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Metoda Gauss de reducere a formelor patratice la forma canonica
I. V = R3
1. Cazul a11 ≠ 0
Exemplu:
h : R3 R, h(x) = x12 + 2x1x2
+ x22 + 2x2x3 + x32
Scriem matricea asociata formei patratice:
2 3
A = 0 1 2
0 2 1
1 2 0
0 1 2 = 1 + 0 + 0 - (0 + 4 +0) = -3
2 1
0 1 =
0 nu se poate aplica metoda Iacobi
0 2
Aplicam metoda Gauss de reducere a formelor patratice la forma canonica: cazul 1 - curs " Matematica p 20420d32u entru economisti ", Ed. Universitara, pg. 19
a11 = 1 ≠ 0
h(x) = x12 + 2x1x2 + x22 - x22 + 2x2x3 + x32 = (x1 + x2)2 - x22 + (x2 + x3)2
unde: ξ1 = x1 + x2
ξ2
= x2 h(x)
= ξ12 - ξ22 + ξ32
ξ = x2 + x3
2. Cazul a11 = 0
Cautam un indice j , astfel incat ajj ≠ 0
Exemplu:
h: R3 R, h(x) = x22 + x32
+ 2x1x2 + 2x1x3
a22 = 1 + 0
Renumerotam variabilele astfel incat variabila j sa devina prima variablia, rezultand astfel cazul 1)
y1 = x2
y2 = x3
y3 = x1
h(x) = y12 + y22
+ 2y1y3 + 2y2y3 = y12
+ 2y1y3 + y32 + y22
+ 2y2y3 - y32 =
(y1 + y3)2 + y22 + 2y2y3 + y32 - 2y32 = (y1 + y3)2 + (y2 + y3)2 - 2y32
Notam :
ξ1 = y1 + y3
ξ2 = y2 + y3
ξ3 = y3
h(x) = ξ12 + ξ22
- 2 ξ32
3. Cazul a11
= 0, oricare j , exista un indice i, astfel incat a11≠ 0
Exemplu :
h: R3 R, h(x) = x1x2 + x2x3
+ x1x3
a11 = a22 = a33 = 0
Facem o transformare liniara astfel incat problema sa se reduca la cazul 1.
y1 = (x1 + x2)/2 x1 + x2 =
2y1
y2 = (x1 - x2)/2 x1
- x2 = 2y2
y3 = x3 2x1 / =
2(y1 + y2) x1 = y1 + y2
x2 = y1 - y2
x3 = y3
h(x) = y12 - y22
+ y1y3 - y2y3 + y1y3
+ y2y3 = y12 - y22
+ 2y1y3
h(x) = y12 - y22
+ 2y1y3
a11
= 1 ≠ 0 caz 1
h(x) = y12 - y22 + 2y1y3 + y32 - y32 = (y1 + y3)2 - y22 - y32
Notam :
ξ1 = y1 + y3
ξ2 = y2 h(x) = ξ12 - ξ22
- ξ32 + 2x2
ξ3 = y3
II. V = R4
Exemplul 1 :
h : R4 R, h(x) = x12
+ x22 + x32 - 2x42
- 2x1x2 + 2x1x3 - 2x1x4
+ 2x1x4 + 2x2x3 - 4x2x4
a11 = 1 ≠ 0
Alegem toti termenii care contin pe x1 si formam un patrat cu acestia.
h(x) = (x12 - 2x1x2 + 2x1x3 - 2x1x4) = x22 + x32 - 2x42 + 2x2x3 - 4x2x4 = (x1 - x2 + x3 - x4)2 - x22 - x32 - x42 + 2x2x3 - 2x2x4 + 2x3x4 + x22 + x32 - 2x42 + 2x2x3 - 4x2x4
h(x) = (x1 - x2 + x3 - x4)2
- 3x42 + 4x2x3 - 6x2x4
+ 2x3x4
Procedam la fel pentru forma patratica h1(x) = -3x42 + 4x2x3 - 6x2x4 + 2x3x4
h(x) = (x1 - x2 + x3 - x4)2
- 3(x42 + 2x2x4 - 2x3x4
/ 3) + 4x2x3 = (x1 - x2 + x3
- x4)2 -3[(x4 + x2 - x3
/ 3)2 - x22 - x32 / 9 +
2x2x3 / 3] + 4x2x3
h(x) = (x1 - x2
+ x3 - x4)2 - 3(x2 - x3
/ 3 + x4)2 + 3(x22 + 2x2x3
/ 3) + x32 / 3
h(x) = (x1 - x2
+ x3 - x4)2 - 3(x2 - x3
/ 3 + x4)2 + 3(x2 + x3 / 3)2
Notam:
ξ = x1 - x2
+ x3 - x4
ξ2 = x2 - x3/3 + x4
ξ3 = x2 + x3/3
ξ4 = x4
h(x) = ξ12 - 3 ξ22
+ 3 ξ32
Exemplul 2 :
Cazul a11 = 0
Cautam un indice j , astfel incat ajj ≠ 0
h : R4 R,
h(x) = 4x22 + x32/4 + 2x42
+ 4x1x2 + 2x3x4
Renumerotam variabilele, astfel incat variabila j sa devina prima variabila, rezultand cazul 1.
y1 = x2
y2 = x3
y3 = x1
y4 = x4
h(x) = 4y12 + y22/4 + 2y42
+ 4y1y3 + 2y2y4 = (2y1 +
y3)2 - y32 + (y2/2 + y4)2
+ y42
Notam :
ξ1 = 2y1 + y3
ξ2 = y3
ξ3 = y2/2 + y4
ξ4 = y4
h(x) = ξ12 + ξ22 - 2 ξ32
Exemplul 3 :
h : R4 R, h(x) =
2x1x2 - x3x4
Notam :
x1
= y1 - y2
x2 = y1 + y2
x3 = y3 - y4
x4 = y3 + y4
h(x) = (y12 - y22) + y32
- y42
Notam:
ξ1 = y1
ξ2 = y2
ξ3 = y3
ξ4 = y4
h(x) = 2 ξ12 - 2 ξ22
+ ξ32 - ξ42
Metoda Gauss de reducere a formelor patratice la forma canonica
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