let A = (((2 1)),((1 3)) ) 1) calculate A^n 2) find e^A ,e^(−A) 3)determine ch(A) and sh(A) is ch^2 A−sh^2 A =1?


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Question Number 100994 by mathmax by abdo last updated on 29/Jun/20

$let A = (\begin{matrix} 2 1 \\ 1 3 \end{matrix})$ $1) calculate A^{n}$ $2) find e^{A}, e^{- A}$ $3) determine ch (A) and sh (A) is {ch}^{2} A - {sh}^{2} A = 1 ?$
	Answered by mathmax by abdo last updated on 01/Jul/20
	$1) the caracteristic polynom of A is p_c (x)=det(A−xI) = determinant (((2−x 1)),((1 3−x))) =(x−2)(x−3)−1 =x^2 −5x+6−1 =x^2 −5x +5 P_c (x)=0 ⇔x^2 −5x +5 =0 →Δ =25−20 =5 ⇒α=((5+(√5))/2) and β=((5−(√5))/2) we have x^(n ) =qP_c (x) +u_n x +v_n ⇒ { ((α^n =u_n α +v_n )),((β^n =u_n β +v_n )) :} ⇒u_n =((α^n −β^n )/(α−β)) =(1/(√5)){ (((5+(√5))/2))^n −(((5−(√5))/2))^n } v_n =α^n −αu_n =α^n −α×((α^n −β^n )/(α−β)) =((α^(n+1) −β α^n −α^(n+1) +αβ^n )/(α−β)) =((αβ^n −βα^n )/(α−β)) =((((5+(√5))/2)(((5−(√5))/2))^n −((5−(√5))/2)(((5+(√5))/2))^n )/(√5)) we have A^n =u_n A +v_n I (by cayley hamilton) ⇒ A^n =(1/(√5)){(((5+(√5))/2))^n −(((5−(√5))/2))^n } (((2 1)),((1 3)) ) +((((5+(√5))/2)(((5−(√5))/2))^n −((5−(√5))/2)(((5+(√5))/2))^n )/(√5)) (((1 0)),((0 1)) )$
	$1) the caracteristic polynom of A is p_{c} (x) = \det (A - xI)$ $= \| \begin{matrix} 2 - x 1 \\ 1 3 - x \end{matrix} \| = (x - 2) (x - 3) - 1 = x^{2} - 5 x + 6 - 1 = x^{2} - 5 x + 5$ $P_{c} (x) = 0 \Leftrightarrow x^{2} - 5 x + 5 = 0 \to Δ = 25 - 20 = 5 \Rightarrow α = \frac{5 + \sqrt{5}}{2} and β = \frac{5 - \sqrt{5}}{2}$ $we have x^{n} = {qP}_{c} (x) + u_{n} x + v_{n} \Rightarrow {\begin{cases} α^{n} = u_{n} α + v_{n} \\ β^{n} = u_{n} β + v_{n} \end{cases}$ $\Rightarrow u_{n} = \frac{α^{n} - β^{n}}{α - β} = \frac{1}{\sqrt{5}} {{(\frac{5 + \sqrt{5}}{2})}^{n} - {(\frac{5 - \sqrt{5}}{2})}^{n}}$ $v_{n} = α^{n} - α u_{n} = α^{n} - α \times \frac{α^{n} - β^{n}}{α - β} = \frac{α^{n + 1} - β α^{n} - α^{n + 1} + α β^{n}}{α - β} = \frac{α β^{n} - β α^{n}}{α - β}$ $= \frac{\frac{5 + \sqrt{5}}{2} {(\frac{5 - \sqrt{5}}{2})}^{n} - \frac{5 - \sqrt{5}}{2} {(\frac{5 + \sqrt{5}}{2})}^{n}}{\sqrt{5}}$ $we have A^{n} = u_{n} A + v_{n} I (by cayley hamilton) \Rightarrow$ $A^{n} = \frac{1}{\sqrt{5}} {{(\frac{5 + \sqrt{5}}{2})}^{n} - {(\frac{5 - \sqrt{5}}{2})}^{n}} (\begin{matrix} 2 1 \\ 1 3 \end{matrix}) + \frac{\frac{5 + \sqrt{5}}{2} {(\frac{5 - \sqrt{5}}{2})}^{n} - \frac{5 - \sqrt{5}}{2} {(\frac{5 + \sqrt{5}}{2})}^{n}}{\sqrt{5}} (\begin{matrix} 1 0 \\ 0 1 \end{matrix})$
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