Question Number 100650 by bobhans last updated on 28/Jun/20

find all 2x2 matrices A such  that A^3 −3A^2  =  (((−2     −2)),((−2      −2)) )

Answered by bramlex last updated on 28/Jun/20

let A =  (((a     b)),((c     d)) )  A^2 (A−3I) =  (((−2     −2)),((−2      −2)) )  (det(A))^2  det(A−3I) = det  (((−2    −2)),((−2    −2)) ) = 0  case 1  det (A)=0 ⇒p(λ) = det(λI−A)=0   determinant (((λ−a     −b)),((−c       λ−d)))= λ^2 −(a+d)λ+ad−bc   λ^2 −TR(A)λ+det(A) = 0  ⇒λ^2 −TR(A)λ=0 ; A^2 = TR(A).A  A^3 = A(TR(A))= (TR(A))^2 A  ⇒A^3 −3A= (((−2    −2)),((−2     −2)) )  TR(A)^3 −3(TR(A))^2 +4 = 0  TR(A) = 2 or TR(A) = −1  case 1a  TR(A)=2 ⇒ −2A =  (((−2   −2)),((−2    −2)) )  A =  (((1    1)),((1    1)) ) . TR(A) = −1   4A =  (((−2    −2)),((−2      −2)) ) ⇒A= (((−(1/2)    −(1/2))),((−(1/2)     −(1/2))) )  case 2  det(A−3I)=0  det(A−2I)^2 (A+I)=0  case2a  det(A−2I) = 0 ⇒p(λ)=λ^2 −5λ+6  A^2 =5A−6I ⇒A^3 −3A^2 =(19A−30I)−3(5A−6I)  = 4A−12I =  (((−2    −2)),((−2     −2)) )  4A =  (((10   −2)),((−2   10)) ) ⇒A= ((((5/2)   −(1/2))),((−(1/2)    (5/2))) )  case2b  det(A+I)=0 ⇒p(λ)=λ^2 −2λ−3  A^2 −2A−3I=0; A^2 =2A+3I  A^3 −3A^2 =7A+6I−3(2A+3I)=A−3I  ⇒A−3I =  (((−2    −2)),((−2     −2)) )  A =  (((1    −2)),((−2    1)) )  ∴ A =  (((1   1)),((1    1)) ) ;  (((−(1/2)     −(1/2))),((−(1/2)      −(1/2))) )  ;  ((((5/2)      −(1/2))),((−(1/2)     (5/2))) ) ;  (((  1     −2)),((−2      1)) )

Commented bybobhans last updated on 28/Jun/20

great===