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Question Number 91268 by mathmax by abdo last updated on 29/Apr/20

let A = (((1 2)),((1 −1)) ) 1) calculste A^n 2) determine e^A and e^(−2A) 3)find cos(A)and sinA is cos^2 A +sin^2 A =I? 4) determine sh(A) and ch(A) is ch^2 A−sh^2 A =I ?

letA=(121−1)1)calculsteAn2)determineeAande−2A3)findcos(A)andsinAiscos2A+sin2A=I?4)determinesh(A)andch(A)isch2A−sh2A=I?

Commented by mathmax by abdo last updated on 29/Apr/20

1) the caracteristic polynom is p(x)=det (A−xI) = determinant (((1−x 2)),((1 −1−x)))=−(1+x)(1−x)−2 =−(1−x^2 )−2 =−1+x^2 −2 =x^2 −3 =(x−(√3))(x+(√3)) p(x)=0 ⇔x =+^− (√3) x^n =Qp(x)+u_n x +v_n ⇒((√3))^n =u_n (√3)+v_n (−(√3))^n =−(√3)u_n +v_n ⇒((√3))^n −(−(√3))^n =2(√3)u_n ⇒ u_n =((((√3))^n −(−(√3))^n )/(2(√3))) and ((√3))^n +(−(√3))^n =2v_n ⇒ v_n =((((√3))^n +(−(√3))^n )/2) A^n =Q(A)p(A)+u_n A +v_n I =u_n A +v_n I (P(A)=0 by cayley hamiton) ⇒A^n =((((√3))^n −(−(√3))^n )/(2(√3))) (((1 2)),((1 −1)) ) +((((√3))^n +(−(√3))^n )/2) (((1 0)),((0 1)) ) A^n = (((((((√3))^n −(−(√3))^n )/(2(√3))) ((((√3))^n −(−(√3))^n )/(√3)))),((((((√3))^n −(−(√3))^n )/(2(√3))) (((−(√3))^n −((√3))^n )/(2(√3))))) ) + (((((((√3))^n +(−(√3))^n )/2) 0)),((0 ((((√3))^n +(−(√3))^n )/2))) ) = (((((((√3))^n −(−(√3))^n +((√3))((√3))^n +(√3)(−(√3))^n )/(2(√3))) ((((√3))^n −(−(√3))^n )/(√3)))),((((((√3))^n −(−(√3))^n )/(2(√3))) (((−(√3))^n −((√3))^n +(√3)((√3))^n +(√3)(−(√3))^n )/(2(√3))))) ) A^n = ((((((1+(√3))((√3))^n +((√3)−1)(−(√3))^n )/(2(√3))) ((((√3))^n −(−(√3))^n )/(√3)))),((((((√3))^n −(−(√3))^n )/(2(√3))) ((((√3)−1)((√3))^n +(1+(√3))(−(√3))^n )/(2(√3))))) )

1)thecaracteristicpolynomisp(x)=det(A−xI)=|1−x21−1−x|=−(1+x)(1−x)−2=−(1−x2)−2=−1+x2−2=x2−3=(x−3)(x+3)p(x)=0⇔x=+−3xn=Qp(x)+unx+vn⇒(3)n=un3+vn(−3)n=−3un+vn⇒(3)n−(−3)n=23un⇒un=(3)n−(−3)n23and(3)n+(−3)n=2vn⇒vn=(3)n+(−3)n2An=Q(A)p(A)+unA+vnI=unA+vnI(P(A)=0bycayleyhamiton)⇒An=(3)n−(−3)n23(121−1)+(3)n+(−3)n2(1001)An=((3)n−(−3)n23(3)n−(−3)n3(3)n−(−3)n23(−3)n−(3)n23)+((3)n+(−3)n200(3)n+(−3)n2)=((3)n−(−3)n+(3)(3)n+3(−3)n23(3)n−(−3)n3(3)n−(−3)n23(−3)n−(3)n+3(3)n+3(−3)n23)An=((1+3)(3)n+(3−1)(−3)n23(3)n−(−3)n3(3)n−(−3)n23(3−1)(3)n+(1+3)(−3)n23)

Commented by mathmax by abdo last updated on 29/Apr/20

A^(2p) = ((((((1+(√3))3^p +((√3)−1)3^p )/(2(√3))) 0)),((0 ((((√3)−1)3^p +(1+(√3))3^p )/(2(√3))))) ) = (((3^p 0)),((0 3^p )) ) A^(2p+1) = ((((((1+(√3))(√3)3^p +((√3)−1)(−(√3))3^p )/(2(√3))) (((√3)3^p −(−(√3))3^p )/(√3)))),(((((√3)3^p −(−(√3))3^p )/(2(√3))) ((((√3)−1)(√3)3^p +(1+(√3))(−(√3))3^p )/(2(√3))))) ) = (((3^p 2.3^p )),((2.3^p −3^p )) )

A2p=((1+3)3p+(3−1)3p2300(3−1)3p+(1+3)3p23)=(3p003p)A2p+1=((1+3)33p+(3−1)(−3)3p2333p−(−3)3p333p−(−3)3p23(3−1)33p+(1+3)(−3)3p23)=(3p2.3p2.3p−3p)

Commented by mathmax by abdo last updated on 29/Apr/20

sorry A^(2p+1) = (((3^p 2 .3^p )),((3^p −3^p )) )

sorryA2p+1=(3p2.3p3p−3p)

Commented by mathmax by abdo last updated on 29/Apr/20

2) e^A =Σ_(n=0) ^∞ (A^n /(n!)) =Σ_(n=0) ^∞ (A^(2n) /((2n)!)) +Σ_(n=0) ^∞ (A^(2n+1) /((2n+1)!)) =Σ_(n=0) ^∞ (1/((2n)!)) (((3^n 0)),((0 3^n )) ) +Σ_(n=0) ^∞ (1/((2n+1)!)) (((3^n 23^n )),((3^n −3^n )) ) = (((Σ_(n=0) ^∞ (3^n /((2n)!)) 0)),((0 Σ_(n=0) ^∞ (3^n /((2n)!)))) ) + (((Σ_(n=0) ^∞ (3^n /((2n+1)!)) 2Σ_(n=0) ^∞ (3^n /((2n+1)!)))),((Σ_(n=0) ^∞ (3^n /((2n+1)!)) −Σ_(n=0) ^∞ (3^n /((2n+1)!)))) )

2)eA=∑n=0∞Ann!=∑n=0∞A2n(2n)!+∑n=0∞A2n+1(2n+1)!=∑n=0∞1(2n)!(3n003n)+∑n=0∞1(2n+1)!(3n23n3n−3n)=(∑n=0∞3n(2n)!00∑n=0∞3n(2n)!)+(∑n=0∞3n(2n+1)!2∑n=0∞3n(2n+1)!∑n=0∞3n(2n+1)!−∑n=0∞3n(2n+1)!)

Commented by mathmax by abdo last updated on 29/Apr/20

we have Σ_(n=0) ^∞ (3^n /((2n)!)) =Σ_(n=0) ^∞ ((((√3))^(2n) )/((2n)!)) =cos((√3)) Σ_(n=0) ^∞ (3^n /((2n+1)!)) =(1/(√3))Σ_(n=0) ^∞ ((((√3))^(2n+1) )/((2n+1)!)) =(1/(√3))sin((√3)) ⇒ e^A = (((cos((√3)) 0)),((0 cos((√3)))) ) + ((((1/(√3))sin((√3)) (2/(√3))sin((√3)))),(((1/(√3))sin((√3)) −(1/(√3))sin((√3)))) ) = (((cos((√3))+(1/(√3))sin((√3)) (2/(√3))sin((√3)))),(((1/(√3))sin((√3)) cos((√3))−(1/((√3) ))sin((√3)))) )

wehave∑n=0∞3n(2n)!=∑n=0∞(3)2n(2n)!=cos(3)∑n=0∞3n(2n+1)!=13∑n=0∞(3)2n+1(2n+1)!=13sin(3)⇒eA=(cos(3)00cos(3))+(13sin(3)23sin(3)13sin(3)−13sin(3))=(cos(3)+13sin(3)23sin(3)13sin(3)cos(3)−13sin(3))

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