let A = (((1 1)),((−2 3)) ) 1) find A^(−1) 2) calculate A^n 3) determine e^A and e^(−2A) .


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Question Number 60500 by prof Abdo imad last updated on 21/May/19

$l e t A = (\begin{matrix} 1 1 \\ - 2 3 \end{matrix})$ $1) f i n d A^{- 1}$ $2) c a l c u l a t e A^{n}$ $3) d e t e r m i n e e^{A} a n d e^{- 2 A} .$
Commented by maxmathsup by imad last updated on 22/May/19
$2) we have P_c (A) =det(A−xI)= determinant (((1−x 1)),((−2 3−x)))=(1−x)(3−x) +2=3−x−3x+x^2 +2 =x^2 −4x+5 P_c (A)=0 ⇒x^2 −4x +5 =0→Δ^′ =4−5 =−1 =i^2 ⇒ x_1 =2+i and x_2 =2−i v(x_1 )=Ker(A−x_1 I) ={u /(A−x_1 )u=0} let u ((x),(y) ) (A−x_1 ) ((x),(y) ) =0 ⇒ (((−1−i 1)),((−2 1−i)) ) ((x),(y) )= ((0),(0) ) ⇒ {_(−2+(1−i)y =0) ^((−1−i)x+y =o) Δ_s = determinant (((−1−i 1)),((−2 1−i)))=−(1+i)(1−i)+2 =−(2) +2 =0 ⇒ y=(1+i)x ⇒(x,y) =(x,(1+i)x) =xe_1 with e_1 ((1),((1+i)) ) v(x_2 ) = Ker (A−x_2 I) ={u/(A−x_2 I)u=0} let u ((x),(y) ) (A−x_2 I)u =0 ⇒ (((−1+i 1)),((−2 1+i)) ) ((x),(y) )= ((0),(0) ) ⇒{_(−2x +(1+i)y =0) ^((−1+i)x +y =0) ⇒ Δ_s = determinant (((−1+i 1)),((−2 1+i)))=−(1−i)(1+i)+2 =0 ⇒y=(1−i)x ⇒ (x,y) =(x,(1−i)x) =x(1,1−i) ⇒e_2 ((1),((1−i )) ) ⇒M_p = (((1 1)),((1+i 1−i)) )=P and D = (((2+i 0)),((0 2−i)) ) we have A =P DP^(−1) ⇒ A^n =PD^n P^(−1) P^(−1) =((t(com(P)))/(det P)) det P =1−i−(1+i) =1−i−1−i =−2i com(P) =(−1)^(i+j) A_(ij) (((α_(11) α_(12) )),((α_2 1 α_(22) )) ) α_(11) =1−i , α_(21) =−(1) , α_(12) =−(1+i) , α_(22) =1 ⇒ com(P) = (((1−i −1−i )),((−1 1)) ) ⇒t(com(P)) = (((1−i −1)),((−1−i 1)) ) ⇒ P^(−1) =−(1/(2i)) (((1−i −1)),((−1−i 1)) ) = (((((−1+i)/(2i)) (1/(2i)))),((((1+i)/(2i)) ((−1)/(2i)))) ) = (((((−i−1)/(−2)) (i/(−2)))),((((i−1)/(−2)) ((−i)/(−2)))) ) = (((((1+i)/2) ((−i)/2))),((((1−i)/2) (i/2))) ) A^n =(1/2) (((1 1)),((1+i 1−i)) ) ((((2+i)^n 0)),((0 (2−i)^n )) ) (((1+i −i)),((1−i i)) ) rest to finish the calculus....$
$2) w e h a v e P_{c} (A) = d e t (A - x I) = \| \begin{matrix} 1 - x 1 \\ - 2 3 - x \end{matrix} \| = (1 - x) (3 - x) + 2 = 3 - x - 3 x + x^{2} + 2$ $= x^{2} - 4 x + 5 P_{c} (A) = 0 \Rightarrow x^{2} - 4 x + 5 = 0 \to Δ^{^{'}} = 4 - 5 = - 1 = i^{2} \Rightarrow$ $x_{1} = 2 + i a n d x_{2} = 2 - i$ $v (x_{1}) = K e r (A - x_{1} I) = {u / (A - x_{1}) u = 0} l e t u (\begin{matrix} x \\ y \end{matrix})$ $(A - x_{1}) (\begin{matrix} x \\ y \end{matrix}) = 0 \Rightarrow (\begin{matrix} - 1 - i 1 \\ - 2 1 - i \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \Rightarrow {_{- 2 + (1 - i) y = 0}^{(- 1 - i) x + y = o}$ $Δ_{s} = \| \begin{matrix} - 1 - i 1 \\ - 2 1 - i \end{matrix} \| = - (1 + i) (1 - i) + 2 = - (2) + 2 = 0 \Rightarrow$ $y = (1 + i) x \Rightarrow (x, y) = (x, (1 + i) x) = {x e}_{1} w i t h e_{1} (\begin{matrix} 1 \\ 1 + i \end{matrix})$ $v (x_{2}) = K e r (A - x_{2} I) = {u / (A - x_{2} I) u = 0} l e t u (\begin{matrix} x \\ y \end{matrix})$ $(A - x_{2} I) u = 0 \Rightarrow (\begin{matrix} - 1 + i 1 \\ - 2 1 + i \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \Rightarrow {_{- 2 x + (1 + i) y = 0}^{(- 1 + i) x + y = 0} \Rightarrow$ $Δ_{s} = \| \begin{matrix} - 1 + i 1 \\ - 2 1 + i \end{matrix} \| = - (1 - i) (1 + i) + 2 = 0 \Rightarrow y = (1 - i) x \Rightarrow$ $(x, y) = (x, (1 - i) x) = x (1, 1 - i) \Rightarrow e_{2} (\begin{matrix} 1 \\ 1 - i \end{matrix}) \Rightarrow M_{p} = (\begin{matrix} 1 1 \\ 1 + i 1 - i \end{matrix}) = P$ $a n d D = (\begin{matrix} 2 + i 0 \\ 0 2 - i \end{matrix})$ $w e h a v e A = P {D P}^{- 1} \Rightarrow A^{n} = {P D}^{n} P^{- 1}$ $P^{- 1} = \frac{t (c o m (P))}{d e t P} d e t P = 1 - i - (1 + i) = 1 - i - 1 - i = - 2 i$ $c o m (P) = {(- 1)}^{i + j} A_{i j} (\begin{matrix} α_{11} α_{12} \\ α_{2} 1 α_{22} \end{matrix})$ $α_{11} = 1 - i, α_{21} = - (1), α_{12} = - (1 + i), α_{22} = 1 \Rightarrow$ $c o m (P) = (\begin{matrix} 1 - i - 1 - i \\ - 1 1 \end{matrix}) \Rightarrow t (c o m (P)) = (\begin{matrix} 1 - i - 1 \\ - 1 - i 1 \end{matrix}) \Rightarrow$ $P^{- 1} = - \frac{1}{2 i} (\begin{matrix} 1 - i - 1 \\ - 1 - i 1 \end{matrix}) = (\begin{matrix} \frac{- 1 + i}{2 i} \frac{1}{2 i} \\ \frac{1 + i}{2 i} \frac{- 1}{2 i} \end{matrix})$ $= (\begin{matrix} \frac{- i - 1}{- 2} \frac{i}{- 2} \\ \frac{i - 1}{- 2} \frac{- i}{- 2} \end{matrix}) = (\begin{matrix} \frac{1 + i}{2} \frac{- i}{2} \\ \frac{1 - i}{2} \frac{i}{2} \end{matrix})$ $A^{n} = \frac{1}{2} (\begin{matrix} 1 1 \\ 1 + i 1 - i \end{matrix}) (\begin{matrix} {(2 + i)}^{n} 0 \\ 0 {(2 - i)}^{n} \end{matrix}) (\begin{matrix} 1 + i - i \\ 1 - i i \end{matrix})$ $r e s t t o f i n i s h t h e c a l c u l u s . . . .$
Commented by maxmathsup by imad last updated on 22/May/19

$1) w e h a v e P_{c} (X) = x^{2} - 4 x + 5 c a y l e y h a m i l t o n t h e o r e m g i v e$ $P_{c} (A) = 0 \Rightarrow A^{2} - 4 A + 5 I = 0 \Rightarrow A^{2} = 4 A - 5 I \Rightarrow A = 4 I - 5 A^{- 1} \Rightarrow$ $5 A^{- 1} = - A + 4 I = - (\begin{matrix} 1 1 \\ - 2 3 \end{matrix}) + (\begin{matrix} 4 0 \\ 0 4 \end{matrix}) = (\begin{matrix} 3 - 1 \\ 2 1 \end{matrix})$ $\Rightarrow A^{- 1} = (\begin{matrix} \frac{3}{5} \frac{- 1}{5} \\ \frac{2}{5} \frac{1}{5} \end{matrix}) b u t f i r s t w d m u s t v e r i f y t h a t d e t A \neq 0!$
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