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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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