If A= [((4 −3)),((1 0)) ]use the fact that A^2 =4A−3I_2 and mathematical induction to prove A^n =(((3^n −1))/2)A +((3−3^n )/2)I if n≥1


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Question Number 40352 by scientist last updated on 20/Jul/18

$I f A = [\begin{matrix} 4 - 3 \\ 1 0 \end{matrix}] u s e t h e f a c t t h a t$ $A^{2} = 4 A - 3 I_{2} a n d m a t h e m a t i c a l i n d u c t i o n t o p r o v e$ $A^{n} = \frac{(3^{n} - 1)}{2} A + \frac{3 - 3^{n}}{2} I i f n ⩾ 1$
Commented by prof Abdo imad last updated on 20/Jul/18
$the caracteristic polynom of A is p_c (A)=det(A−xI)= determinant (((4−x −3)),((1 −x))) =−x(4−x)+3 =x^2 −4x +3 kayley hamilton theorem give A^2 −4A +3I =0 ⇒A^2 =4A−3I ⇒ let prove by recurrence that A^(2n) =((3^(2n) −1)/2) A +((3−3^(2n) )/2) I the equality is true for n=0 A^(2n+2) =A^2 {((3^(2n) −1)/2) A +((3−3^(2n) )/2) I} =(4A−3I){((3^(2n) −1)/2) A +((3−3^(2n) )/2) I} =2(3^(2n) −1)A^2 +2(3−3^(2n) )A −3((3^(2n) −1)/2) A −3((3−3^(2n) )/2)I =2(3^(2n) −1)( 4A−3I) + ((4(3−3^(2n) )−33^(2n) +3)/2) A −3((3−3^(2n) )/2) I =8(3^(2n) −1)A −6(3^(2n) −1)I + ((15−7 3^(2n) )/2) A −3 ((3−3^(2n) )/2) I = ((16(3^(2n) −1)+15−73^(2n) )/2) A −((12(3^(2n) −1)+3(3−3^(2n) ))/2) A =((9.3^(2n) −1)/2) A −((9 3^(2n) −3)/2) I =((3^(2n+2) −1)/2) A+((3−3^(2n+2) )/2) I the relation is true at term (n+1) also we must prove by recurrence that A^(2n+1) =((3^(2n+1) −1)/2) A +((3−3^(2n+1) )/2) I .$
$t h e c a r a c t e r i s t i c p o l y n o m o f A i s$ $p_{c} (A) = d e t (A - x I) = \| \begin{matrix} 4 - x - 3 \\ 1 - x \end{matrix} \|$ $= - x (4 - x) + 3 = x^{2} - 4 x + 3$ $k a y l e y h a m i l t o n t h e o r e m g i v e$ $A^{2} - 4 A + 3 I = 0 \Rightarrow A^{2} = 4 A - 3 I \Rightarrow$ $l e t p r o v e b y r e c u r r e n c e t h a t$ $A^{2 n} = \frac{3^{2 n} - 1}{2} A + \frac{3 - 3^{2 n}}{2} I$ $t h e e q u a l i t y i s t r u e f o r n = 0$ $A^{2 n + 2} = A^{2} {\frac{3^{2 n} - 1}{2} A + \frac{3 - 3^{2 n}}{2} I}$ $= (4 A - 3 I) {\frac{3^{2 n} - 1}{2} A + \frac{3 - 3^{2 n}}{2} I}$ $= 2 (3^{2 n} - 1) A^{2} + 2 (3 - 3^{2 n}) A - 3 \frac{3^{2 n} - 1}{2} A$ $- 3 \frac{3 - 3^{2 n}}{2} I$ $= 2 (3^{2 n} - 1) (4 A - 3 I) + \frac{4 (3 - 3^{2 n}) - 33^{2 n} + 3}{2} A$ $- 3 \frac{3 - 3^{2 n}}{2} I$ $= 8 (3^{2 n} - 1) A - 6 (3^{2 n} - 1) I + \frac{15 - 7 3^{2 n}}{2} A$ $- 3 \frac{3 - 3^{2 n}}{2} I$ $= \frac{16 (3^{2 n} - 1) + 15 - 73^{2 n}}{2} A - \frac{12 (3^{2 n} - 1) + 3 (3 - 3^{2 n})}{2} A$ $= \frac{9 . 3^{2 n} - 1}{2} A - \frac{9 3^{2 n} - 3}{2} I$ $= \frac{3^{2 n + 2} - 1}{2} A + \frac{3 - 3^{2 n + 2}}{2} I t h e r e l a t i o n i s t r u e$ $a t t e r m (n + 1) a l s o w e m u s t p r o v e b y r e c u r r e n c e$ $t h a t$ $A^{2 n + 1} = \frac{3^{2 n + 1} - 1}{2} A + \frac{3 - 3^{2 n + 1}}{2} I .$
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