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Culegere de probleme de Analiz˘a numeric˘a

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66 Rezolvarea numerică a sistemelor algebrice liniare<br />

Γ11 = (−1) 1+1<br />

Γ13 = (−1) 1+3<br />

Γ22 = (−1) 1+2<br />

Γ31 = (−1) 3+1<br />

TGS = (D−L) −1 U =<br />

<br />

<br />

5 1 <br />

<br />

0 5 = 25, Γ12 = (−1) 1+2<br />

<br />

<br />

0 1 <br />

<br />

0 5 = 0<br />

<br />

<br />

0 5 <br />

<br />

0 0 = 0, Γ21 = (−1) 2+1<br />

<br />

<br />

1 1 <br />

<br />

0 5 = −5<br />

<br />

<br />

5 1 <br />

<br />

0 5 = 25, Γ23 = (−1) 2+3<br />

<br />

<br />

5 1 <br />

<br />

0 0 = 0<br />

<br />

<br />

1 1 <br />

<br />

5 1 = −5, Γ32 = (−1) 3+2<br />

<br />

<br />

5 1 <br />

<br />

0 1 = −5<br />

Γ33 = (−1) 3+3<br />

<br />

<br />

5 1 <br />

<br />

0 5 = 25<br />

E −1 ⎛<br />

1 0 0 5<br />

= ⎝ − 1 1 0 25 5<br />

− 1<br />

⎞<br />

⎠<br />

1 1 − 25 25 5<br />

⎛<br />

1 0 0 5<br />

⎝ − 1<br />

⎞⎛<br />

⎞ ⎛<br />

0 −1 −1 0 −<br />

1 0 ⎠⎝<br />

0 0 −1 ⎠<br />

25 5 = ⎝<br />

0 0 0<br />

1<br />

1 0 25<br />

− 1<br />

25<br />

− 1<br />

25<br />

1<br />

5<br />

TGSn = 2<br />

5<br />

<br />

2 k<br />

5<br />

3<br />

5<br />

· 7<br />

5<br />

< ε<br />

Problema 4.3.5 Arătat¸i că pentruA ∈ Mn,n(R)<br />

A2 = [ρ(A t A)] 1/2<br />

0 1<br />

25<br />

5 −1<br />

5<br />

− 4<br />

25<br />

2<br />

25<br />

⎞<br />

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