4.30)
d d Matrix 5, 5, k ;
d :=
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
Rank d ;2
Trace d ;5
NullSpace d ;1
0
0
0
1
,
0
1
0
1
0
,
1
0
1
0
0
Determinant d ;0
l d i , j / piecewise i =1 or j = n , 1 ;l := i, j / piecewise i = 1 or j = n, 1
n d 7; E d Matrix 7, 7, l ;n := 7
E :=
1 1 1 1 1 1 1
0 0 0 0 0 0 1
0 0 0 0 0 0 1
0 0 0 0 0 0 1
0 0 0 0 0 0 1
0 0 0 0 0 0 1
0 0 0 0 0 0 1
n d 5; E d Matrix 5, 5, l ;n := 5
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(4.34)(4.34)
O O
O O
O O
O O
O O
(4.38)(4.38)
(4.40)(4.40)
(4.36)(4.36)
O O (4.33)(4.33)
(4.42)(4.42)
(4.32)(4.32)
O O
O O
(4.39)(4.39)
(4.37)(4.37)
(4.35)(4.35)
O O
O O
(4.41)(4.41)
E :=
1 1 1 1 1
0 0 0 0 1
0 0 0 0 1
0 0 0 0 1
0 0 0 0 1
Rank E ;2
Trace E ;2
NullSpace E ;1
1
0
0
0
,
1
0
0
1
0
,
1
0
1
0
0
Determinant E ;0
m d i , j / piecewise i % j , j K i C 1 ;m := i, j / piecewise i % j, j K i C 1
F d Matrix 5, 5, m ;
F :=
1 2 3 4 5
0 1 2 3 4
0 0 1 2 3
0 0 0 1 2
0 0 0 0 1
Rank F ;5
Trace F ;5
NullSpace F ;
Determinant F ;1
MatrixInverse F ;
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(4.43)(4.43)
(5.1)(5.1)
1 2 1 0 0
0 1 2 1 0
0 0 1 2 1
0 0 0 1 2
0 0 0 0 1
Exercice 05 :Construire les lments de la base canonique de M
35= .
restart ; with LinearAlgebra :base d NULL : for i from 1 to 3 do for j from 1 to 5 do
f d k
, l / piecewise i
=k
and
j
=l
, 1 ; base d base , Matrix 3, 5, f ; end do : end do :Base d base ;
Base :=
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
,
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
,
0 0 1 0 0
0 0 0 0 0
0 0 0 0 0
,
0 0 0 1 0
0 0 0 0 0
0 0 0 0 0
,
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0
,
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
,
0 0 0 0 0
0 1 0 0 0
0 0 0 0 0
,
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0
,
0 0 0 0 0
0 0 0 1 0
0 0 0 0 0
,
0 0 0 0 0
0 0 0 0 1
0 0 0 0 0
,
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
,
0 0 0 0 0
0 0 0 0 0
0 1 0 0 0
,
0 0 0 0 0
0 0 0 0 0
0 0 1 0 0
,
0 0 0 0 0
0 0 0 0 0
0 0 0 1 0
,
0 0 0 0 0
0 0 0 0 0
0 0 0 0 1
base d NULL : for i from 1 to 3 do for j from 1 to 5 do
f d k , l / f l oor1
abs i K k C abs j K l C 1;
base d base , Matrix 3, 5, f ;
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(6.1)(6.1)
2.2.
O O
3.3.
(5.2)(5.2)
O O
1.1.
end do : end do :Base d base ;
Base :=
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
,
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
,
0 0 1 0 0
0 0 0 0 0
0 0 0 0 0
,
0 0 0 1 0
0 0 0 0 0
0 0 0 0 0
,
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0
,
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
,
0 0 0 0 0
0 1 0 0 0
0 0 0 0 0
,
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0
,
0 0 0 0 0
0 0 0 1 0
0 0 0 0 0
,
0 0 0 0 0
0 0 0 0 1
0 0 0 0 0
,
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
,
0 0 0 0 0
0 0 0 0 0
0 1 0 0 0
,
0 0 0 0 00 0 0 0 0
0 0 1 0 0
,0 0 0 0 00 0 0 0 0
0 0 0 1 0
,0 0 0 0 00 0 0 0 0
0 0 0 0 1
Exercice 06 :Soit n 2.
Calculer les rangs des matrices cares d'ordre n : ( | i - j |)ij et ( i + j )
ij pour n
variant de 2 10.Quelles conjectures peut-on dduire?Montrer que ces conjectures restent vraies pour n 2 { 2,...,100 }.restart ; with LinearAlgebra :f d i , j / abs i K j ; g d i , j / i C j ;
f := i, j / i K jg := i, j / i C j
for n from 2 to 10 do
n , Rank Matrix n , n , f , Rank Matrix n , n , g ; end do ;
2, 2, 23, 3, 24, 4, 25, 5, 26, 6, 27, 7, 2
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O O
(7.2)(7.2)
(6.3)(6.3)
O O
4.4.
1.1.
(7.4)(7.4)
2.2.
(6.2)(6.2)
(7.3)(7.3)
3.3.
(7.1)(7.1)
O O
O O
8, 8, 29, 9, 2
10, 10, 2
for n from 10 to 100 while n = Rank Matrix n , n , f and 2= Rank Matrix n , n , g do
end do : n ;
101
Exercice 07 :
Soit f 2 L (= 4} dfini par f ( x, y, z, t ) = ( x - y - z - 3t, y + 3 z + t, -x + y + z + 3t, -3 x + y+ 3 z - 2 t ).
Donner la matrice de f dans la base canonique ( e1, e
2, e
3, e
3) de = 4.
Donner la dimension de Im f , ainsi qu'une base.Donner la dimension de ker f , ainsi qu'une base.Les sous-espaces Im f et ker f sont-ils supplmentaires ?
restart ; with LinearAlgebra :f d x , y , z , t / x y z 3 $ t , y C 3 $ z C t , x
C y C z C 3 $ t , 3 $ x C y C 3 $ z 2 $ t ;
f := x, y, z, t / x K y K z K 3 t , y C 3 z C t , x C y C z C 3 t , 3 x C y C 3 z K 2 t
A d GenerateMatrix f x , y , z , t , x , y , z , t 1 ;
A :=
1 1 1 3
0 1 3 1
1 1 1 3
3 1 3 2
Rank A ; F d ColumnSpace A ;3
F :=
1
0
10
,
0
1
00
,
0
0
01
G d NullSpace A ;
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(8.4)(8.4)
O O
S := x1, 12 C x1, 2 x2, 1 C x1, 1 K 1, x1, 1 x1, 2 C x1, 2 x2, 2 C x1, 2 K 1, x1, 2 x2, 1 C x2, 2
2 C x2, 2 K 1,
x2, 1 x1, 1 C x2, 2 x2, 1 C x2, 1 K 1
sol d solve S ; n d nops sol : for k from 1 to n do x d ' x ' :X d Matrix 2, 2, symbol =' x ' :assi gn sol k : eval m X ; end do ;
sol := x1, 1 = 0, x1, 2 = 1, x2, 1 = 1, x2, 2 = 0 , x1, 1 =32
, x1, 2 =12
, x2, 1 =12
, x2, 2 =
32
, x1, 1 =12
, x1, 2 =12
, x2, 1 =12
, x2, 2 =12
, x1, 1 = 1, x1, 2 = 1, x2, 1 = 1, x2, 2
= 1
x := x
X := x1, 1 x1, 2
x2, 1 x2, 2
0 1
1 0
x := x
X := x1, 1 x1, 2
x2, 1 x2, 2
32
12
12
32
x := x
X := x
1, 1 x
1, 2
x2, 1 x2, 2
12
12
12
12
x := x
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(9.6)(9.6)
(9.5)(9.5)
O O
O O
(10.1)(10.1)
O O O O
O O
C 3 x2, 2 C x3, 2 K 3 x1, 1 K 3 x1, 3, 2 x1, 2 C 3 x2, 2 K 3 x3, 1 K x3, 2 K 3 x3, 3, x1, 3 C x2, 3K x3, 3 K x2, 1 C x2, 2, 2 x1, 3 C 3 x2, 3 C x3, 3 K x1, 1 C x1, 2
sol ut i ons d solve S ;
solutions := x1, 1 = x1, 1, x1, 2 =152
x2, 1 K92
x1, 3, x1, 3 = x1, 3, x2, 1 = x2, 1, x2, 2 = x1, 1
K 92
x2, 1 C 72 x1, 3, x2, 3 = 12
x2, 1 K 12 x1, 3, x3, 1 = 92
x2, 1 K 52 x1, 3, x3, 2 = 6 x2, 1
K 3 x1, 3, x3, 3 = 4 x1, 3 K 6 x2, 1 C x1, 1
M d subs solutions 1 , evalm X ;
M :=
x1, 1152
x2, 1 K92
x1, 3 x1, 3
x2, 1 x1, 1 K92
x2, 1 C72
x1, 312
x2, 1 K12
x1, 3
92
x2, 1 K 52 x1, 3 6 x2, 1 K 3 x1, 3 4 x1, 3 K 6 x2, 1 C x1, 1
Exercice 10 :
Rsoudre le systme
2 x C 3 y C z = 1
x C 2 y C 3 z = 2
3 x C y C 2 z = 1
.
restart ;sol ve 2 $x C 3$y C z =1, x C 2$y C 3$z =2, 3 $x C y C 2$z = 1 ;
x = 89
, y = 79
, z = 49
Exercice 11 :
Donner une base du sous-espace de = 5 dfini par : x
1 C 2 x
2 C x
3 C 3 x
4 C x
5 = 0
x2 C x
3 2 x
4 C 2 x
5 = 0
2 x1 C x
2 5 x
3 4 x
5 = 0
.
restart ; with LinearAlgebra :
sys d x 1
C 2 x 2 C x 3 C 3 x 4 C x 5 =0, x 2 C x 3 K 2 x 4 C 2 x 5 =0 , 2 x 1
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1.1.
O O
2.2.
O O
(11.2)(11.2)
(11.1)(11.1)
(11.3)(11.3)
C x 2 K 5 x 3 K 4 x 5 =0 ;
vars d x 1, x 2, x 3, x 4, x 5 ;
sys := x1 C 2 x2 C x3 C 3 x4 C x5 = 0, x2 C x3 K 2 x4 C 2 x5 = 0, 2 x1 C x2 K 5 x3 K 4 x5 = 0
vars := x1, x2, x3, x4, x5
Ad
GenerateMatrix sys , vars 1 ;
A :=
1 2 1 3 1
0 1 1 2 2
2 1 5 0 4
NullSpace A ;3
2
00
1
,
10
5
31
0
Exercice 12 :On considre les vecteurs v
1= (2, 4, 5, 6), v
2 = (1, 2, 5, 3), v
3 = (3, 1, - 1, 0) et v
4 = (4,
3, 4, 3).Donner une base du sous-espace vectoriel engendr par v
1, v
2 , v
3 et v
4 .
Le vecteur w = (4, 3, - 1, 3) est-il dans ce sous-espace vectoriel? Si oui,exprimez-le comme combinaison linaire des vecteurs de la base.restart ; with LinearAlgebra :v1 d Vector 2, 4, 5, 6 ; v2 d Vector 1, 2, 5, 3 ; v3
d Vector 3, 1, 1, 0 ; v4 d Vector 4, 3, 4, 3 ;
v1 :=
2
4
5
6
v2 :=
1
2
5
3
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O O
O O
O O
(12.2)(12.2)
(12.6)(12.6)O O
(12.1)(12.1)
(12.7)(12.7)
(12.8)(12.8)
(12.3)(12.3)
O O
(12.5)(12.5)
(12.4)(12.4)
O O
v3 :=
3
1
1
0
v4 :=
43
4
3
Basis v1 , v2 , v3 , v4 ;1
2
5
3
,
3
1
1
0
,
2
4
5
6
w d Vector 4, 3, 1, 3 ;
w :=
4
3
1
3
Determinant Matrix v1 , v2 , v3 , w ;0
U d x $v1 C y $v2 C z $v3 K w ;
U :=
2 x C y C 3 z K 4
4 x C 2 y C z K 3
5 x C 5 y K z C 1
6 x C 3 y K 3
Sys d convert U , set ;Sys := 6 x C 3 y K 3, 2 x C y C 3 z K 4, 4 x C 2 y C z K 3, 5 x C 5 y K z C 1
solve Sys ; x = 1, y = 1, z = 1
LinearSolve Matrix v1 , v2 , v3 , w ; # Autre mthode1
1
1
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5.5.
(13.2)(13.2)
O O
(13.1)(13.1)
O O
O O
O O
(13.6)(13.6)
2.2.
O O
3.3.
(13.4)(13.4)
(13.5)(13.5)
O O
1.1.
O O
O O
O O
(13.9)(13.9)
(13.8)(13.8)
(13.3)(13.3)
O O
4.4.
(13.7)(13.7)
Exercice 13 :Soit lapplication f : =
2 X / =
2 X dfinie par f P = le reste de la division
euclidienne de X 3P par X K 1 X K 2 X K 3 .
Construire f et calculer f 1 , f X , f X 2 et gnralement f x C yX C zX 2 .Montrer que f est un endomorphisme de =
2 X .
Donner la matrice de f .Montrer que f est un automorphisme de =
2 X .
Dterminer f 1.restart ; with LinearAlgebra :
f d P / rem X 3
$P , X K 1 $ X K 2 $ X K 3 , X ; f := P / rem X 3 P , X K 1 X K 2 X K 3 , X
sort f 1 ;6 X 2 K 11 X C 6
sort f X ;25 X 2 K 60 X C 36
sort f X 2
;90 X 2 K 239 X C 150
f x C y $X C z $X 2
;90 z C 25 y C 6 x X 2 C 60 y K 239 z K 11 x X C 6 x C 150 z C 36 y
expand f $P C $Q K $ f P K $ f Q ;0
M d GenerateMatrix seq coeff f x C y $X C z $X 2
, X , k , k =0 . . 2 ,
x , y , z 1 ;
M :=
6 36 150
11 60 239
6 25 90
Determinant M ;216
N d MatrixInverse M ;
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O O
(13.13)(13.13)
(13.10)(13.10)
O O
(13.11)(13.11)
O O
O O
(13.9)(13.9)
(13.12)(13.12)
N :=
575216
8536
116
3718
53
1
85
216
11
36
1
6V d Vector x, y, z ;
V :=
x
y
z
K d MatrixVectorMultiply N , V ;
K :=
575216
x C 8536
y C 116
z
3718
x K 53
y K z
85216
x C 1136
y C 16
z
U d Matrix 1, X , X 2
;
U := 1 X X 2
sort MatrixVectorMultiply U , K 1 ;85
216 xC
1136 y
C16 z X
2 C3718 x
K53 y
K z X
C575216 x
C8536 y
C116 z