let-A-2-1-1-2-1-calculate-A-n-2-determine-cosA-and-sinA-3-find-chA-and-shA-

Question Number 99828 by mathmax by abdo last updated on 23/Jun/20

let A = (\begin{matrix} 2 1 \\ 1 2 \end{matrix})

1) calculate A^{n}

2) determine cosA and sinA

3) find chA and shA

Commented by bachamohamed last updated on 23/Jun/20

1)

A = (\begin{matrix} 2 1 \\ 1 2 \end{matrix})

p_{A} (λ) = d e t (A - λ I_{2}) = (λ - 1) (λ - 2) d o n c l e s v a l e u r s p r o p r e s s o n t {_{λ_{2} = 2}^{λ_{1} = 1}

E_{λ_{1}} =< (- 1.1) > E_{λ_{2}} =< (1.1) > o u l e s v e c t e u r s p r o p r e s s o n t {_{v_{2} = (1.1)}^{v_{1} = (- 1.1)}

e t d o n c \exists p \in M_{2 \times 2} (R) t e l l e q u e B = p^{- 1} A . p

o n a {i d}_{E} (R^{2} . (v_{1} . v_{2})) \overset{p m a t r i x d e {i d}_{E}}{\to} (R^{2} . (e_{1} . e_{2})) e n d e d u i r e q u e p = (\begin{matrix} - 1 1 \\ 1 1 \end{matrix}) e t p^{- 1} = (\begin{matrix} - \frac{1}{2} \frac{1}{2} \\ \frac{1}{2} \frac{1}{2} \end{matrix})

B = p^{- 1} A p = (\begin{matrix} 1 0 \\ 0 3 \end{matrix}) e t B^{n} = {(p^{- 1} A p)}^{n} = (p^{- 1} A p) (p^{- 1} A p) \dots . . \overset{n f o i s}{.} \dots \dots (p^{- 1} A p) = p^{- 1} A^{n} . p \Rightarrow A^{n} = p . B^{n} . p^{- 1}

\Rightarrow A^{n} = \frac{1}{2} (\begin{matrix} (3^{n} + 1) 3^{n} - 1 \\ 3^{n} - 1 3^{n} + 1 \end{matrix})

Commented by mathmax by abdo last updated on 23/Jun/20

thankx sir .

Answered by mathmax by abdo last updated on 23/Jun/20

1) the caracteristic polynome is P_{c} (A) = \det (A - xI) = | \begin{matrix} 2 - x 1 \\ 1 2 - x \end{matrix} |

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

\Rightarrow λ_{1} = 1 and λ_{2} = 3 we have x^{n} = p_{c} (x) q + u_{n} x + v_{n} \Rightarrow

1 = u_{n} + v_{n} and 3^{n} = {3 u}_{n} + v_{n} \Rightarrow {2 u}_{n} = 3^{n} - 1 \Rightarrow u_{n} = \frac{3^{n} - 1}{2}

v_{n} = 1 - u_{n} = 1 - \frac{3^{n} - 1}{2} = \frac{2 - 3^{n} + 1}{2} = \frac{3 - 3^{n}}{2}

we have A^{n} = u_{n} A + v_{n} I = \frac{3^{n} - 1}{2} (\begin{matrix} 2 1 \\ 1 2 \end{matrix}) + \frac{3 - 3^{n}}{2} (\begin{matrix} 1 0 \\ 0 1 \end{matrix})

= (\begin{matrix} 3^{n} - 1 \frac{3^{n} - 1}{2} \\ \frac{3^{n} - 1}{2} 3^{n} - 1 \end{matrix}) + (\begin{matrix} \frac{3 - 3^{n}}{2} 0 \\ 0 \frac{3 - 3^{n}}{2} \end{matrix})

= (\begin{matrix} \frac{2 . 3^{n} - 2 + 3 - 3^{n}}{2} \frac{3^{n} - 1}{2} \\ \frac{3^{n} - 1}{2} \frac{2 . 3^{n} - 2 + 3 - 3^{n}}{2} \end{matrix})

= (\begin{matrix} \frac{3^{n} + 1}{2} \frac{3^{n} - 1}{2} \\ \frac{3^{n} - 1}{2} \frac{3^{n} + 1}{2} \end{matrix})

Commented by mathmax by abdo last updated on 24/Jun/20

2) \cos A = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} A^{2 n}}{(2 n)!} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 (2 n)!} (\begin{matrix} 3^{2 n} + 1 3^{2 n} - 1 \\ 3^{2 n} - 1 3^{2 n} + 1 \end{matrix})

= \frac{1}{2} (\begin{matrix} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} 3^{2 n} + \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} 3^{2 n}}{(2 n)!} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} \\ \dots . . \dots . \end{matrix})

= \frac{1}{2} (\begin{matrix} \cos (3) + \cos (1) \cos 3 - \cos (1) \\ \cos (3) - \cos (1) \cos (3) + \cos (1) \end{matrix})

sinA = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} A^{2 n + 1}}{(2 n + 1)!} = \dots . .