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Question Number 99580 by mathmax by abdo last updated on 21/Jun/20

let A = (\begin{matrix} 2 - 1 \\ 3 1 \end{matrix})

1) prove that A is inversible and calculste A^{- 1}

2) calculate A^{n}

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

4) calculate \cos A and sinA is \cos^{2} A + \sin^{2} A = I ?

Commented by john santu last updated on 22/Jun/20

(1) \det (A) \neq 0 \Rightarrow A^{- 1} = \frac{1}{5} [\begin{matrix} 1 1 \\ - 3 2 \end{matrix}]

Answered by MWSuSon last updated on 22/Jun/20

1) ∣ A ∣= 2 + 3 = 5 \neq 0

2) A^{n} = {PD}^{n} P^{- 1}

∣ A - λ I ∣= | \begin{matrix} 2 - λ & - 1 \\ 3 & 1 - λ \end{matrix} | = 0

(2 - λ) (1 - λ) + 3 = 0

λ = \frac{1}{2} (3 \pm i \sqrt{11})

when λ = \frac{1}{2} (3 + i \sqrt{11})

let B = (\begin{matrix} \frac{1}{2} - i \frac{\sqrt{11}}{2} & - 1 \\ 3 & - \frac{1}{2} - \frac{i \sqrt{11}}{2} \end{matrix})

B \vec{X} = (\begin{matrix} \frac{1}{2} - i \frac{\sqrt{11}}{2} & - 1 \\ 3 & - \frac{1}{2} - \frac{i \sqrt{11}}{2} \end{matrix}) [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]

augmented matrix

(\begin{matrix} \frac{1}{2} - \frac{i \sqrt{11}}{2} & - 1 & 0 \\ 3 & - \frac{1}{2} - \frac{i \sqrt{11}}{2} & 0 \end{matrix}) = [\begin{matrix} x_{1} \\ x_{2} \end{matrix}]

by gaussian elimination

R 2 \to R 2 - \frac{3}{\frac{1}{2} - \frac{i \sqrt{11}}{2}} R 1 and R 1 \to \frac{2}{1 - i \sqrt{11}}

= (\begin{matrix} 1 & - \frac{2}{1 - i \sqrt{11}} \\ 0 & 0 \end{matrix}) [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]

x_{1} - \frac{{2 x}_{2}}{1 - i \sqrt{11}} = 0

x_{1} = \frac{{2 x}_{2}}{1 - i \sqrt{11}}

\vec{X} = [\begin{matrix} \frac{{2 ix}_{2}}{i + \sqrt{11}} \\ x_{2} \end{matrix}], let x_{2} = 1

\vec{X} = [\begin{matrix} \frac{2 i}{i + \sqrt{11}} \\ 1 \end{matrix}]

fellowing same procedure second

eigenvector \vec{X} = [\begin{matrix} - \frac{2 i}{\sqrt{11} - i} \\ 1 \end{matrix}]

P = [\begin{matrix} \frac{2 i}{\sqrt{11} + i} & - \frac{2 i}{\sqrt{11} - i} \\ 1 & 1 \end{matrix}]

P^{- 1} = [\begin{matrix} \frac{3 i}{\sqrt{11}} & \frac{1}{2} - \frac{i}{2 \sqrt{11}} \\ - \frac{3 i}{\sqrt{11}} & \frac{1}{2} + \frac{i}{2 \sqrt{11}} \end{matrix}]

D = P^{- 1} AP

D = [\begin{matrix} \frac{1}{2} (3 + i \sqrt{11}) & 0 \\ 0 & \frac{1}{2} (3 - i \sqrt{11}) \end{matrix}]

Hence A^{n} = [\begin{matrix} \frac{2 i}{\sqrt{11} + i} & \frac{2 i}{i - \sqrt{11}} \\ 1 & 1 \end{matrix}] [\begin{matrix} {(\frac{1}{2})}^{n} {(3 + i \sqrt{11})}^{n} & 0 \\ 0 & {(\frac{1}{2})}^{n} {(3 - \sqrt{11})}^{n} \end{matrix}] [\begin{matrix} \frac{3 i}{\sqrt{11}} & \frac{1}{2} - \frac{i}{2 \sqrt{11}} \\ - \frac{3 i}{\sqrt{11}} & \frac{1}{2} + \frac{i}{2 \sqrt{11}} \end{matrix}]

Commented by MWSuSon last updated on 22/Jun/20

you are welcome sir.

Commented by abdomathmax last updated on 22/Jun/20

thanks sir .

Answered by MWSuSon last updated on 22/Jun/20

if A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] = [\begin{matrix} 2 & - 1 \\ 3 & 1 \end{matrix}]

then e^{A} = [\begin{matrix} e^{a_{11}} & e^{a_{12}} \\ e^{a 21} & e^{a_{22}} \end{matrix}] = [\begin{matrix} e^{2} & \frac{1}{e} \\ e^{3} & e \end{matrix}]

e^{- A} = [\begin{matrix} \frac{1}{e^{2}} & e \\ \frac{1}{e^{3}} & \frac{1}{e} \end{matrix}]

4) cosA = [\begin{matrix} \cos (a_{11}) & \cos (a_{12}) \\ \cos (a_{21}) & \cos (a_{22}) \end{matrix}] = [\begin{matrix} \cos 2 & \cos 1 \\ \cos 3 & \cos 1 \end{matrix}]

sinA = [\begin{matrix} \sin 2 & - \sin 1 \\ \sin 3 & \sin 1 \end{matrix}] .

yes \cos^{2} A + \sin^{2} A = I

Answered by MWSuSon last updated on 22/Jun/20

e^{A} = {Pe}^{D} P^{- 1}

e^{- A} = {Pe}^{- D} P^{- 1}

\Rightarrow e^{A} = [\begin{matrix} \frac{2 i}{\sqrt{11} + i} & \frac{- 2 i}{\sqrt{11} - i} \\ 1 & 1 \end{matrix}] [\begin{matrix} e^{\frac{1}{2} (3 + i \sqrt{11})} & 0 \\ 0 & e^{\frac{1}{2} (3 - i \sqrt{11})} \end{matrix}] [\begin{matrix} \frac{3 i}{\sqrt{11}} & \frac{1}{2} - \frac{i}{2 \sqrt{11}} \\ - \frac{3 i}{\sqrt{11}} & \frac{1}{2} + \frac{i}{2 \sqrt{11}} \end{matrix}]

\Rightarrow e^{- A} = [\begin{matrix} \frac{2 i}{\sqrt{11} + i} & \frac{- 2 i}{\sqrt{11} - i} \\ 1 & 1 \end{matrix}] [\begin{matrix} e^{- \frac{1}{2} (3 + i \sqrt{11})} & 0 \\ 0 & e^{- \frac{1}{2} (3 - i \sqrt{11})} \end{matrix}] [\begin{matrix} \frac{3 i}{\sqrt{11}} & \frac{1}{2} - \frac{i}{2 \sqrt{11}} \\ - \frac{3 i}{\sqrt{11}} & \frac{1}{2} + \frac{i}{2 \sqrt{11}} \end{matrix}]

Answered by mathmax by abdo last updated on 22/Jun/20

1) P_{c} (x) = \det (A - xI) = | \begin{matrix} 2 - x - 1 \\ 3 1 - x \end{matrix} | = (2 - x) (1 - x) + 3

= (x - 2) (x - 1) + 3 = x^{2} - 3 x + 2 + 3 = x^{2} - 3 x + 5

P_{c} (A) = 0 \Rightarrow A^{2} - 3 A + 5 I = 0 \Rightarrow A^{2} - 3 A = - 5 I \Rightarrow

(- \frac{1}{5}) A (A - 3 I) = I \Rightarrow A is inversible and A^{- 1} = \frac{1}{5} (3 I - A)

= \frac{3}{5} I - \frac{1}{5} A = (\begin{matrix} \frac{3}{5} 0 \\ 0 \frac{3}{5} \end{matrix}) - (\begin{matrix} \frac{2}{5} - \frac{1}{5} \\ \frac{3}{5} \frac{1}{5} \end{matrix}) = (\begin{matrix} \frac{1}{5} \frac{1}{5} \\ - \frac{3}{5} \frac{2}{5} \end{matrix})

Commented by mathmax by abdo last updated on 22/Jun/20

2) we have P_{c} (x) = x^{2} - 3 x + 5

P_{c} (x) = 0 \Rightarrow x^{2} - 3 x + 5 = 0 \to Δ = 9 - 20 = - 11 \Rightarrow λ_{1} = \frac{3 + i \sqrt{11}}{2}

and λ_{2} = \frac{3 - i \sqrt{11}}{2} we divide x^{n} by p_{c} (x) \Rightarrow x^{n} = {qp}_{c} (x) + u_{n} x + v_{n}

\Rightarrow λ_{1}^{n} = u_{n} λ_{1} + v_{n} and λ_{2}^{n} = u_{n} λ_{2} + v_{n} \Rightarrow

\frac{λ_{1}^{n} - λ_{2}^{n}}{λ_{1} - λ_{2}} = u_{n} \Rightarrow v_{n} = λ_{1}^{n} - λ_{1} u_{n} = λ_{1}^{n} - λ_{1} \times \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{- 5 λ_{1}^{n - 1} + 5 λ_{2}^{n - 1}}{i \sqrt{11}} = \frac{- 5}{i \sqrt{11}} (λ_{1}^{n - 1} - λ_{2}^{n - 1})

∣ λ_{1} ∣ = \frac{1}{2} (2 \sqrt{5}) = \sqrt{5} \Rightarrow λ_{1} = \sqrt{5} e^{iarctan (\frac{\sqrt{11}}{3})} and λ_{2} = \sqrt{5} e^{- iarctan (\frac{\sqrt{11}}{3})} \Rightarrow

u_{n} = \frac{{(\sqrt{5})}^{n} {e^{inarctan (\frac{\sqrt{11}}{3})} - e^{- inarctan (\frac{\sqrt{11}}{3})}}}{i \sqrt{11}} = \frac{{(\sqrt{5})}^{n}}{i \sqrt{11}} {2 isin (narctan (\frac{\sqrt{11}}{3}))}

= \frac{2 {(\sqrt{5})}^{n}}{\sqrt{11}} \sin (narctan (\frac{\sqrt{11}}{3}))

v_{n} = \frac{- 5}{i \sqrt{11}} {2 i \sin (n - 1) \arctan (\frac{\sqrt{11}}{3}))}

= \frac{- 10}{\sqrt{11}} {\sin (n - 1) \arctan (\frac{\sqrt{11}}{3}))}

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

= (\begin{matrix} {2 u}_{n} + v_{n} - u_{n} \\ {3 u}_{n} u_{n} + v_{n} \end{matrix})