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Question-185013




Question Number 185013 by Shrinava last updated on 15/Jan/23
Commented by a.lgnaoui last updated on 16/Jan/23
question inteligente  (Clever question)
$${question}\:{inteligente} \\ $$$$\left({Clever}\:{question}\right) \\ $$
Commented by Shrinava last updated on 16/Jan/23
but a,b∈Z for please
$$\mathrm{but}\:\mathrm{a},\mathrm{b}\in\mathbb{Z}\:\mathrm{for}\:\mathrm{please} \\ $$
Commented by Shrinava last updated on 16/Jan/23
cool dear professor thank you
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by a.lgnaoui last updated on 16/Jan/23
  A= (a^2 −2b^2 )   9A=(9a^2 −18b^2 )     6A^2 =6 (a^2 −2b^2 )^2  ;   A^3 =(a^2 −2b^2 )^3   27I_2  =27  A^3 −6A^2 +9A−27=0    A=5,2646329  a^2 −2b^2 =5,2646329  a^2 =2b^2 +5,264633    a^6 −8b^6 −3(2a^2 b^2 (a^2 −2b^2 )−6(a^4 +4b^4 −4a^2 b^2 )+9(a^2 −2b^2 )−27=0  a^6 −(a^2 −x_0 )^3 −6a^2 ×(((a^2 −x_0 )/2))x_0 −6[a^4 +4(((a^2 −x_0 ))/2))(((a^2 −x_0 ))/2)−a^2 )]+9x_0 −27=0  a^6 −(a^2 −x_0 )^3 −a^2 x_0 (a^2 −x_0 )−6(a^2 −x_0 )(((a^2 −x_0 )/2)−a^2 )+9x_0 −27=0  =3a^4 x_0 −3a^2 x_0 ^2 +x_0 ^3 −(a^2 −x_0 )[a^2 x_0 +3(a^2 −x_0 )−6a^2 )]+9x_0 −27=0  =3a^2 x_0 (a^2 −x_0 )+x_0 ^3 −(a^2 −x_0 )[a^2 (x_0 −3)−3x_0 ]+9x_0 −27=0  (a^2 −x_0 )[3a^2 x_0 −a^2 (x_0 −3)−3x_0 ]+x_0 ^3 +9x_0 −27=0  (a^2 −x_0 )[a^2 (2x_0 −3)−3x_0 ]+x_0 ^3 +9(x_0 −3)=0  (a^2 −x_0 )[2x_0 (a^2 −x_0 )−3(a^2 −x_0 )+2x_0 ^2 −6x_0 ]+x_0 ^3 +9(x_0 −3)  a^2 −x_0 =z    x_0 −3=y_0   z(2x_0 z−3+2x_0 y_0 )]+(y_0 +3)^3 +9y_0   2x_0 z^2 +(2x_0 y_0 −3)z+(y_0 +3)^2 +9y_0 =0  z^2 +(y_0 −(3/(2x_0 ))−(3/(2x_0 )))z+(y_0 +3)^2 +9y_0 =0  (z+((x_0 y_0 −3)/2))^2 −(((x_0 y_0 −3)^2 )/4)+x_0 ^3 +9y_0 =0  x_0 y_0 =5,264633×2,264633=11,922461  2x_0 =          y_0 =2,264633  (z+((8,922461)/2))^2 −(((8,922461)^2 )/4)+253,218  (z+4,46123)^2 −273,12057  (z+4,46123−(√(273,12057)) )=0  z=16,52636−4,46123=12,065129    a^2 −x_0 =12,065129  a^2 =12,065129+5,264633      =17,329762       a=4,163..  ;   b=2,277
$$\:\:\mathrm{A}=\:\left(\mathrm{a}^{\mathrm{2}} −\mathrm{2b}^{\mathrm{2}} \right)\:\:\:\mathrm{9A}=\left(\mathrm{9}{a}^{\mathrm{2}} −\mathrm{18}{b}^{\mathrm{2}} \right)\:\:\: \\ $$$$\mathrm{6A}^{\mathrm{2}} =\mathrm{6}\:\left({a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:;\:\:\:\mathrm{A}^{\mathrm{3}} =\left({a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\mathrm{27I}_{\mathrm{2}} \:=\mathrm{27} \\ $$$$\mathrm{A}^{\mathrm{3}} −\mathrm{6A}^{\mathrm{2}} +\mathrm{9A}−\mathrm{27}=\mathrm{0} \\ $$$$\:\:\mathrm{A}=\mathrm{5},\mathrm{2646329} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} =\mathrm{5},\mathrm{2646329} \\ $$$${a}^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} +\mathrm{5},\mathrm{264633} \\ $$$$ \\ $$$$\mathrm{a}^{\mathrm{6}} −\mathrm{8b}^{\mathrm{6}} −\mathrm{3}\left(\mathrm{2a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} −\mathrm{2b}^{\mathrm{2}} \right)−\mathrm{6}\left(\mathrm{a}^{\mathrm{4}} +\mathrm{4b}^{\mathrm{4}} −\mathrm{4a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \right)+\mathrm{9}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{2b}^{\mathrm{2}} \right)−\mathrm{27}=\mathrm{0}\right. \\ $$$$\mathrm{a}^{\mathrm{6}} −\left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)^{\mathrm{3}} −\mathrm{6a}^{\mathrm{2}} ×\left(\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} }{\mathrm{2}}\right)\mathrm{x}_{\mathrm{0}} −\mathrm{6}\left[\mathrm{a}^{\mathrm{4}} +\mathrm{4}\left(\frac{\left.\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)}{\mathrm{2}}\right)\left(\frac{\left.\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)}{\mathrm{2}}−\mathrm{a}^{\mathrm{2}} \right)\right]+\mathrm{9x}_{\mathrm{0}} −\mathrm{27}=\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{6}} −\left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)^{\mathrm{3}} −\mathrm{a}^{\mathrm{2}} \mathrm{x}_{\mathrm{0}} \left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)−\mathrm{6}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)\left(\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} }{\mathrm{2}}−\mathrm{a}^{\mathrm{2}} \right)+\mathrm{9x}_{\mathrm{0}} −\mathrm{27}=\mathrm{0} \\ $$$$\left.=\mathrm{3a}^{\mathrm{4}} \mathrm{x}_{\mathrm{0}} −\mathrm{3a}^{\mathrm{2}} \mathrm{x}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{0}} ^{\mathrm{3}} −\left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)\left[\mathrm{a}^{\mathrm{2}} \mathrm{x}_{\mathrm{0}} +\mathrm{3}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)−\mathrm{6a}^{\mathrm{2}} \right)\right]+\mathrm{9x}_{\mathrm{0}} −\mathrm{27}=\mathrm{0} \\ $$$$=\mathrm{3a}^{\mathrm{2}} \mathrm{x}_{\mathrm{0}} \left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)+\mathrm{x}_{\mathrm{0}} ^{\mathrm{3}} −\left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)\left[\mathrm{a}^{\mathrm{2}} \left(\mathrm{x}_{\mathrm{0}} −\mathrm{3}\right)−\mathrm{3x}_{\mathrm{0}} \right]+\mathrm{9x}_{\mathrm{0}} −\mathrm{27}=\mathrm{0} \\ $$$$\left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)\left[\mathrm{3a}^{\mathrm{2}} \mathrm{x}_{\mathrm{0}} −\mathrm{a}^{\mathrm{2}} \left(\mathrm{x}_{\mathrm{0}} −\mathrm{3}\right)−\mathrm{3x}_{\mathrm{0}} \right]+\mathrm{x}_{\mathrm{0}} ^{\mathrm{3}} +\mathrm{9x}_{\mathrm{0}} −\mathrm{27}=\mathrm{0} \\ $$$$\left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)\left[\mathrm{a}^{\mathrm{2}} \left(\mathrm{2x}_{\mathrm{0}} −\mathrm{3}\right)−\mathrm{3x}_{\mathrm{0}} \right]+\mathrm{x}_{\mathrm{0}} ^{\mathrm{3}} +\mathrm{9}\left(\mathrm{x}_{\mathrm{0}} −\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)\left[\mathrm{2x}_{\mathrm{0}} \left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)−\mathrm{3}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} \right)+\mathrm{2x}_{\mathrm{0}} ^{\mathrm{2}} −\mathrm{6x}_{\mathrm{0}} \right]+\mathrm{x}_{\mathrm{0}} ^{\mathrm{3}} +\mathrm{9}\left(\mathrm{x}_{\mathrm{0}} −\mathrm{3}\right) \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{x}_{\mathrm{0}} =\mathrm{z}\:\:\:\:\mathrm{x}_{\mathrm{0}} −\mathrm{3}=\mathrm{y}_{\mathrm{0}} \\ $$$$\left.\mathrm{z}\left(\mathrm{2x}_{\mathrm{0}} \mathrm{z}−\mathrm{3}+\mathrm{2x}_{\mathrm{0}} \mathrm{y}_{\mathrm{0}} \right)\right]+\left(\mathrm{y}_{\mathrm{0}} +\mathrm{3}\right)^{\mathrm{3}} +\mathrm{9y}_{\mathrm{0}} \\ $$$$\mathrm{2x}_{\mathrm{0}} \mathrm{z}^{\mathrm{2}} +\left(\mathrm{2x}_{\mathrm{0}} \mathrm{y}_{\mathrm{0}} −\mathrm{3}\right)\mathrm{z}+\left(\mathrm{y}_{\mathrm{0}} +\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9y}_{\mathrm{0}} =\mathrm{0} \\ $$$$\boldsymbol{\mathrm{z}}^{\mathrm{2}} +\left(\boldsymbol{\mathrm{y}}_{\mathrm{0}} −\frac{\mathrm{3}}{\mathrm{2}\boldsymbol{\mathrm{x}}_{\mathrm{0}} }−\frac{\mathrm{3}}{\mathrm{2}\boldsymbol{\mathrm{x}}_{\mathrm{0}} }\right)\boldsymbol{\mathrm{z}}+\left(\boldsymbol{\mathrm{y}}_{\mathrm{0}} +\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}\boldsymbol{\mathrm{y}}_{\mathrm{0}} =\mathrm{0} \\ $$$$\left(\boldsymbol{\mathrm{z}}+\frac{\boldsymbol{\mathrm{x}}_{\mathrm{0}} \boldsymbol{\mathrm{y}}_{\mathrm{0}} −\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\left(\boldsymbol{\mathrm{x}}_{\mathrm{0}} \boldsymbol{\mathrm{y}}_{\mathrm{0}} −\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{4}}+{x}_{\mathrm{0}} ^{\mathrm{3}} +\mathrm{9}\boldsymbol{\mathrm{y}}_{\mathrm{0}} =\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}_{\mathrm{0}} \boldsymbol{\mathrm{y}}_{\mathrm{0}} =\mathrm{5},\mathrm{264633}×\mathrm{2},\mathrm{264633}=\mathrm{11},\mathrm{922461} \\ $$$$\mathrm{2x}_{\mathrm{0}} =\:\:\:\:\:\:\:\:\:\:\mathrm{y}_{\mathrm{0}} =\mathrm{2},\mathrm{264633} \\ $$$$\left({z}+\frac{\mathrm{8},\mathrm{922461}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\left(\mathrm{8},\mathrm{922461}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{253},\mathrm{218} \\ $$$$\left({z}+\mathrm{4},\mathrm{46123}\right)^{\mathrm{2}} −\mathrm{273},\mathrm{12057} \\ $$$$\left({z}+\mathrm{4},\mathrm{46123}−\sqrt{\mathrm{273},\mathrm{12057}}\:\right)=\mathrm{0} \\ $$$${z}=\mathrm{16},\mathrm{52636}−\mathrm{4},\mathrm{46123}=\mathrm{12},\mathrm{065129} \\ $$$$ \\ $$$${a}^{\mathrm{2}} −{x}_{\mathrm{0}} =\mathrm{12},\mathrm{065129} \\ $$$${a}^{\mathrm{2}} =\mathrm{12},\mathrm{065129}+\mathrm{5},\mathrm{264633} \\ $$$$\:\:\:\:=\mathrm{17},\mathrm{329762} \\ $$$$ \\ $$$$\:\:\:\boldsymbol{{a}}=\mathrm{4},\mathrm{163}..\:\:;\:\:\:\boldsymbol{{b}}=\mathrm{2},\mathrm{277} \\ $$$$\:\:\:\:\: \\ $$$$ \\ $$
Commented by aba last updated on 16/Jan/23
why A=a^2 −2b^2 ?
$$\mathrm{why}\:\mathrm{A}=\mathrm{a}^{\mathrm{2}} −\mathrm{2b}^{\mathrm{2}} ?\: \\ $$
Commented by aba last updated on 16/Jan/23
A is a matrice
$$\mathrm{A}\:\mathrm{is}\:\mathrm{a}\:\mathrm{matrice} \\ $$
Commented by aba last updated on 16/Jan/23
A^3 +9A=6A^2 +27I_2  ⇒ A^3 −9A^2 +27A−27I_2 +3A^2 −18A=0                                          ⇒(A−3I_2 )^3 +3A(A−6I2)=0
$$\mathrm{A}^{\mathrm{3}} +\mathrm{9A}=\mathrm{6A}^{\mathrm{2}} +\mathrm{27I}_{\mathrm{2}} \:\Rightarrow\:\mathrm{A}^{\mathrm{3}} −\mathrm{9A}^{\mathrm{2}} +\mathrm{27A}−\mathrm{27I}_{\mathrm{2}} +\mathrm{3A}^{\mathrm{2}} −\mathrm{18A}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left(\mathrm{A}−\mathrm{3I}_{\mathrm{2}} \right)^{\mathrm{3}} +\mathrm{3A}\left(\mathrm{A}−\mathrm{6I2}\right)=\mathrm{0} \\ $$
Commented by a.lgnaoui last updated on 16/Jan/23
A= determinant (((a+2b      2b)),((−b      a−2b)))=(a+2b)(a−2b)+2b^2 =a^2 −4b^2 +2b^2   =a^2 −2b^2     A=5,264633
$$\mathrm{A}=\begin{vmatrix}{\mathrm{a}+\mathrm{2b}\:\:\:\:\:\:\mathrm{2b}}\\{−\mathrm{b}\:\:\:\:\:\:\mathrm{a}−\mathrm{2b}}\end{vmatrix}=\left(\mathrm{a}+\mathrm{2b}\right)\left({a}−\mathrm{2}{b}\right)+\mathrm{2}{b}^{\mathrm{2}} ={a}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} \:\:\:\:\mathrm{A}=\mathrm{5},\mathrm{264633} \\ $$
Commented by Frix last updated on 16/Jan/23
A matrix is a matrix and a determinant is  a determinant. They are not interchangeable.
$$\mathrm{A}\:\mathrm{matrix}\:\mathrm{is}\:\mathrm{a}\:\mathrm{matrix}\:\mathrm{and}\:\mathrm{a}\:\mathrm{determinant}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{determinant}.\:\mathrm{They}\:\mathrm{are}\:{not}\:\mathrm{interchangeable}. \\ $$
Answered by Frix last updated on 16/Jan/23
I think there′s no solution for a, b ∈Z.  There′s not even one for a, b ∈R.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{a},\:{b}\:\in\mathbb{Z}. \\ $$$$\mathrm{There}'\mathrm{s}\:\mathrm{not}\:\mathrm{even}\:\mathrm{one}\:\mathrm{for}\:{a},\:{b}\:\in\mathbb{R}. \\ $$
Answered by aleks041103 last updated on 17/Jan/23
characteristic polynomial of A:  p(x)= determinant (((a+2b−x),(2b)),((−b),(a−2b−x)))=  =(a+2b−x)(a−2b−x)+2b^2 =  =x^2 −2ax+(a^2 −2b^2 )  Hamilton−Cayley theorem:  p(A)=0  ⇒A^2 −2aA+(a^2 −2b^2 )I=0  ⇒A^2 =2aA+(2b^2 −a^2 )I  ⇒A^3 =AA^2 =A(2aA+(2b^2 −a^2 )I)=  =2aA^2 +(2b^2 −a^2 )A=  =2a(2aA+(2b^2 −a^2 )I)+(2b^2 −a^2 )A=  =(3a^2 +2b^2 )A+2a(2b^2 −a^2 )I  We want   A^3 −6A^2 +9A−27I=0  ⇒(3a^2 +2b^2 )A+2a(2b^2 −a^2 )I−6(2aA+(2b^2 −a^2 )I)+9A−27I=0  (3a^2 +2b^2 −12a+9)A=(2a^3 −4ab^2 −6a^2 +12b^2 +27)I  This means that either  A∼I, i.e. A is diagonal iff  3a^2 +2b^2 −12a+9≠0 and 2a^3 −4ab^2 −6a^2 +12b^2 +27≠0  or we get no info if both are 0.  but for a,b∈Z,  2a^3 −4ab^2 −6a^2 +12b^2 +27 is odd and cannot  be 0  ⇒A is diagonal, i.e.  A= (((a+2b),(2b)),((−b),(a−2b)) ) =  ((c,0),(0,c) )  ⇒b=0 and A=  ((a,0),(0,a) )  now if A^3 −6A^2 +9A−27I=0, then  a^3 −6a^2 +9a−27=0  the only solution in R is not in Z.  ⇒∄a,b∈Z, such that for  A= (((a+2b),(2b)),((−b),(a−2b)) )  is satisfied  A^3 +9A=6A^2 +27I
$${characteristic}\:{polynomial}\:{of}\:{A}: \\ $$$${p}\left({x}\right)=\begin{vmatrix}{{a}+\mathrm{2}{b}−{x}}&{\mathrm{2}{b}}\\{−{b}}&{{a}−\mathrm{2}{b}−{x}}\end{vmatrix}= \\ $$$$=\left({a}+\mathrm{2}{b}−{x}\right)\left({a}−\mathrm{2}{b}−{x}\right)+\mathrm{2}{b}^{\mathrm{2}} = \\ $$$$={x}^{\mathrm{2}} −\mathrm{2}{ax}+\left({a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} \right) \\ $$$${Hamilton}−{Cayley}\:{theorem}: \\ $$$${p}\left({A}\right)=\mathrm{0} \\ $$$$\Rightarrow{A}^{\mathrm{2}} −\mathrm{2}{aA}+\left({a}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} \right){I}=\mathrm{0} \\ $$$$\Rightarrow{A}^{\mathrm{2}} =\mathrm{2}{aA}+\left(\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){I} \\ $$$$\Rightarrow{A}^{\mathrm{3}} ={AA}^{\mathrm{2}} ={A}\left(\mathrm{2}{aA}+\left(\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){I}\right)= \\ $$$$=\mathrm{2}{aA}^{\mathrm{2}} +\left(\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){A}= \\ $$$$=\mathrm{2}{a}\left(\mathrm{2}{aA}+\left(\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){I}\right)+\left(\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){A}= \\ $$$$=\left(\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \right){A}+\mathrm{2}{a}\left(\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){I} \\ $$$${We}\:{want}\: \\ $$$${A}^{\mathrm{3}} −\mathrm{6}{A}^{\mathrm{2}} +\mathrm{9}{A}−\mathrm{27}{I}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \right){A}+\mathrm{2}{a}\left(\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){I}−\mathrm{6}\left(\mathrm{2}{aA}+\left(\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){I}\right)+\mathrm{9}{A}−\mathrm{27}{I}=\mathrm{0} \\ $$$$\left(\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −\mathrm{12}{a}+\mathrm{9}\right){A}=\left(\mathrm{2}{a}^{\mathrm{3}} −\mathrm{4}{ab}^{\mathrm{2}} −\mathrm{6}{a}^{\mathrm{2}} +\mathrm{12}{b}^{\mathrm{2}} +\mathrm{27}\right){I} \\ $$$${This}\:{means}\:{that}\:{either} \\ $$$${A}\sim{I},\:{i}.{e}.\:{A}\:{is}\:{diagonal}\:{iff} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −\mathrm{12}{a}+\mathrm{9}\neq\mathrm{0}\:{and}\:\mathrm{2}{a}^{\mathrm{3}} −\mathrm{4}{ab}^{\mathrm{2}} −\mathrm{6}{a}^{\mathrm{2}} +\mathrm{12}{b}^{\mathrm{2}} +\mathrm{27}\neq\mathrm{0} \\ $$$${or}\:{we}\:{get}\:{no}\:{info}\:{if}\:{both}\:{are}\:\mathrm{0}. \\ $$$${but}\:{for}\:{a},{b}\in\mathbb{Z}, \\ $$$$\mathrm{2}{a}^{\mathrm{3}} −\mathrm{4}{ab}^{\mathrm{2}} −\mathrm{6}{a}^{\mathrm{2}} +\mathrm{12}{b}^{\mathrm{2}} +\mathrm{27}\:{is}\:{odd}\:{and}\:{cannot} \\ $$$${be}\:\mathrm{0} \\ $$$$\Rightarrow{A}\:{is}\:{diagonal},\:{i}.{e}. \\ $$$${A}=\begin{pmatrix}{{a}+\mathrm{2}{b}}&{\mathrm{2}{b}}\\{−{b}}&{{a}−\mathrm{2}{b}}\end{pmatrix}\:=\:\begin{pmatrix}{{c}}&{\mathrm{0}}\\{\mathrm{0}}&{{c}}\end{pmatrix} \\ $$$$\Rightarrow{b}=\mathrm{0}\:{and}\:{A}=\:\begin{pmatrix}{{a}}&{\mathrm{0}}\\{\mathrm{0}}&{{a}}\end{pmatrix} \\ $$$${now}\:{if}\:{A}^{\mathrm{3}} −\mathrm{6}{A}^{\mathrm{2}} +\mathrm{9}{A}−\mathrm{27}{I}=\mathrm{0},\:{then} \\ $$$${a}^{\mathrm{3}} −\mathrm{6}{a}^{\mathrm{2}} +\mathrm{9}{a}−\mathrm{27}=\mathrm{0} \\ $$$${the}\:{only}\:{solution}\:{in}\:\mathbb{R}\:{is}\:{not}\:{in}\:\mathbb{Z}. \\ $$$$\Rightarrow\nexists{a},{b}\in\mathbb{Z},\:{such}\:{that}\:{for} \\ $$$${A}=\begin{pmatrix}{{a}+\mathrm{2}{b}}&{\mathrm{2}{b}}\\{−{b}}&{{a}−\mathrm{2}{b}}\end{pmatrix} \\ $$$${is}\:{satisfied} \\ $$$${A}^{\mathrm{3}} +\mathrm{9}{A}=\mathrm{6}{A}^{\mathrm{2}} +\mathrm{27}{I} \\ $$
Commented by Frix last updated on 17/Jan/23
Great!
$$\mathrm{Great}! \\ $$
Commented by Shrinava last updated on 20/Jan/23
perfect dear professor Aleks thank you
$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{Aleks}\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Frix last updated on 17/Jan/23
Just to complete it:  A^3 −6A^2 +9A−27I_2 = [((term_1 ),(term_2 )),((term_3 ),(term_4 )) ]  All terms =0  (1) a^3 +6a^2 b+6ab^2 +4b^3 −6a^2 −24ab−12b^2 +9a+18b−27=0  (2) 6a^2 b+4b^3 −24ab+18b=0  (3) −3a^2 b−2b^3 +12ab−9b=0  (4) a^3 −6a^2 b+6ab^2 −4b^3 −6a^2 +24ab−12b^2 +9a−18b−27=0  From (2) or (3) we get  a=2+((√(3−2b^2 ))/( (√3)))s with s=±1  From (1) or (4) we get  ((2(8b^2 −3)(√(9−6b^2 )))/9)s−25=0  We get  s=−1∧b≈±1.58009i∧a≈.367684  s=+1∧b≈±(1.73133±.790045i)∧a≈2.81616±1.11729i
$$\mathrm{Just}\:\mathrm{to}\:\mathrm{complete}\:\mathrm{it}: \\ $$$${A}^{\mathrm{3}} −\mathrm{6}{A}^{\mathrm{2}} +\mathrm{9}{A}−\mathrm{27}{I}_{\mathrm{2}} =\begin{bmatrix}{{term}_{\mathrm{1}} }&{{term}_{\mathrm{2}} }\\{{term}_{\mathrm{3}} }&{{term}_{\mathrm{4}} }\end{bmatrix} \\ $$$$\mathrm{All}\:\mathrm{terms}\:=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:{a}^{\mathrm{3}} +\mathrm{6}{a}^{\mathrm{2}} {b}+\mathrm{6}{ab}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{3}} −\mathrm{6}{a}^{\mathrm{2}} −\mathrm{24}{ab}−\mathrm{12}{b}^{\mathrm{2}} +\mathrm{9}{a}+\mathrm{18}{b}−\mathrm{27}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{6}{a}^{\mathrm{2}} {b}+\mathrm{4}{b}^{\mathrm{3}} −\mathrm{24}{ab}+\mathrm{18}{b}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:−\mathrm{3}{a}^{\mathrm{2}} {b}−\mathrm{2}{b}^{\mathrm{3}} +\mathrm{12}{ab}−\mathrm{9}{b}=\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:{a}^{\mathrm{3}} −\mathrm{6}{a}^{\mathrm{2}} {b}+\mathrm{6}{ab}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{3}} −\mathrm{6}{a}^{\mathrm{2}} +\mathrm{24}{ab}−\mathrm{12}{b}^{\mathrm{2}} +\mathrm{9}{a}−\mathrm{18}{b}−\mathrm{27}=\mathrm{0} \\ $$$$\mathrm{From}\:\left(\mathrm{2}\right)\:\mathrm{or}\:\left(\mathrm{3}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$${a}=\mathrm{2}+\frac{\sqrt{\mathrm{3}−\mathrm{2}{b}^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}}{s}\:\mathrm{with}\:{s}=\pm\mathrm{1} \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\:\mathrm{or}\:\left(\mathrm{4}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{2}\left(\mathrm{8}{b}^{\mathrm{2}} −\mathrm{3}\right)\sqrt{\mathrm{9}−\mathrm{6}{b}^{\mathrm{2}} }}{\mathrm{9}}{s}−\mathrm{25}=\mathrm{0} \\ $$$$\mathrm{We}\:\mathrm{get} \\ $$$${s}=−\mathrm{1}\wedge{b}\approx\pm\mathrm{1}.\mathrm{58009i}\wedge{a}\approx.\mathrm{367684} \\ $$$$\mathrm{s}=+\mathrm{1}\wedge{b}\approx\pm\left(\mathrm{1}.\mathrm{73133}\pm.\mathrm{790045i}\right)\wedge{a}\approx\mathrm{2}.\mathrm{81616}\pm\mathrm{1}.\mathrm{11729i} \\ $$

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