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Question Number 60500 by prof Abdo imad last updated on 21/May/19
let A = (((1 1)),((−2 3)) ) 1) find A^(−1) 2) calculate A^n 3) determine e^A and e^(−2A) .
$${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\mathrm{3}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{A}^{−\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{3}\right)\:{determine}\:{e}^{{A}} \:\:\:{and}\:{e}^{−\mathrm{2}{A}} \:. \\ $$
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....
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{P}_{{c}} \left({A}\right)\:={det}\left({A}−{xI}\right)=\begin{vmatrix}{\mathrm{1}−{x}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{3}−{x}}\end{vmatrix}=\left(\mathrm{1}−{x}\right)\left(\mathrm{3}−{x}\right)\:+\mathrm{2}=\mathrm{3}−{x}−\mathrm{3}{x}+{x}^{\mathrm{2}} \:+\mathrm{2} \\ $$$$={x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\:\:\:\:\:{P}_{{c}} \left({A}\right)=\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{5}\:=\mathrm{0}\rightarrow\Delta^{'} =\mathrm{4}−\mathrm{5}\:=−\mathrm{1}\:={i}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\mathrm{2}+{i}\:\:\:\:{and}\:{x}_{\mathrm{2}} =\mathrm{2}−{i} \\ $$$${v}\left({x}_{\mathrm{1}} \right)={Ker}\left({A}−{x}_{\mathrm{1}} {I}\right)\:=\left\{{u}\:/\left({A}−{x}_{\mathrm{1}} \right){u}=\mathrm{0}\right\}\:\:{let}\:{u}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix} \\ $$$$\left({A}−{x}_{\mathrm{1}} \right)\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\mathrm{0}\:\Rightarrow\begin{pmatrix}{−\mathrm{1}−{i}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−{i}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\Rightarrow\:\left\{_{−\mathrm{2}+\left(\mathrm{1}−{i}\right){y}\:=\mathrm{0}} ^{\left(−\mathrm{1}−{i}\right){x}+{y}\:={o}} \right. \\ $$$$\Delta_{{s}} =\begin{vmatrix}{−\mathrm{1}−{i}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−{i}}\end{vmatrix}=−\left(\mathrm{1}+{i}\right)\left(\mathrm{1}−{i}\right)+\mathrm{2}\:=−\left(\mathrm{2}\right)\:+\mathrm{2}\:=\mathrm{0}\:\Rightarrow \\ $$$${y}=\left(\mathrm{1}+{i}\right){x}\:\Rightarrow\left({x},{y}\right)\:=\left({x},\left(\mathrm{1}+{i}\right){x}\right)\:={xe}_{\mathrm{1}} \:\:\:{with}\:\:{e}_{\mathrm{1}} \begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}+{i}}\end{pmatrix} \\ $$$${v}\left({x}_{\mathrm{2}} \right)\:=\:{Ker}\:\left({A}−{x}_{\mathrm{2}} {I}\right)\:=\left\{{u}/\left({A}−{x}_{\mathrm{2}} {I}\right){u}=\mathrm{0}\right\}\:\:{let}\:{u}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix} \\ $$$$\left({A}−{x}_{\mathrm{2}} {I}\right){u}\:=\mathrm{0}\:\Rightarrow\begin{pmatrix}{−\mathrm{1}+{i}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}+{i}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:\Rightarrow\left\{_{−\mathrm{2}{x}\:+\left(\mathrm{1}+{i}\right){y}\:=\mathrm{0}} ^{\left(−\mathrm{1}+{i}\right){x}\:+{y}\:=\mathrm{0}} \:\:\:\Rightarrow\right. \\ $$$$\Delta_{{s}} =\begin{vmatrix}{−\mathrm{1}+{i}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}+{i}}\end{vmatrix}=−\left(\mathrm{1}−{i}\right)\left(\mathrm{1}+{i}\right)+\mathrm{2}\:=\mathrm{0}\:\Rightarrow{y}=\left(\mathrm{1}−{i}\right){x}\:\Rightarrow \\ $$$$\left({x},{y}\right)\:=\left({x},\left(\mathrm{1}−{i}\right){x}\right)\:={x}\left(\mathrm{1},\mathrm{1}−{i}\right)\:\Rightarrow{e}_{\mathrm{2}} \begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}−{i}\:}\end{pmatrix}\:\Rightarrow{M}_{{p}} =\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}+{i}\:\:\:\:\:\:\:\mathrm{1}−{i}}\end{pmatrix}={P} \\ $$$${and}\:\:{D}\:=\begin{pmatrix}{\mathrm{2}+{i}\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}−{i}}\end{pmatrix} \\ $$$${we}\:{have}\:{A}\:={P}\:{DP}^{−\mathrm{1}} \:\:\Rightarrow\:{A}^{{n}} \:={PD}^{{n}} {P}^{−\mathrm{1}} \:\: \\ $$$${P}^{−\mathrm{1}} \:=\frac{{t}\left({com}\left({P}\right)\right)}{{det}\:{P}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{det}\:{P}\:=\mathrm{1}−{i}−\left(\mathrm{1}+{i}\right)\:=\mathrm{1}−{i}−\mathrm{1}−{i}\:=−\mathrm{2}{i} \\ $$$${com}\left({P}\right)\:=\left(−\mathrm{1}\right)^{{i}+{j}} {A}_{{ij}} \:\:\begin{pmatrix}{\alpha_{\mathrm{11}} \:\:\:\:\:\:\:\:\:\:\:\alpha_{\mathrm{12}} }\\{\alpha_{\mathrm{2}} \mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\alpha_{\mathrm{22}} }\end{pmatrix} \\ $$$$\alpha_{\mathrm{11}} =\mathrm{1}−{i}\:\:\:\:\:,\:\:\:\:\alpha_{\mathrm{21}} =−\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:,\:\:\:\alpha_{\mathrm{12}} =−\left(\mathrm{1}+{i}\right)\:\:\:\:\:,\:\:\alpha_{\mathrm{22}} =\mathrm{1}\:\Rightarrow \\ $$$${com}\left({P}\right)\:=\begin{pmatrix}{\mathrm{1}−{i}\:\:\:\:\:\:\:\:\:−\mathrm{1}−{i}\:}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\Rightarrow{t}\left({com}\left({P}\right)\right)\:=\:\begin{pmatrix}{\mathrm{1}−{i}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}−{i}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\Rightarrow \\ $$$${P}^{−\mathrm{1}} \:=−\frac{\mathrm{1}}{\mathrm{2}{i}}\begin{pmatrix}{\mathrm{1}−{i}\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}−{i}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:=\:\:\begin{pmatrix}{\frac{−\mathrm{1}+{i}}{\mathrm{2}{i}}\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}{i}}}\\{\frac{\mathrm{1}+{i}}{\mathrm{2}{i}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{2}{i}}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\frac{−{i}−\mathrm{1}}{−\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{{i}}{−\mathrm{2}}}\\{\frac{{i}−\mathrm{1}}{−\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{−{i}}{−\mathrm{2}}}\end{pmatrix}\:=\begin{pmatrix}{\frac{\mathrm{1}+{i}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{−{i}}{\mathrm{2}}}\\{\frac{\mathrm{1}−{i}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{i}}{\mathrm{2}}}\end{pmatrix} \\ $$$${A}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{2}}\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}+{i}\:\:\:\:\:\:\:\:\mathrm{1}−{i}}\end{pmatrix}\begin{pmatrix}{\left(\mathrm{2}+{i}\right)^{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}−{i}\right)^{{n}} }\end{pmatrix}\begin{pmatrix}{\mathrm{1}+{i}\:\:\:\:\:\:\:−{i}}\\{\mathrm{1}−{i}\:\:\:\:\:\:\:\:\:{i}}\end{pmatrix} \\ $$$${rest}\:{to}\:{finish}\:\:{the}\:{calculus}…. \\ $$
Commented by maxmathsup by imad last updated on 22/May/19
1) we have P_c (X) = x^2 −4x +5 cayley hamilton theorem give P_c (A) =0 ⇒A^2 −4A +5I =0 ⇒A^2 =4A −5I ⇒A =4I −5 A^(−1) ⇒ 5A^(−1) =−A +4I =− (((1 1)),((−2 3)) ) + (((4 0)),((0 4)) ) = (((3 −1)),((2 1)) ) ⇒ A^(−1) = ((((3/5) ((−1)/5))),(((2/5) (1/5))) ) but first wd must verify that detA≠0 !
$$\left.\mathrm{1}\right)\:\:{we}\:{have}\:{P}_{{c}} \left({X}\right)\:=\:{x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{5}\:\:\:\:\:{cayley}\:{hamilton}\:{theorem}\:{give} \\ $$$${P}_{{c}} \left({A}\right)\:=\mathrm{0}\:\Rightarrow{A}^{\mathrm{2}} −\mathrm{4}{A}\:+\mathrm{5}{I}\:=\mathrm{0}\:\:\Rightarrow{A}^{\mathrm{2}} \:=\mathrm{4}{A}\:−\mathrm{5}{I}\:\Rightarrow{A}\:=\mathrm{4}{I}\:−\mathrm{5}\:{A}^{−\mathrm{1}} \:\Rightarrow \\ $$$$\mathrm{5}{A}^{−\mathrm{1}} \:=−{A}\:+\mathrm{4}{I}\:\:\:=−\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{4}\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{4}}\end{pmatrix}\:\:=\:\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\Rightarrow\:{A}^{−\mathrm{1}} \:=\:\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{5}}}\\{\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{5}}}\end{pmatrix}\:\:\:\:\:{but}\:{first}\:{wd}\:{must}\:{verify}\:\:{that}\:\:{detA}\neq\mathrm{0}\:! \\ $$

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