A= [(1,2),(2,5) ],B= [(3,(−1)),(4,(+2)) ] determine A^B


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Question Number 2564 by Rasheed Soomro last updated on 22/Nov/15

$A = [\begin{matrix} 1 & 2 \\ 2 & 5 \end{matrix}], B = [\begin{matrix} 3 & - 1 \\ 4 & + 2 \end{matrix}]$ $d e t e r m i n e A^{B}$
	Answered by Yozzi last updated on 22/Nov/15
	$∣A∣=5−4=1≠0 ⇒A is non−singular. A^B =e^((lnA)B) =Σ_(k=0) ^∞ (1/(k!)){(lnA)B}^k For eigenvalues k of A ∣A−kI∣=0 ∴ determinant (((1−k 2)),((2 5−k)))=0 (1−k)(5−k)−4=0 5−6k+k^2 −4=0 k^2 −6k+1=0 (k−3)^2 −9+1=0 (k−3)^2 =8⇒k=3±2(√2) For k=3+2(√2), (∵ Ae=ke⇒(A−kI)e=0) [((1−3−2(√2) 2)),((2 5−3−2(√2))) ] [(x),(y) ]= [(0),(0) ] (−2−2(√2))x+2y=0⇒y=(1+(√2))x.....(i) 2x+(2−2(√2))y=0 x+(1−(√2))y=0......(ii) y from (i) leads to x+(1−(√2))(1+(√2))x=0 x+(1−2)x=0 x−x=0 (true) ∴ We can choose eigenvector e_1 = [(1),((1+(√2))) ] for k=3+2(√2) For k=3−2(√2) [((−2+2(√2) 2)),((2 2+2(√2))) ] [(x),(y) ]= [(0),(0) ] ⇒(−1+(√2))x+y=0⇒y=(1−(√2))x ..............(iii) and 2x+(2+2(√2))y=0⇒x+(1+(√2))y=0........(iv) y from (iii) into (iv) gives x+(1−2)x=0⇒x−x=0 (true) So let e_2 = [(1),((1−(√2))) ] for k=3−2(√2). ∴ Let V=[e_1 e_2 ]= [((1 1)),((1+(√2) 1−(√2))) ] and A^′ = [((3+2(√2) 0)),((0 3−2(√2))) ]⇒lnA^′ = [((ln(3+2(√2)) 0)),((0 ln(3−2(√2)))) ] V^(−1) =(1/(∣V∣))adj(V) adj(V)= [((1−(√2) −1)),((−1−(√2) 1)) ] ∣V∣=1−(√2)−1−(√2)=−2(√2) V^(−1) =((−1)/(2(√2))) [((1−(√2) −1)),((−1−(√2) 1)) ] ∴ lnA=V(lnA^′ )V^(−1) lnA=((−1)/(2(√2))) [((1 1)),((1+(√2) 1−(√2))) ] [((ln(3+2(√2)) 0)),((0 ln(3−2(√2)))) ] [((1−(√2) −1)),((−1−(√2) 1)) ] lnA=((−1)/(2(√2))) [((ln(3+2(√2)) ln(3−2(√2)))),(((1+(√2))ln(3+2(√2)) (1−(√2))ln(3−2(√2)))) ] [((1−(√2) −1)),((−1−(√2) 1)) ] lnA=((−1)/(2(√2))) [(((1−(√2))ln(3+2(√2))−(1+(√2))ln(3−2(√2)) ln(((3−2(√2))/(3+2(√2)))) )),(( ln(((3−2(√2))/(3+2(√2)))) (1−(√2))ln(3−2(√2))−(1+(√2))ln(3+2(√2)))) ] ∴(lnA)B=((−1)/(2(√2))) [(((1−(√2))ln(3+2(√2))−(1+(√2))ln(3−2(√2)) ln(((3−2(√2))/(3+2(√2)))))),(( ln(((3−2(√2))/(3+2(√2)))) (1−(√2))ln(3−2(√2))−(1+(√2))ln(3+2(√2)))) ]×B$
	$∣ A ∣= 5 - 4 = 1 \neq 0 \Rightarrow A i s n o n - s i n g u l a r .$ $A^{B} = e^{(l n A) B} = \underset{k = 0}{\sum^{\infty}} \frac{1}{k!} {(l n A) B}^{k}$ $F o r e i g e n v a l u e s k o f A$ $∣ A - k I ∣= 0$ $∴ \| \begin{matrix} 1 - k 2 \\ 2 5 - k \end{matrix} \| = 0$ $(1 - k) (5 - k) - 4 = 0$ $5 - 6 k + k^{2} - 4 = 0$ $k^{2} - 6 k + 1 = 0$ ${(k - 3)}^{2} - 9 + 1 = 0$ ${(k - 3)}^{2} = 8 \Rightarrow k = 3 \pm 2 \sqrt{2}$ $F o r k = 3 + 2 \sqrt{2}, (∵ A e = k e \Rightarrow (A - k I) e = 0)$ $[\begin{matrix} 1 - 3 - 2 \sqrt{2} 2 \\ 2 5 - 3 - 2 \sqrt{2} \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$ $(- 2 - 2 \sqrt{2}) x + 2 y = 0 \Rightarrow y = (1 + \sqrt{2}) x . . . . . (i)$ $2 x + (2 - 2 \sqrt{2}) y = 0$ $x + (1 - \sqrt{2}) y = 0 . . . . . . (i i)$ $y f r o m (i) l e a d s t o$ $x + (1 - \sqrt{2}) (1 + \sqrt{2}) x = 0$ $x + (1 - 2) x = 0$ $x - x = 0 (t r u e)$ $∴ W e c a n c h o o s e e i g e n v e c t o r e_{1} = [\begin{matrix} 1 \\ 1 + \sqrt{2} \end{matrix}]$ $f o r k = 3 + 2 \sqrt{2}$ $F o r k = 3 - 2 \sqrt{2}$ $[\begin{matrix} - 2 + 2 \sqrt{2} 2 \\ 2 2 + 2 \sqrt{2} \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$ $\Rightarrow (- 1 + \sqrt{2}) x + y = 0 \Rightarrow y = (1 - \sqrt{2}) x . . . . . . . . . . . . . . (i i i)$ $a n d 2 x + (2 + 2 \sqrt{2}) y = 0 \Rightarrow x + (1 + \sqrt{2}) y = 0 . . . . . . . . (i v)$ $y f r o m (i i i) i n t o (i v) g i v e s$ $x + (1 - 2) x = 0 \Rightarrow x - x = 0 (t r u e)$ $S o l e t e_{2} = [\begin{matrix} 1 \\ 1 - \sqrt{2} \end{matrix}] f o r k = 3 - 2 \sqrt{2} .$ $∴ L e t V = [e_{1} e_{2}] = [\begin{matrix} 1 1 \\ 1 + \sqrt{2} 1 - \sqrt{2} \end{matrix}]$ $a n d A^{^{'}} = [\begin{matrix} 3 + 2 \sqrt{2} 0 \\ 0 3 - 2 \sqrt{2} \end{matrix}] \Rightarrow {l n A}^{^{'}} = [\begin{matrix} l n (3 + 2 \sqrt{2}) 0 \\ 0 l n (3 - 2 \sqrt{2}) \end{matrix}]$ $V^{- 1} = \frac{1}{∣ V ∣} a d j (V)$ $a d j (V) = [\begin{matrix} 1 - \sqrt{2} - 1 \\ - 1 - \sqrt{2} 1 \end{matrix}] ∣ V ∣= 1 - \sqrt{2} - 1 - \sqrt{2} = - 2 \sqrt{2}$ $V^{- 1} = \frac{- 1}{2 \sqrt{2}} [\begin{matrix} 1 - \sqrt{2} - 1 \\ - 1 - \sqrt{2} 1 \end{matrix}]$ $∴ l n A = V ({l n A}^{^{'}}) V^{- 1}$ $l n A = \frac{- 1}{2 \sqrt{2}} [\begin{matrix} 1 1 \\ 1 + \sqrt{2} 1 - \sqrt{2} \end{matrix}] [\begin{matrix} l n (3 + 2 \sqrt{2}) 0 \\ 0 l n (3 - 2 \sqrt{2}) \end{matrix}] [\begin{matrix} 1 - \sqrt{2} - 1 \\ - 1 - \sqrt{2} 1 \end{matrix}]$ $l n A = \frac{- 1}{2 \sqrt{2}} [\begin{matrix} l n (3 + 2 \sqrt{2}) l n (3 - 2 \sqrt{2}) \\ (1 + \sqrt{2}) l n (3 + 2 \sqrt{2}) (1 - \sqrt{2}) l n (3 - 2 \sqrt{2}) \end{matrix}] [\begin{matrix} 1 - \sqrt{2} - 1 \\ - 1 - \sqrt{2} 1 \end{matrix}]$ $l n A = \frac{- 1}{2 \sqrt{2}} [\begin{matrix} (1 - \sqrt{2}) l n (3 + 2 \sqrt{2}) - (1 + \sqrt{2}) l n (3 - 2 \sqrt{2}) l n (\frac{3 - 2 \sqrt{2}}{3 + 2 \sqrt{2}}) \\ l n (\frac{3 - 2 \sqrt{2}}{3 + 2 \sqrt{2}}) (1 - \sqrt{2}) l n (3 - 2 \sqrt{2}) - (1 + \sqrt{2}) l n (3 + 2 \sqrt{2}) \end{matrix}]$ $∴ (l n A) B = \frac{- 1}{2 \sqrt{2}} [\begin{matrix} (1 - \sqrt{2}) l n (3 + 2 \sqrt{2}) - (1 + \sqrt{2}) l n (3 - 2 \sqrt{2}) l n (\frac{3 - 2 \sqrt{2}}{3 + 2 \sqrt{2}}) \\ l n (\frac{3 - 2 \sqrt{2}}{3 + 2 \sqrt{2}}) (1 - \sqrt{2}) l n (3 - 2 \sqrt{2}) - (1 + \sqrt{2}) l n (3 + 2 \sqrt{2}) \end{matrix}] \times B$
	Commented by RasheedAhmad last updated on 22/Nov/15

	$I^{'} v e l e a r n t s o m e t h i n g f r o m y o u$ $t h a t I {d i d n}^{'} t k n o w b e f o r e!$
	Commented by Yozzi last updated on 22/Nov/15

	$H o n e s t l y, I {h a d n}^{'} t k n o w n h o w t o f i n d$ $A^{B} u n t i l t o d a y . S o m e t i m e a g o$ $I l o o k e d a t W i k i p e d i a t o o b t a i n i n f o r m a t i o n$ $o n s o l v i n g a m a t r i x d i f f e r e n t i a l$ $e q u a t i o n o f t h e f o r m \frac{d r}{d t} + A r = o$ $w h e r e i t w a s s u g g e s t e d t h a t a$ $s u i t a b l e m a t r i x i n t e g r a t i n g f a c t o r$ $b e e m p l o y e d . I n (i n c o r r e c t l y) g u e s s i n g t h a t t h e$ $r e q u i r e d f a c t o r i s o f t h e f o r m e^{A}$ $I w a s m e t w i t h a c o o l i d e a c o n c e r n i n g$ $t h e e x p r e s s i o n e^{X} f o r X b e i n g a m a t r i x$ $a n d e b e i n g {E u l e r}^{'} s c o n s t a n t . S o$ $w h e n I s a w y o u r q u e s t i o n o n A^{B} I t h o u g h t$ $a b o u t t h e e q u i v a l e n t e x p r e s s i o n,$ $w h i c h w i k i c o r r e c t e d m e o n, e^{(l n A) B} .$ $W e h a v e B_{A} \equiv e^{B (l n A)} d u e t o t h e g e n e r a l$ $n o n - c o m m u t a t i v e n a t u r e o f m a t r i c e s .$ $L u c k i l y I s t u d i e d s o m e l i n e a r a l g e b r a$ $b e f o r e s o a p p l y i n g t h e m e t h o d I$ $f o u n d o n W i k i p e d i a {w a s n}^{'} t t o o h a r d .$
	Commented by Rasheed Soomro last updated on 23/Nov/15

	$TH a N k S s s s f o r S h a r i n g t h o u g h t s w i t h o p e n m i n d!$ $I t c a n b e s a i d t h a t m y q u e s t i o n h a s b e c o m e c a u s e t o$ $i n c r e a s e y o u r k n o w l e d g e a n d i n t e r e s t AND y o u r$ $S h a r i n g - K n o w l e d g e h a s b e c o m e c a u s e t o i n c r e a s e$ $m y k n o w l e d g e!$ $A c t u a l l y t h e q u e s t i o n i s r e s u l t o f m y i m a g i n a t i o n,$ $o t h e r w i s e m y k n o w l e d g e r e g a r d i n g M a t r i c e s i s o f$ $e l e m e n t a r y l e v e l .$ $Y o u r c o m m e n t c o n t a i n s a n e x p r e s s i o n^{'} B_{A} \equiv e^{B (l n A)} . . . d u e t o$ $n o n - c o m m u t a t i v e n a t u r e . . .^{'} . I w o u l d l i k e t o k n o w$ $t h e m e a n i n g o f B_{A} . C o n s i d e r i n g n o n - c o m m u t a t i v e$ $n a t u r e . . . {s h o u l d n}^{'} t w e w r i t e B^{A} a s n o n - c o m m u t a t i v e$ $p a r t o f A^{B} ?$
	Commented by Yozzi last updated on 23/Nov/15

	$T h e w i k i i n d i c a t e d t h a t w h i l e$ $A^{B} = e^{(l n A) B},^{A} B = e^{B (l n A)} .$ $I j u s t f i g u r e d o u t h o w t o o b t a i n$ $l e f t - h a n d e x p o n e n t s . I w a s t r y i n g$ $t o d o s o b y w r i t t i n g B_{A} . . .$
	Commented by Rasheed Soomro last updated on 23/Nov/15

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