let-A-1-1-1-1-1-calculate-A-n-2-find-e-A-e-A-3-find-sinA-and-cosA-4-find-ch-A-and-sh-A-

Question Number 76361 by mathmax by abdo last updated on 26/Dec/19

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

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

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

3) f i n d s i n A a n d c o s A

4) f i n d c h (A) a n d s h (A)

Commented by mathmax by abdo last updated on 27/Dec/19

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

= (1 - x - 1) (1 - x + 1) = - x (2 - x) = x (x - 2) s o t h e p r o p e r v a l u e s

a r e λ_{1} = 0 a n d λ_{2} = 2 l e t d i v i d e x^{n} b y P_{c} (x) \Rightarrow

x^{n} = Q (x) P_{c} (x) + u_{n} x + v_{n} \Rightarrow 0 = v_{n}

2^{n} = 2 u_{n} \Rightarrow u_{n} = 2^{n - 1} w e h a v e A^{n} = u_{n} A = 2^{n - 1} (\begin{matrix} 1 1 \\ 1 1 \end{matrix})

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

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

= \frac{1}{2} (\begin{matrix} e^{2} e^{2} \\ e^{2} e^{2} \end{matrix})

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

= (\begin{matrix} \frac{1}{2} e^{- 2} \frac{1}{2} e^{- 2} \\ \frac{1}{2} e^{- 2} \frac{1}{2} e^{- 2} \end{matrix})

Commented by mathmax by abdo last updated on 27/Dec/19

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

= (\begin{matrix} \frac{1}{2} s i n (2) \frac{1}{2} s i n (2) \\ \frac{1}{2} s i n (2) \frac{1}{2} s i n (2) \end{matrix})

c o s A = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} A^{2 n} = (\begin{matrix} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} 2^{2 n - 1} \dots \\ \dots \dots \end{matrix})

= (\begin{matrix} \frac{1}{2} c o s (2) \frac{1}{2} c o s (2) \\ \frac{1}{2} c o s (2) \frac{1}{2} c o s (2) \end{matrix})

Commented by mathmax by abdo last updated on 27/Dec/19

4) s h (A) = \frac{1}{2} (e^{A} - e^{- A}) = \dots .

Answered by john santu last updated on 27/Dec/19

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