Problem

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4. Gasiti polinomul caracteristic prin metoda lui Fadeev. A= ( 1 2 1 −1 1 0 2 1 2 1 −1 3 4 −5 0 4 ) Pentru k=1,2,3,4,calculam: A k =AB k−1 q k = −1 k trace ( A k ) A 1 = ( 1 2 1 −1 1 0 2 1 2 1 −1 3 4 −5 0 4 ) q 1 =-4 A 2 = ( −3 0 0 4 5 −1 −9 5 5 −16 9 −4 −1 8 −6 −9 ) q 2 =2 A 3 = ( 15 −22 −1 17 8 −24 16 −11 −5 41 −38 −4 −33 27 21 −37 ) q 3 =28 A 4 = ( 87 0 0 0 0 87 0 0 0 0 87 0 0 0 0 87 ) q 4 =-87 Asadar,polinomul caracteristic corespunzator matricii A,va fi: K A ( λ ) =λ 4 -4 λ 3 +2 λ 2 +28-87

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Metode numerice

Transcript of Problem

4. Gasiti polinomul caracteristic prin metoda lui Fadeev.

A=

Pentru k=1,2,3,4,calculam:

= =-4

= =2

= =28

= =-87

Asadar,polinomul caracteristic corespunzator matricii A,va fi:

-4+2+28-87


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