Menu Close

let-A-1-1-1-1-1-calculate-A-1-and-A-2-2-calculate-A-n-3-find-e-A-and-e-A-

Tinku Tara 0 Comments
Pin




Question Number 93415 by abdomathmax last updated on 13/May/20
let A = (((1 −1)),((1 1)) ) 1) calculate A^(−1) and A^(−2) 2) calculate A^n 3) find e^A and e^(−A)
letA=(1−111)1)calculateA−1andA−22)calculateAn3)findeAande−A
Commented by prakash jain last updated on 13/May/20
det(A−λI)= determinant (((1−λ),(−1)),(1,(1−λ))) =(1−λ)^2 +1=0 λ=1−i,1+i (i=(√(−1))) Eigenvector for λ=1−i (A−λI) [(x_1 ),(x_2 ) ]=0 ( [(1,(−1)),(1,1) ]− [((1−i),0),(0,(1−i)) ]) [(x_1 ),(x_2 ) ]=0 [(i,(−1)),(1,i) ] [(x_1 ),(x_2 ) ]=0 ix_1 −x_2 =0 x_1 +ix_2 =0 x_1 =1,x_2 =i ( [(1,(−1)),(1,1) ]− [((1+i),0),(0,(1+i)) ]) [(y_1 ),(y_2 ) ]=0 [((−i),(−1)),(1,(−i)) ] [(y_1 ),(y_2 ) ]=0 −iy_1 −y_2 =0 y_1 −iy_2 =0 y_1 =1, y_2 =−i S= [(1,1),((−i),i) ] S^(−1) = [((1/2),(i/2)),((1/2),(−i/2)) ] S^(−1) AS= [((1+i),0),(0,(1−i)) ] A=S [((1+i),0),(0,(1−i)) ]S^(−1) A^n =S [(((1+i)^n ),0),(0,((1−i)^n )) ]S^(−1) A^n = [(1,1),((−i),i) ] [(((1−i)^n ),0),(0,((1+i)^n )) ] [((1/2),(i/2)),((1/2),(−i/2)) ] =(1/2) [(((1−i)^n +(1+i)^n ),(−i((1−i)^n −(1+i))^n )),((i((1−i)^n −(1+i)^n )),((1−i)^n +(1+i)^n )) ]
det(A−λI)=|1−λ−111−λ|=(1−λ)2+1=0λ=1−i,1+i(i=−1)Eigenvectorforλ=1−i(A−λI)[x1x2]=0([1−111]−[1−i001−i])[x1x2]=0[i−11i][x1x2]=0ix1−x2=0x1+ix2=0x1=1,x2=i([1−111]−[1+i001+i])[y1y2]=0[−i−11−i][y1y2]=0−iy1−y2=0y1−iy2=0y1=1,y2=−iS=[11−ii]S−1=[1/2i/21/2−i/2]S−1AS=[1+i001−i]A=S[1+i001−i]S−1An=S[(1+i)n00(1−i)n]S−1An=[11−ii][(1−i)n00(1+i)n][1/2i/21/2−i/2]=12[(1−i)n+(1+i)n−i((1−i)n−(1+i))ni((1−i)n−(1+i)n)(1−i)n+(1+i)n]
Commented by mathmax by abdo last updated on 13/May/20
thank you sir.
thankyousir.
Commented by mathmax by abdo last updated on 13/May/20
1) p_c (x) =det(A−xI) = determinant (((1−x −1)),((1 1−x)))=(1−x)^2 +1 =x^2 −2x+2 csyley hamilton theorem ⇒A^2 −2A +2I =0 ⇒ A(A−2I) =−2I ⇒A×(−(1/2))(A−2I) =I ⇒ A is inversible and A^(−1) =−(1/2)A +I = (((−(1/2) (1/2))),((−(1/2) −(1/2))) )+ (((1 0)),((0 1)) ) A^(−1) = ((((1/2) (1/2))),((−(1/2) (1/2))) ) and A^(−2) =(A^(−1) )^2
1)pc(x)=det(A−xI)=|1−x−111−x|=(1−x)2+1=x2−2x+2csyleyhamiltontheorem⇒A2−2A+2I=0⇒A(A−2I)=−2I⇒A×(−12)(A−2I)=I⇒AisinversibleandA−1=−12A+I=(−1212−12−12)+(1001)A−1=(1212−1212)andA−2=(A−1)2
Commented by mathmax by abdo last updated on 13/May/20
2) we have P_c (x) =det(A−xI) =x^2 −2x +2 P_c (x)=0 ⇔(x−1)^2 +1 =0 ⇔(x−1)^2 =−1 ⇒x =1+i or x=1−i let divide x^n by P_c (x) ⇒x^n =Q(x)P_c (x) +u_n x +v_n ⇒(1+i)^n =u_n (1+i)+v_n ⇒(1+i−1+i)u_n =(1+i)^n −(1−i)^n (1−i)^n =u_n (1−i) +v_n ⇒u_n =(1/(2i)){(1+i)^n −(1−i)^n } =(1/(2i)){ ((√2))^n e^((inπ)/4) −((√2))^n e^(−i((nπ)/4)) } =((√2))^n sin(((nπ)/4)) (1+i)^n +(1−i)^n =2u_n +2v_n ⇒u_n +v_n =(1/2){ (1+i)^n +(1−i)^n } =((√2))^(n ) cos(((nπ)/4)) ⇒v_n =((√2))^n cos(((nπ)/4))−((√2))^n sin(((nπ)/4)) we have Pc(A)=0 ⇒A^n =u_(n ) A +v_n I =((√2))^n sin(((nπ)/4)) (((1 −1)),((1 1)) ) +((√2))^n (cos(((nπ)/4))−sin(((nπ)/4)) (((1 0)),((0 1)) ) A^n = (((((√2))^n cos(((nπ)/4)) −((√2))^(n ) sin(((nπ)/4)))),((((√2))^n sin(((nπ)/4)) ((√2))^n cos(((nπ)/4) )) )
2)wehavePc(x)=det(A−xI)=x2−2x+2Pc(x)=0⇔(x−1)2+1=0⇔(x−1)2=−1⇒x=1+iorx=1−iletdividexnbyPc(x)⇒xn=Q(x)Pc(x)+unx+vn⇒(1+i)n=un(1+i)+vn⇒(1+i−1+i)un=(1+i)n−(1−i)n(1−i)n=un(1−i)+vn⇒un=12i{(1+i)n−(1−i)n}=12i{(2)neinπ4−(2)ne−inπ4}=(2)nsin(nπ4)(1+i)n+(1−i)n=2un+2vn⇒un+vn=12{(1+i)n+(1−i)n}=(2)ncos(nπ4)⇒vn=(2)ncos(nπ4)−(2)nsin(nπ4)wehavePc(A)=0⇒An=unA+vnI=(2)nsin(nπ4)(1−111)+(2)n(cos(nπ4)−sin(nπ4)(1001)An=((2)ncos(nπ4)−(2)nsin(nπ4)(2)nsin(nπ4)(2)ncos(nπ4)
Commented by mathmax by abdo last updated on 13/May/20
⇒ A^n = (((((√2))^n cos(((nπ)/4)) −((√2))^n sin(((nπ)/4)))),((((√2))^n sin(((nπ)/4)) ((√2))^n cos(((nπ)/4)))) )
⇒An=((2)ncos(nπ4)−(2)nsin(nπ4)(2)nsin(nπ4)(2)ncos(nπ4))

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *