Culegere de probleme de Analiz˘a numeric˘a

64 Rezolvarea numerică a sistemelor algebrice liniare
x (2)
1
x (k)
1 = 1
5 (7−x(k−1) 2 −x (k−1)
3 )
x (k)
2 = 1
5 (7−x(k−1) 1 −x (k−1)
3
x (k)
3
= 1
x (2)
1 = 1
5
x (2)
2
= 1
5
5 (7−x(k−1) 1 −x (k−1)
2
7− 7
7
− =
5 5
21
25
7− 7
7
− =
5 5
21
25
x (2)
3
= 21
25
x (k)
1 = 1
5 (7−x(k−1) 2 −x (k−1)
3 )
x (k)
2 = 1
5 (7−x(k) 1 −x (k−1)
3 )
x (k)
3
x (1)
1 = 7
5
x (1)
3
1
=
5 (7−x(k) 1 −x(k) 2
, x(1) 2 = 7 7
−
5 5
7 7 21
= − =
5 25 25
= 0
7 21 175−21 154
= − = =
5 125 125 125
x (2)
2 = 7 154 21
− − , x(3) 3 =
5 625 125 7 154
−
5 125 −x(2) 2
Pentru a rezolva a doua parte a problemei vom scrie sistemul sub forma
x = Tx+c ⇒ x−x (k) ≤ Tk
1−T x(1) −x (0)
Pentru Jacobi ⎧ ⎨
x1 = 1
5 (7−x2 −x3)
x2 = 1
5 (7−x1 −x3)
x3 = 1
⎩
5 (7−x1 −x2)
⎡
0 −
x = ⎣
1
5 −1
5
− 1
⎤ ⎡
⎦x+ ⎣
0 − 5 1
5
−1 5 −1 0 5
7
5
7
5
7
5
⎤
⎦