let-A-1-2-2-1-1-find-e-A-and-e-A-2-calculate-ch-A-and-sh-A-

Question Number 130771 by mathmax by abdo last updated on 28/Jan/21

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

1) find e^{A} and e^{- A}

2) calculate ch (A) and sh (A)

Answered by Olaf last updated on 28/Jan/21

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

A^{2} = (\begin{matrix} 5 4 \\ 4 5 \end{matrix}) = 2 A + 3 I

A^{3} = {AA}^{2} = A (2 A + 3 I) = 7 A + 6 I

\dots

Let A^{n} = a_{n} A + b_{n} I, a_{1} = 1, b_{1} = 0

\Rightarrow A^{n + 1} = {AA}^{n} = A (a_{n} A + b_{n} I)

A^{n + 1} = a_{n} A^{2} + b_{n} A

A^{n + 1} = (2 a_{n} + b_{n}) A + 3 a_{n} I

\Rightarrow {\begin{cases} a_{n + 1} = 2 a_{n} + b_{n} (1) \\ b_{n + 1} = 3 a_{n} (2) \end{cases}

(1) : a_{n + 2} = 2 a_{n + 1} + b_{n + 1}

a_{n + 2} = 2 a_{n + 1} + 3 a_{n}, a_{1} = 1, a_{2} = 2

\Rightarrow a_{n} = \frac{1}{4} [3^{n} + {(- 1)}^{n + 1}]

(2) : b_{n} = 3 a_{n - 1} = \frac{1}{4} [3^{n} + 3 {(- 1)}^{n}]

A^{n} = a_{n} A + b_{n} I

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

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

e^{A} = \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} A^{n}

\underset{n = 0}{\sum^{\infty}} \frac{3^{n}}{n!} = e^{3}, \underset{n = 0}{\sum^{\infty}} \frac{1}{3^{n} n!} = \frac{1}{e^{3}}, \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n!} = \frac{1}{e}

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

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

coshA = \frac{1}{2} (e^{A} + e^{- A})

coshA = \frac{1}{4} (\begin{matrix} e^{3} + \frac{1}{e^{3}} + \frac{2}{e} & e^{3} + \frac{1}{e^{3}} \\ e^{3} + \frac{1}{e^{3}} & e^{3} + \frac{1}{e^{3}} + \frac{2}{e} \end{matrix})

sinhA = \frac{1}{2} (e^{A} - e^{- A})

sinhA = \frac{1}{4} (\begin{matrix} e^{3} - \frac{1}{e^{3}} & e^{3} - \frac{1}{e^{3}} - \frac{2}{e} \\ e^{3} - \frac{1}{e^{3}} - \frac{2}{e} & e^{3} - \frac{1}{e^{3}} \end{matrix})

Commented by mathmax by abdo last updated on 29/Jan/21

thank you sir Olaf

Answered by mathmax by abdo last updated on 29/Jan/21

1) p_{c} (A) = \det (A - xI) = | \begin{matrix} 1 - x 2 \\ 2 1 - x \end{matrix} | = {(1 - x)}^{2} - 4

= (1 - x - 2) (1 - x + 2) = (- x - 1) (3 - x) = (x + 1) (x - 3)

\Rightarrow λ_{1} = - 1 and λ_{2} = 3

x^{n} = q (x) p (x) + u_{n} x + v_{n} we get {(- 1)}^{n} = - u_{n} + v_{n} and 3^{n} = {3 u}_{n} + v_{n} \Rightarrow

{\begin{cases} - u_{n} + v_{n} = {(- 1)}^{n} \\ {3 u}_{n} + v_{n} = 3^{n} \Rightarrow {\begin{cases} {4 u}_{n} = 3^{n} - {(- 1)}^{n} \Rightarrow {\begin{cases} u_{n} = \frac{3^{n} - {(- 1)}^{n}}{4} \\ v_{n} = \frac{3^{n} - {(- 1)}^{n}}{4} + {(- 1)}^{n} \end{cases} \\ v_{n} = u_{n} + {(- 1)}^{n} \end{cases} \end{cases}

\Rightarrow {\begin{cases} u_{n} = \frac{3^{n} - {(- 1)}^{n}}{4} \\ v_{n} = \frac{3^{n} + 3 {(- 1)}^{n}}{4} \end{cases}

A^{n} = u_{n} A + v_{n} I = \frac{3^{n} - {(- 1)}^{n}}{4} (\begin{matrix} 1 2 \\ 2 1 \end{matrix}) + \frac{3^{n} + 3 {(- 1)}^{n}}{4} (\begin{matrix} 1 0 \\ 0 1 \end{matrix})

= (\begin{matrix} \frac{3^{n} - {(- 1)}^{n}}{4} + \frac{3^{n} + 3 {(- 1)}^{n}}{4} \frac{3^{n} - {(- 1)}^{n}}{2} \\ \frac{3^{n} - {(- 1)}^{n}}{2} \frac{3^{n} - {(- 1)}^{n}}{4} + \frac{3^{n} + 3 {(- 1)}^{n}}{4} \end{matrix})

= (\begin{matrix} \frac{2 . 3^{n} + 2 {(- 1)}^{n}}{4} \frac{3^{n} - {(- 1)}^{n}}{2} \\ \frac{3^{n} - {(- 1)}^{n}}{2} \frac{2 . 3^{n} + 2 {(- 1)}^{n}}{4} \end{matrix})

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

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

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

Commented by mathmax by abdo last updated on 29/Jan/21

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

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

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

Commented by mathmax by abdo last updated on 29/Jan/21

ch (A) = \frac{1}{2} (e^{A} + e^{- A}) = \frac{1}{4} (\begin{matrix} e^{3} + e^{- 1} e^{3} - e^{- 1} \\ e^{3} - e^{- 1} e^{3} + e^{- 1} \end{matrix})

+ \frac{1}{4} (\begin{matrix} e^{- 3} + e e^{- 3} - e \\ e^{- 3} - e e^{- 3} + e \end{matrix})

= \frac{1}{4} (\begin{matrix} e^{3} + e^{- 1} + e^{- 3} + e e^{3} - e^{- 1} + e^{- 3} - e \\ e^{3} - e^{- 1} + e^{- 3} - e e^{3} + e^{- 1} + e^{- 3} + e \end{matrix})