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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)) )
$$\mathrm{find}\:\mathrm{all}\:\mathrm{2×2}\:\mathrm{matrices}\:\mathrm{A}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{A}^{\mathrm{3}} −\mathrm{3A}^{\mathrm{2}} \:=\:\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\:\:−\mathrm{2}}\end{pmatrix} \\ $$
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)) )
$${let}\:{A}\:=\:\begin{pmatrix}{{a}\:\:\:\:\:{b}}\\{{c}\:\:\:\:\:{d}}\end{pmatrix} \\ $$$${A}^{\mathrm{2}} \left({A}−\mathrm{3}{I}\right)\:=\:\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\:\:−\mathrm{2}}\end{pmatrix} \\ $$$$\left({det}\left({A}\right)\right)^{\mathrm{2}} \:{det}\left({A}−\mathrm{3}{I}\right)\:=\:{det}\:\begin{pmatrix}{−\mathrm{2}\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:−\mathrm{2}}\end{pmatrix}\:=\:\mathrm{0} \\ $$$${case}\:\mathrm{1} \\ $$$${det}\:\left({A}\right)=\mathrm{0}\:\Rightarrow{p}\left(\lambda\right)\:=\:{det}\left(\lambda{I}−{A}\right)=\mathrm{0} \\ $$$$\begin{vmatrix}{\lambda−{a}\:\:\:\:\:−{b}}\\{−{c}\:\:\:\:\:\:\:\lambda−{d}}\end{vmatrix}=\:\lambda^{\mathrm{2}} −\left({a}+{d}\right)\lambda+{ad}−{bc}\: \\ $$$$\lambda^{\mathrm{2}} −{TR}\left({A}\right)\lambda+{det}\left({A}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\lambda^{\mathrm{2}} −{TR}\left({A}\right)\lambda=\mathrm{0}\:;\:{A}^{\mathrm{2}} =\:{TR}\left({A}\right).{A} \\ $$$${A}^{\mathrm{3}} =\:{A}\left({TR}\left({A}\right)\right)=\:\left({TR}\left({A}\right)\right)^{\mathrm{2}} {A} \\ $$$$\Rightarrow{A}^{\mathrm{3}} −\mathrm{3}{A}=\begin{pmatrix}{−\mathrm{2}\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\:−\mathrm{2}}\end{pmatrix} \\ $$$${TR}\left({A}\right)^{\mathrm{3}} −\mathrm{3}\left({TR}\left({A}\right)\right)^{\mathrm{2}} +\mathrm{4}\:=\:\mathrm{0} \\ $$$${TR}\left({A}\right)\:=\:\mathrm{2}\:{or}\:{TR}\left({A}\right)\:=\:−\mathrm{1} \\ $$$${case}\:\mathrm{1}{a} \\ $$$${TR}\left({A}\right)=\mathrm{2}\:\Rightarrow\:−\mathrm{2}{A}\:=\:\begin{pmatrix}{−\mathrm{2}\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:−\mathrm{2}}\end{pmatrix} \\ $$$${A}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\mathrm{1}}\end{pmatrix}\:.\:{TR}\left({A}\right)\:=\:−\mathrm{1}\: \\ $$$$\mathrm{4}{A}\:=\:\begin{pmatrix}{−\mathrm{2}\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\:\:−\mathrm{2}}\end{pmatrix}\:\Rightarrow{A}=\begin{pmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\ $$$${case}\:\mathrm{2} \\ $$$${det}\left({A}−\mathrm{3}{I}\right)=\mathrm{0} \\ $$$${det}\left({A}−\mathrm{2}{I}\right)^{\mathrm{2}} \left({A}+{I}\right)=\mathrm{0} \\ $$$${case}\mathrm{2}{a} \\ $$$${det}\left({A}−\mathrm{2}{I}\right)\:=\:\mathrm{0}\:\Rightarrow{p}\left(\lambda\right)=\lambda^{\mathrm{2}} −\mathrm{5}\lambda+\mathrm{6} \\ $$$${A}^{\mathrm{2}} =\mathrm{5}{A}−\mathrm{6}{I}\:\Rightarrow{A}^{\mathrm{3}} −\mathrm{3}{A}^{\mathrm{2}} =\left(\mathrm{19}{A}−\mathrm{30}{I}\right)−\mathrm{3}\left(\mathrm{5}{A}−\mathrm{6}{I}\right) \\ $$$$=\:\mathrm{4}{A}−\mathrm{12}{I}\:=\:\begin{pmatrix}{−\mathrm{2}\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\:−\mathrm{2}}\end{pmatrix} \\ $$$$\mathrm{4}{A}\:=\:\begin{pmatrix}{\mathrm{10}\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\mathrm{10}}\end{pmatrix}\:\Rightarrow{A}=\begin{pmatrix}{\frac{\mathrm{5}}{\mathrm{2}}\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\frac{\mathrm{5}}{\mathrm{2}}}\end{pmatrix} \\ $$$${case}\mathrm{2}{b} \\ $$$${det}\left({A}+{I}\right)=\mathrm{0}\:\Rightarrow{p}\left(\lambda\right)=\lambda^{\mathrm{2}} −\mathrm{2}\lambda−\mathrm{3} \\ $$$${A}^{\mathrm{2}} −\mathrm{2}{A}−\mathrm{3}{I}=\mathrm{0};\:{A}^{\mathrm{2}} =\mathrm{2}{A}+\mathrm{3}{I} \\ $$$${A}^{\mathrm{3}} −\mathrm{3}{A}^{\mathrm{2}} =\mathrm{7}{A}+\mathrm{6}{I}−\mathrm{3}\left(\mathrm{2}{A}+\mathrm{3}{I}\right)={A}−\mathrm{3}{I} \\ $$$$\Rightarrow{A}−\mathrm{3}{I}\:=\:\begin{pmatrix}{−\mathrm{2}\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\:−\mathrm{2}}\end{pmatrix} \\ $$$${A}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\therefore\:{A}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\mathrm{1}}\end{pmatrix}\:;\:\begin{pmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\ $$$$;\:\begin{pmatrix}{\frac{\mathrm{5}}{\mathrm{2}}\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{2}}}\end{pmatrix}\:;\:\begin{pmatrix}{\:\:\mathrm{1}\:\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$ \\ $$
Commented by bobhans last updated on 28/Jun/20
great===
$$\mathrm{great}=== \\ $$

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