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Question Number 74019 by mathmax by abdo last updated on 17/Nov/19
let the matrix A = (((1 2)),((0 −3)) ) 1) calculate A^n for n integr 2) find e^A and e^(−A) .
$${let}\:{the}\:{matrix}\:\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:−\mathrm{3}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}^{{n}} \:\:{for}\:{n}\:{integr} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{e}^{{A}} \:\:{and}\:{e}^{−{A}} . \\ $$
Commented by mathmax by abdo last updated on 18/Nov/19
the caracteristic polynom of A is p(x)=det(A−xI) = determinant (((1−x 2)),((0 −3−x)))=−(3+x)(1−x) =(x+3)(x−1) the propers value is λ_1 =−3 and λ_2 =1 V(λ_1 )=ker(A+3I)={u/(A+3I)u=0} u ((x),(y) ) (A+3I)u=0 ⇔ (((4 2)),((0 0)) ) ((x),(y) ) =0 ⇒4x+2y =0 ⇒2x+y=0 ⇒ y=−2x ⇒(x,y)=(x,−2x)=x(1,−2) ⇒V(−3)=vet(e_1 ) with e_1 (1,−2) V(λ_1 )=Ker(A−I)={u /(A−I)u=0} (A−I)u=0 ⇒ (((0 2)),((0 −4)) ) ((x),(y) )=0 ⇒y=0 ⇒(x,y)=x(1,0) ⇒P = (((1 1)),((−2 0)) ) and D = (((−3 0)),((0 1)) ) A =P DP^(−1) ⇒ A^n =PD^n P^(−1) we have p_c (P)= determinant (((1−x 1)),((−2 −x)))=−x(1−x)+2 =x^2 −x +2 cayley hamilton ⇒p^2 −p +2I =0 ⇒p(p−I) =−2I ⇒ p{−(1/2)(p−I)} =I ⇒p^(−1) =(1/2){ I−p} =(1/2){ (((1 0)),((0 1)) ) − (((1 1)),((−2 0)) )} =(1/2) (((0 −1)),((2 1)) ) A^n =(1/2) (((1 1)),((−2 0)) ) ((((−3)^n 0)),((0 1)) ) (((0 −1)),((2 1)) ) =(1/2) ((((−3)^(n ) 1)),((−2(−3)^n 0)) ) (((0 −1)),((2 1)) ) =(1/2) ((( 2 1−(−3)^n )),((0 2(−3)^n )) ) = (((1 ((1−(−3)^n )/2))),((0 (−3)^n )) )
$${the}\:{caracteristic}\:{polynom}\:{of}\:{A}\:{is} \\ $$$${p}\left({x}\right)={det}\left({A}−{xI}\right)\:=\begin{vmatrix}{\mathrm{1}−{x}\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:−\mathrm{3}−{x}}\end{vmatrix}=−\left(\mathrm{3}+{x}\right)\left(\mathrm{1}−{x}\right) \\ $$$$=\left({x}+\mathrm{3}\right)\left({x}−\mathrm{1}\right)\:{the}\:{propers}\:{value}\:{is}\:\lambda_{\mathrm{1}} =−\mathrm{3}\:\:{and}\:\lambda_{\mathrm{2}} =\mathrm{1} \\ $$$${V}\left(\lambda_{\mathrm{1}} \right)={ker}\left({A}+\mathrm{3}{I}\right)=\left\{{u}/\left({A}+\mathrm{3}{I}\right){u}=\mathrm{0}\right\} \\ $$$${u}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\:\:\:\left({A}+\mathrm{3}{I}\right){u}=\mathrm{0}\:\:\Leftrightarrow\begin{pmatrix}{\mathrm{4}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\:=\mathrm{0}\:\Rightarrow\mathrm{4}{x}+\mathrm{2}{y}\:=\mathrm{0}\:\Rightarrow\mathrm{2}{x}+{y}=\mathrm{0}\:\Rightarrow \\ $$$${y}=−\mathrm{2}{x}\:\Rightarrow\left({x},{y}\right)=\left({x},−\mathrm{2}{x}\right)={x}\left(\mathrm{1},−\mathrm{2}\right)\:\Rightarrow{V}\left(−\mathrm{3}\right)={vet}\left({e}_{\mathrm{1}} \right)\:{with} \\ $$$${e}_{\mathrm{1}} \left(\mathrm{1},−\mathrm{2}\right) \\ $$$${V}\left(\lambda_{\mathrm{1}} \right)={Ker}\left({A}−{I}\right)=\left\{{u}\:/\left({A}−{I}\right){u}=\mathrm{0}\right\} \\ $$$$\left({A}−{I}\right){u}=\mathrm{0}\:\Rightarrow\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:−\mathrm{4}}\end{pmatrix}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\mathrm{0}\:\:\:\Rightarrow{y}=\mathrm{0}\:\Rightarrow\left({x},{y}\right)={x}\left(\mathrm{1},\mathrm{0}\right) \\ $$$$\Rightarrow{P}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\:\:\:{and}\:\:{D}\:=\begin{pmatrix}{−\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${A}\:={P}\:{DP}^{−\mathrm{1}} \:\Rightarrow\:{A}^{{n}} ={PD}^{{n}} {P}^{−\mathrm{1}} \: \\ $$$${we}\:{have}\:\:{p}_{{c}} \left({P}\right)=\begin{vmatrix}{\mathrm{1}−{x}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:−{x}}\end{vmatrix}=−{x}\left(\mathrm{1}−{x}\right)+\mathrm{2}\:={x}^{\mathrm{2}} −{x}\:+\mathrm{2} \\ $$$${cayley}\:{hamilton}\:\Rightarrow{p}^{\mathrm{2}} −{p}\:+\mathrm{2}{I}\:=\mathrm{0}\:\Rightarrow{p}\left({p}−{I}\right)\:=−\mathrm{2}{I}\:\Rightarrow \\ $$$${p}\left\{−\frac{\mathrm{1}}{\mathrm{2}}\left({p}−{I}\right)\right\}\:={I}\:\Rightarrow{p}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{I}−{p}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:−\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${A}^{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\:\begin{pmatrix}{\left(−\mathrm{3}\right)^{{n}} \:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\left(−\mathrm{3}\right)^{{n}\:\:\:\:} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\left(−\mathrm{3}\right)^{{n}} \:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\:\:\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\begin{pmatrix}{\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−\left(−\mathrm{3}\right)^{{n}} }\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\left(−\mathrm{3}\right)^{{n}} }\end{pmatrix}\:\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}−\left(−\mathrm{3}\right)^{{n}} }{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(−\mathrm{3}\right)^{{n}} }\end{pmatrix} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 18/Nov/19
2) e^A =Σ_(n=0) ^∞ (A^n /(n!)) = (((Σ_(n=0) ^∞ (1/(n!)) Σ_(n=0) ((1−(−3)^n )/(2n!)))),((0 Σ_(n=0) ^∞ (((−3)^n )/(n!)) )) ) = ((( e (1/2)(e−e^(−3) ))),((0 e^(−3) )) )
$$\left.\mathrm{2}\right)\:{e}^{{A}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{A}^{{n}} }{{n}!}\:=\:\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} \:\frac{\mathrm{1}−\left(−\mathrm{3}\right)^{{n}} }{\mathrm{2}{n}!}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{3}\right)^{{n}} }{{n}!}\:}\end{pmatrix} \\ $$$$=\:\begin{pmatrix}{\:\:{e}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left({e}−{e}^{−\mathrm{3}} \right)}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{−\mathrm{3}} }\end{pmatrix} \\ $$
Commented by mathmax by abdo last updated on 18/Nov/19
e^(−A) =Σ_(n=0) ^∞ (((−1)^n )/(n!)) A^n =Σ_(n=0) ^∞ (((−1)^n )/(n!)) (((1 ((1−(−3)^n )/2))),((0 (−3)^n )) ) = ((( Σ_(n=0) ^∞ (((−1)^n )/(n!)) Σ_(n=0) ^∞ (((−1)^n )/(n!)) ((1−(−3)^n )/2))),((0 Σ_(n=0) ^∞ (((−1)^n (−3)^n )/(n!)) )) ) = ((( e^(−1) (1/2)( e^(−1) −e^3 ) )),((0 e^3 )) )
$${e}^{−{A}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{A}^{{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}−\left(−\mathrm{3}\right)^{{n}} }{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(−\mathrm{3}\right)^{{n}} }\end{pmatrix} \\ $$$$=\begin{pmatrix}{\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\:\frac{\mathrm{1}−\left(−\mathrm{3}\right)^{{n}} }{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} \left(−\mathrm{3}\right)^{{n}} }{{n}!}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{pmatrix} \\ $$$$=\:\begin{pmatrix}{\:\:\:{e}^{−\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\:\:{e}^{−\mathrm{1}} −{e}^{\mathrm{3}} \right)\:\:\:\:}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{pmatrix} \\ $$
Answered by mind is power last updated on 17/Nov/19
det(A−xI_2 )=0 let f(x,y)=(x+2y,−3y) ⇒(1−x)(−3−x)=0⇒x∈{1,−3} x=1 f(x,y)=(x,y)⇒y=0 e=(1,0) f(x,y)=−3(x,y)⇒4x+2y=0⇒y=−2x e_2 =(1,−2) A=PDP^(−1) P= (((1 1)),((0 −2)) ),P^− =−(1/2) (((−2 −1)),((0 1)) ) A=−(1/2) (((1 1)),((0 −2)) ). (((1 0)),((0 −3)) ). (((−2 −1)),(( 0 1)) ) A^n =ePD^n P^− e^A =Σ(A^k /(k!))=P(Σ_(k=0) ^(+∞) (D^k /(k!)).)P^− =P. (((e 0)),((0 e^(−3) )) ).P^− e^(−A) =P(Σ_(k=0) ^(+∞) (((−D)^k )/(k!)))P^− =P. (((e^(−1) 0)),((0 e^3 )) ) P^−
$$\mathrm{det}\left(\mathrm{A}−\mathrm{xI}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{x}+\mathrm{2y},−\mathrm{3y}\right) \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{x}\right)\left(−\mathrm{3}−\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{x}\in\left\{\mathrm{1},−\mathrm{3}\right\} \\ $$$$\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{x},\mathrm{y}\right)\Rightarrow\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{e}=\left(\mathrm{1},\mathrm{0}\right) \\ $$$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=−\mathrm{3}\left(\mathrm{x},\mathrm{y}\right)\Rightarrow\mathrm{4x}+\mathrm{2y}=\mathrm{0}\Rightarrow\mathrm{y}=−\mathrm{2x} \\ $$$$\mathrm{e}_{\mathrm{2}} =\left(\mathrm{1},−\mathrm{2}\right) \\ $$$$\mathrm{A}=\mathrm{PDP}^{−\mathrm{1}} \\ $$$$\mathrm{P}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\:−\mathrm{2}}\end{pmatrix},\mathrm{P}^{−} =−\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{A}=−\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\:−\mathrm{2}}\end{pmatrix}.\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:−\mathrm{3}}\end{pmatrix}.\begin{pmatrix}{−\mathrm{2}\:\:\:−\mathrm{1}}\\{\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{A}^{\mathrm{n}} =\mathrm{ePD}^{\mathrm{n}} \mathrm{P}^{−} \\ $$$$\mathrm{e}^{\mathrm{A}} =\Sigma\frac{\mathrm{A}^{\mathrm{k}} }{\mathrm{k}!}=\mathrm{P}\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{D}^{\mathrm{k}} }{\mathrm{k}!}.\right)\mathrm{P}^{−} =\mathrm{P}.\begin{pmatrix}{\mathrm{e}\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{e}^{−\mathrm{3}} }\end{pmatrix}.\mathrm{P}^{−} \\ $$$$\mathrm{e}^{−\mathrm{A}} =\mathrm{P}\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{D}\right)^{\mathrm{k}} }{\mathrm{k}!}\right)\mathrm{P}^{−} =\mathrm{P}.\begin{pmatrix}{\mathrm{e}^{−\mathrm{1}} \:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{e}^{\mathrm{3}} }\end{pmatrix}\:\mathrm{P}^{−} \\ $$
Commented by abdomathmax last updated on 18/Nov/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$

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