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Question Number 74794 by mathmax by abdo last updated on 30/Nov/19

l e t A = (\begin{matrix} 1 2 \\ - 1 1 \end{matrix})

1) c a l c u l a t e A^{n}

2) f i n d e^{A} a n d e^{- A}

3) c a l c u l a t e c o s A a n d s i n A

4) c a l c u l a t e c h (A) a n d s h (A)

Commented by Cmr 237 last updated on 01/Dec/19

1) c a l c u l d e A^{n}

o n a l e p o l y n o m e c a r a c t e r i s t i q u e e s t

P (λ) = d e t (A - λ I_{2})

= λ^{2} - 2 λ + 3

d e p l u s o n a :

P (λ) = 0 \Rightarrow λ_{1} = 1 - i \sqrt{2}, λ_{2} = 1 + i \sqrt{2}

o n r e m a r q u e q u e P (A) = 0

d o n c \forall X \in K [X] o n a :

X^{n} = Q (X) P (X) + a_{n} X + b_{n}

o u Q (X) e s t u n p o l y n o m e q u o t i e n t . .

λ_{1}^{n} = Q (λ_{1}) P (λ_{1}) + a_{n} λ_{1} + b_{n}

λ_{2}^{n} = Q (λ_{2}) P (λ_{2}) + a_{n} λ_{2} + b_{n}

{\begin{cases} a_{n} λ_{1} + b_{n} = λ_{1}^{n} \\ a_{n} λ_{2} + b_{n} = λ_{2}^{n} \end{cases}

a_{n} = \frac{λ_{2}^{n} - λ_{1}^{n}}{λ_{2} - λ_{1}} = \frac{{(1 + i \sqrt{2})}^{n} - {(1 - i \sqrt{2})}^{n}}{2 i \sqrt{2}}

b_{n} = \frac{3 (λ_{1}^{n - 1} - λ_{2}^{n - 1})}{2 i \sqrt{2}}

A^{n} = Q (A) P (A) + a_{n} A + b_{n} I_{2}

= a_{n} A + b_{n} I_{2}

A^{n} = (\begin{matrix} a_{n} 2 a_{n} \\ - a_{n} a_{n} \end{matrix}) + (\begin{matrix} b_{n} 0 \\ 0 b_{n} \end{matrix})

= (\begin{matrix} a_{n} + b_{n} 2 a_{n} \\ - a_{n} a_{n} + b_{n} \end{matrix}) o u a_{n} e t b_{n} s o n t l e s s u i t e s d o n n e e s e n h a u t

2) c a l c u l d e e^{A} e t e^{- A}

e^{A} = \underset{k = 0}{\sum^{n}} \frac{A^{k}}{k!}

e^{- A} = \underset{k = 0}{\sum^{n}} \frac{A^{- k}}{k!}

3) c a l c u l d e c o s A e t s i n A

c o s A = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k} A^{2 k}}{(2 k)!}

s i n A = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k} A^{2 k + 1}}{(2 k + 1)!}

4) c a l c u l d e c h A e t s h A

c h A = \underset{k = 0}{\sum^{n}} \frac{A^{2 k}}{(2 k)!}

s h A = \underset{k = 0}{\sum^{n}} \frac{A^{2 k + 1}}{(2 k + 1)!}

N B : e^{A}, e^{- A}, c o s A, s i n A, c h A, s h A

s o n t s u r l a f o r m e g e n e r a l e c a r \forall n \in N^{*}, A^{n} \neq 0

Commented by mathmax by abdo last updated on 01/Dec/19

1) w e h a v e P_{c} (x) = d e t (A - x I) = | \begin{matrix} 1 - x 2 \\ - 1 1 - x \end{matrix} |

= {(1 - x)}^{2} + 2 = x^{2} - 2 x + 3

Δ^{^{'}} = {(- 1)}^{2} - 3 = 1 - 3 = - 2 \Rightarrow λ_{1} = 1 + i \sqrt{2} a n d λ_{2} = 1 - i \sqrt{2}

c a y l e y h a m i l t o n g i v e P_{c} (A) = 0 l e t d i v i d e x^{n} b y x^{2} - 2 x + 3 \Rightarrow

x^{n} = q (x) P_{c} (x) + a_{n} x + b_{n} \Rightarrow λ_{1}^{n} = a_{n} λ_{1} + b_{n}

λ_{2}^{n} = a_{n} λ_{2} + b_{n} s o w e g e t t h e s y s t e m e {\begin{cases} λ_{1} a_{n} + b_{n} = λ_{1}^{n} \\ λ_{2} a_{n} + b_{n} = λ_{2}^{n} \end{cases}

\Rightarrow (λ_{1} - λ_{2}) a_{n} = λ_{1}^{n} - λ_{2}^{n} \Rightarrow a_{n} = \frac{λ_{1}^{n} - {({\bar{λ}}_{1})}^{n}}{λ_{1} - λ_{2}} = \frac{2 i I m (λ_{1}^{n})}{2 i \sqrt{2}}

= \frac{1}{\sqrt{2}} I m (λ_{1}^{n}) w e h a v e ∣ λ_{1} ∣ = \sqrt{1 + 2} = \sqrt{3} \Rightarrow λ_{1} = \sqrt{3} e^{i a r c t a n (\sqrt{2})} \Rightarrow

a_{n} = \frac{{(\sqrt{3})}^{n}}{\sqrt{2}} s i n (n a r c t a n (\sqrt{2}) a n d b_{n} = λ_{1}^{n} - λ_{1} a_{n}

= λ_{1}^{n} - λ_{1} \frac{λ_{1}^{n} - λ_{2}^{n}}{λ_{1} - λ_{2}} = \frac{λ_{1}^{n + 1} - λ_{2} λ_{1}^{n} - λ_{1}^{n + 1} + λ_{1} λ_{2}^{n}}{λ_{1} - λ_{2}}

= \frac{λ 1 λ_{2}}{λ_{1} - λ_{2}} {λ_{1}^{n - 1} - {({\bar{λ}}_{1})}^{n - 1}} = \frac{3}{2 i \sqrt{2}} (2 i {(\sqrt{3})}^{n - 1} s i n ((n - 1) a r c t a n (\sqrt{2}))

= \frac{{(\sqrt{3})}^{n + 1}}{\sqrt{2}} s i n {(n - 1) a r c t a n (\sqrt{2}))

w e h a v e A^{n} = a_{n} A + b_{n} I

= \frac{{(\sqrt{3})}^{n}}{\sqrt{2}} s i n (n a r c t a n (\sqrt{2})) (\begin{matrix} 1 2 \\ - 1 1 \end{matrix}) + \frac{{(\sqrt{3})}^{n + 1}}{\sqrt{2}} s i n {(n - 1) a r c t a n (\sqrt{2})) (\begin{matrix} 1 0 \\ 0 1 \end{matrix})