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let-the-matrix-A-1-2-0-3-1-calculate-A-n-for-n-integr-2-find-e-A-and-e-A-

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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) .
letthematrixA=(120−3)1)calculateAnfornintegr2)findeAande−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 )) )
thecaracteristicpolynomofAisp(x)=det(A−xI)=|1−x20−3−x|=−(3+x)(1−x)=(x+3)(x−1)thepropersvalueisλ1=−3andλ2=1V(λ1)=ker(A+3I)={u/(A+3I)u=0}u(xy)(A+3I)u=0⇔(4200)(xy)=0⇒4x+2y=0⇒2x+y=0⇒y=−2x⇒(x,y)=(x,−2x)=x(1,−2)⇒V(−3)=vet(e1)withe1(1,−2)V(λ1)=Ker(A−I)={u/(A−I)u=0}(A−I)u=0⇒(020−4)(xy)=0⇒y=0⇒(x,y)=x(1,0)⇒P=(11−20)andD=(−3001)A=PDP−1⇒An=PDnP−1wehavepc(P)=|1−x1−2−x|=−x(1−x)+2=x2−x+2cayleyhamilton⇒p2−p+2I=0⇒p(p−I)=−2I⇒p{−12(p−I)}=I⇒p−1=12{I−p}=12{(1001)−(11−20)}=12(0−121)An=12(11−20)((−3)n001)(0−121)=12((−3)n1−2(−3)n0)(0−121)=12(21−(−3)n02(−3)n)=(11−(−3)n20(−3)n)
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) )) )
2)eA=∑n=0∞Ann!=(∑n=0∞1n!∑n=01−(−3)n2n!0∑n=0∞(−3)nn!)=(e12(e−e−3)0e−3)
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=∑n=0∞(−1)nn!An=∑n=0∞(−1)nn!(11−(−3)n20(−3)n)=(∑n=0∞(−1)nn!∑n=0∞(−1)nn!1−(−3)n20∑n=0∞(−1)n(−3)nn!)=(e−112(e−1−e3)0e3)
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^−
det(A−xI2)=0letf(x,y)=(x+2y,−3y)⇒(1−x)(−3−x)=0⇒x∈{1,−3}x=1f(x,y)=(x,y)⇒y=0e=(1,0)f(x,y)=−3(x,y)⇒4x+2y=0⇒y=−2xe2=(1,−2)A=PDP−1P=(110−2),P−=−12(−2−101)A=−12(110−2).(100−3).(−2−101)An=ePDnP−eA=ΣAkk!=P(∑+∞k=0Dkk!.)P−=P.(e00e−3).P−e−A=P(∑+∞k=0(−D)kk!)P−=P.(e−100e3)P−
Commented by abdomathmax last updated on 18/Nov/19
thank you sir.
thankyousir.

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