let] A = (((1 1 0)),((1 1 1)) ) (0 1 1 1)find the caractetistic polynom of A 2) calculate A^n


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Question Number 34305 by abdo mathsup 649 cc last updated on 03/May/18

$l e t] A = (\begin{matrix} 1 1 0 \\ 1 1 1 \end{matrix})$ $(0 1 1$ $1) f i n d t h e c a r a c t e t i s t i c p o l y n o m o f A$ $2) c a l c u l a t e A^{n}$
Commented by prof Abdo imad last updated on 06/May/18
$p_c (A)=det(A −XI)= determinant (((1−X 1 0 )),((1 1−X 1))) ∣ 0 1−X 1−X∣∣ =(1−X)((1−X)^2 −1+X) −1(1−X) +0 =(1−X)(X^2 −2X +1−1+X) −1 +X =(1−X)(X^2 −X) +X−1 =−(X−1)^2 X +X −1 =(X−1)( 1−X(X−1)) =(X−1)( −X^2 +X +1)=(1−X)(X^2 −X−1) p_c (A)=0 ⇔ X =1 or X^2 −X −1 =0 Δ= 1−4(−1)=5 ⇒ X_1 = ((1+(√5))/2) X_2 =((1−(√5))/2) .the proper value are λ_1 =1 λ_2 =((1+(√5))/2) and λ_3 = ((1−(√5))/2) v(λ_1 )=ker( A −I) ={ u/(A−I)u =0 } be continued.....$
$p_{c} (A) = d e t (A - X I) = \| \begin{matrix} 1 - X 1 0 \\ 1 1 - X 1 \end{matrix} \|$ $∣ 0 1 - X 1 - X ∣∣$ $= (1 - X) ({(1 - X)}^{2} - 1 + X) - 1 (1 - X) + 0$ $= (1 - X) (X^{2} - 2 X + 1 - 1 + X) - 1 + X$ $= (1 - X) (X^{2} - X) + X - 1$ $= - {(X - 1)}^{2} X + X - 1$ $= (X - 1) (1 - X (X - 1))$ $= (X - 1) (- X^{2} + X + 1) = (1 - X) (X^{2} - X - 1)$ $p_{c} (A) = 0 \Leftrightarrow X = 1 o r X^{2} - X - 1 = 0$ $Δ = 1 - 4 (- 1) = 5 \Rightarrow X_{1} = \frac{1 + \sqrt{5}}{2}$ $X_{2} = \frac{1 - \sqrt{5}}{2} . t h e p r o p e r v a l u e a r e λ_{1} = 1$ $λ_{2} = \frac{1 + \sqrt{5}}{2} a n d λ_{3} = \frac{1 - \sqrt{5}}{2}$ $v (λ_{1}) = k e r (A - I) = {u / (A - I) u = 0}$ $b e c o n t i n u e d . . . . .$
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