If X= [(3,(−4)),(1,(−1)) ], the value of X^n is


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Question Number 41141 by Kishan Daroga last updated on 02/Aug/18

$If X = [\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}], the value of X^{n} is$
Commented by math khazana by abdo last updated on 04/Aug/18
$let A = (((3 −4)),((1 −1)) ) the caracteristic polynome of A is p(x)=det(A−xI) = determinant (((3−x −4)),((1 −1−x))) =(x−3)(x+1) +4 =x^2 +x−3x−3 +4 =x^2 −2x +1 =(x−1)^2 so 1 is double proper value for A v(1)=ker(A−I)={u /(A−I)u=0} let u ((x),(y) ) ⇒ (((2 −4)),((1 −2)) ) ((x),(y) ) =0 ⇒ {_(x−2y=0) ^(2x−4y=0) ⇒x=2y ⇒(x,y)=(2y,y)=y(2,1) so v(1)=D_e with vector e(2,1) its clear that A is not diagonalisable by cayley hamilton theorem give A^2 −2A +I =0⇒ A^2 =2A −I ⇒ A^3 =(2A−I)A=2A^2 −A =2(2A−I)−A =3A −2I for that we must determine the sequences u_n and v_n with verify A^n =u_n A +v_n I ⇒ A^(n+1) =u_(n+1) A +v_(n+1) I =(u_n A +v_n I)A =u_n A^2 +v_n A =u_n (2A−I) +v_n A =(2u_n +v_n )A −u_n I ⇒ u_(n+1) = 2u_n +v_n and v_(n+1) =−u_n ...be continued...$
$l e t A = (\begin{matrix} 3 - 4 \\ 1 - 1 \end{matrix}) t h e c a r a c t e r i s t i c p o l y n o m e o f$ $A i s p (x) = d e t (A - x I) = \| \begin{matrix} 3 - x - 4 \\ 1 - 1 - x \end{matrix} \|$ $= (x - 3) (x + 1) + 4 = x^{2} + x - 3 x - 3 + 4$ $= x^{2} - 2 x + 1 = {(x - 1)}^{2} s o 1 i s d o u b l e p r o p e r$ $v a l u e f o r A$ $v (1) = k e r (A - I) = {u / (A - I) u = 0}$ $l e t u (\begin{matrix} x \\ y \end{matrix}) \Rightarrow (\begin{matrix} 2 - 4 \\ 1 - 2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = 0 \Rightarrow$ ${_{x - 2 y = 0}^{2 x - 4 y = 0} \Rightarrow x = 2 y \Rightarrow (x, y) = (2 y, y) = y (2, 1)$ $s o v (1) = D_{e} w i t h v e c t o r e (2, 1) i t s c l e a r t h a t$ $A i s n o t d i a g o n a l i s a b l e b y c a y l e y h a m i l t o n$ $t h e o r e m g i v e A^{2} - 2 A + I = 0 \Rightarrow$ $A^{2} = 2 A - I \Rightarrow A^{3} = (2 A - I) A = 2 A^{2} - A$ $= 2 (2 A - I) - A = 3 A - 2 I f o r t h a t w e m u s t$ $d e t e r m i n e t h e s e q u e n c e s u_{n} a n d v_{n} w i t h$ $v e r i f y A^{n} = u_{n} A + v_{n} I \Rightarrow$ $A^{n + 1} = u_{n + 1} A + v_{n + 1} I = (u_{n} A + v_{n} I) A$ $= u_{n} A^{2} + v_{n} A = u_{n} (2 A - I) + v_{n} A$ $= (2 u_{n} + v_{n}) A - u_{n} I \Rightarrow$ $u_{n + 1} = 2 u_{n} + v_{n} a n d v_{n + 1} = - u_{n} . . . b e c o n t i n u e d . . .$
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