let A = ((( 1 1)),((1 1)) ) 1)calculate A^n 2) determine e^A and e^(−A) .


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Question Number 60501 by prof Abdo imad last updated on 21/May/19

$l e t A = (\begin{matrix} 1 1 \\ 1 1 \end{matrix})$ $1) c a l c u l a t e A^{n}$ $2) d e t e r m i n e e^{A} a n d e^{- A} .$
Commented by maxmathsup by imad last updated on 21/May/19
$we P_c (A) = determinant (((1−x 1)),((1 1−x)))=(1−x)^2 −1 =x^2 −2x +1−1 =x^2 −2x cayley hamilton theorem give P_c (A)=0 ⇒A^2 −2A =0 ⇒A^2 =2A ⇒ A^3 =2A^2 =2^2 A ⇒ A^n =2^(n−1) A ⇒ A^n = (((2^(n−1) 2^(n−1) )),((2^(n−1) 2^(n−1) )) ) n≥1 2) we have e^A =Σ_(n=0) ^∞ (A^n /(n!)) =I +Σ_(n=1) ^∞ (1/(n!)) (((2^(n−1) 2^(n−1) )),((2^(n−1) 2^(n−1) )) ) =I + (((Σ_(n=1) ^∞ (2^(n−1) /(n!)) Σ_(n=1) ^∞ (2^(n−1) /(n!)))),((Σ_(n=1) ^∞ (2^(n−1) /(n!)) Σ_(n=1) ^∞ (2^(n−1) /(n!)))) ) Σ_(n=1) ^∞ (2^(n−1) /(n!)) =(1/2) Σ_(n=1) ^∞ (2^n /(n!)) =(1/2){ Σ_(n=0) ^∞ (2^n /(n!)) −1} =(1/2){e^2 −1}⇒ e^A = (((1 0)),((0 1)) ) + (((((e^2 −1)/2) ((e^2 −1)/2))),((((e^2 −1)/2) ((e^2 −1)/2))) ) e^A = (((((1+e^2 )/2) ((e^2 −1)/2))),((((e^2 −1)/2) ((1+e^2 )/2))) ) also we calculate e^(−A) by folowing the same steps and use e^(−A) =Σ_(n=0) ^∞ (((−1)^n A^n )/(n!))$
$w e P_{c} (A) = \| \begin{matrix} 1 - x 1 \\ 1 1 - x \end{matrix} \| = {(1 - x)}^{2} - 1 = x^{2} - 2 x + 1 - 1 = x^{2} - 2 x$ $c a y l e y h a m i l t o n t h e o r e m g i v e P_{c} (A) = 0 \Rightarrow A^{2} - 2 A = 0 \Rightarrow A^{2} = 2 A \Rightarrow$ $A^{3} = 2 A^{2} = 2^{2} A \Rightarrow A^{n} = 2^{n - 1} A \Rightarrow A^{n} = (\begin{matrix} 2^{n - 1} 2^{n - 1} \\ 2^{n - 1} 2^{n - 1} \end{matrix}) n ⩾ 1$ $2) w e h a v e e^{A} = \sum_{n = 0}^{\infty} \frac{A^{n}}{n!} = I + \sum_{n = 1}^{\infty} \frac{1}{n!} (\begin{matrix} 2^{n - 1} 2^{n - 1} \\ 2^{n - 1} 2^{n - 1} \end{matrix})$ $= I + (\begin{matrix} \sum_{n = 1}^{\infty} \frac{2^{n - 1}}{n!} \sum_{n = 1}^{\infty} \frac{2^{n - 1}}{n!} \\ \sum_{n = 1}^{\infty} \frac{2^{n - 1}}{n!} \sum_{n = 1}^{\infty} \frac{2^{n - 1}}{n!} \end{matrix})$ $\sum_{n = 1}^{\infty} \frac{2^{n - 1}}{n!} = \frac{1}{2} \sum_{n = 1}^{\infty} \frac{2^{n}}{n!} = \frac{1}{2} {\sum_{n = 0}^{\infty} \frac{2^{n}}{n!} - 1} = \frac{1}{2} {e^{2} - 1} \Rightarrow$ $e^{A} = (\begin{matrix} 1 0 \\ 0 1 \end{matrix}) + (\begin{matrix} \frac{e^{2} - 1}{2} \frac{e^{2} - 1}{2} \\ \frac{e^{2} - 1}{2} \frac{e^{2} - 1}{2} \end{matrix})$ $e^{A} = (\begin{matrix} \frac{1 + e^{2}}{2} \frac{e^{2} - 1}{2} \\ \frac{e^{2} - 1}{2} \frac{1 + e^{2}}{2} \end{matrix})$ $a l s o w e c a l c u l a t e e^{- A} b y f o l o w i n g t h e s a m e s t e p s a n d u s e$ $e^{- A} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} A^{n}}{n!}$
	Answered by Prithwish sen last updated on 21/May/19

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