A-1-2-2-5-B-3-1-4-2-determine-A-B-

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 \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 \dots . . (i)

2 x + (2 - 2 \sqrt{2}) y = 0

x + (1 - \sqrt{2}) y = 0 \dots \dots (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 \dots \dots \dots \dots . . (i i i)

a n d 2 x + (2 + 2 \sqrt{2}) y = 0 \Rightarrow x + (1 + \sqrt{2}) y = 0 \dots \dots . . (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)} \dots 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 \dots^{'} . 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 \dots {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} \dots

Commented by Rasheed Soomro last updated on 23/Nov/15

THANKS!