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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)
questioninteligente(Cleverquestion)
Commented by Shrinava last updated on 16/Jan/23
but a,b∈Z for please
buta,b∈Zforplease
Commented by Shrinava last updated on 16/Jan/23
cool dear professor thank you
cooldearprofessorthankyou
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
A=(a2−2b2)9A=(9a2−18b2)6A2=6(a2−2b2)2;A3=(a2−2b2)327I2=27A3−6A2+9A−27=0A=5,2646329a2−2b2=5,2646329a2=2b2+5,264633a6−8b6−3(2a2b2(a2−2b2)−6(a4+4b4−4a2b2)+9(a2−2b2)−27=0a6−(a2−x0)3−6a2×(a2−x02)x0−6[a4+4(a2−x0)2)(a2−x0)2−a2)]+9x0−27=0a6−(a2−x0)3−a2x0(a2−x0)−6(a2−x0)(a2−x02−a2)+9x0−27=0=3a4x0−3a2x02+x03−(a2−x0)[a2x0+3(a2−x0)−6a2)]+9x0−27=0=3a2x0(a2−x0)+x03−(a2−x0)[a2(x0−3)−3x0]+9x0−27=0(a2−x0)[3a2x0−a2(x0−3)−3x0]+x03+9x0−27=0(a2−x0)[a2(2x0−3)−3x0]+x03+9(x0−3)=0(a2−x0)[2x0(a2−x0)−3(a2−x0)+2x02−6x0]+x03+9(x0−3)a2−x0=zx0−3=y0z(2x0z−3+2x0y0)]+(y0+3)3+9y02x0z2+(2x0y0−3)z+(y0+3)2+9y0=0z2+(y0−32x0−32x0)z+(y0+3)2+9y0=0(z+x0y0−32)2−(x0y0−3)24+x03+9y0=0x0y0=5,264633×2,264633=11,9224612x0=y0=2,264633(z+8,9224612)2−(8,922461)24+253,218(z+4,46123)2−273,12057(z+4,46123−273,12057)=0z=16,52636−4,46123=12,065129a2−x0=12,065129a2=12,065129+5,264633=17,329762a=4,163..;b=2,277
Commented by aba last updated on 16/Jan/23
why A=a^2 −2b^2 ?
whyA=a2−2b2?
Commented by aba last updated on 16/Jan/23
A is a matrice
Aisamatrice
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
A3+9A=6A2+27I2⇒A3−9A2+27A−27I2+3A2−18A=0⇒(A−3I2)3+3A(A−6I2)=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
A=|a+2b2b−ba−2b|=(a+2b)(a−2b)+2b2=a2−4b2+2b2=a2−2b2A=5,264633
Commented by Frix last updated on 16/Jan/23
A matrix is a matrix and a determinant is a determinant. They are not interchangeable.
Amatrixisamatrixandadeterminantisadeterminant.Theyarenotinterchangeable.
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.
Ithinkthere′snosolutionfora,b∈Z.There′snotevenonefora,b∈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
characteristicpolynomialofA:p(x)=|a+2b−x2b−ba−2b−x|==(a+2b−x)(a−2b−x)+2b2==x2−2ax+(a2−2b2)Hamilton−Cayleytheorem:p(A)=0⇒A2−2aA+(a2−2b2)I=0⇒A2=2aA+(2b2−a2)I⇒A3=AA2=A(2aA+(2b2−a2)I)==2aA2+(2b2−a2)A==2a(2aA+(2b2−a2)I)+(2b2−a2)A==(3a2+2b2)A+2a(2b2−a2)IWewantA3−6A2+9A−27I=0⇒(3a2+2b2)A+2a(2b2−a2)I−6(2aA+(2b2−a2)I)+9A−27I=0(3a2+2b2−12a+9)A=(2a3−4ab2−6a2+12b2+27)IThismeansthateitherA∼I,i.e.Aisdiagonaliff3a2+2b2−12a+9≠0and2a3−4ab2−6a2+12b2+27≠0orwegetnoinfoifbothare0.butfora,b∈Z,2a3−4ab2−6a2+12b2+27isoddandcannotbe0⇒Aisdiagonal,i.e.A=(a+2b2b−ba−2b)=(c00c)⇒b=0andA=(a00a)nowifA3−6A2+9A−27I=0,thena3−6a2+9a−27=0theonlysolutioninRisnotinZ.⇒∄a,b∈Z,suchthatforA=(a+2b2b−ba−2b)issatisfiedA3+9A=6A2+27I
Commented by Frix last updated on 17/Jan/23
Great!
Great!
Commented by Shrinava last updated on 20/Jan/23
perfect dear professor Aleks thank you
perfectdearprofessorAleksthankyou
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
Justtocompleteit:A3−6A2+9A−27I2=[term1term2term3term4]Allterms=0(1)a3+6a2b+6ab2+4b3−6a2−24ab−12b2+9a+18b−27=0(2)6a2b+4b3−24ab+18b=0(3)−3a2b−2b3+12ab−9b=0(4)a3−6a2b+6ab2−4b3−6a2+24ab−12b2+9a−18b−27=0From(2)or(3)wegeta=2+3−2b23swiths=±1From(1)or(4)weget2(8b2−3)9−6b29s−25=0Wegets=−1∧b≈±1.58009i∧a≈.367684s=+1∧b≈±(1.73133±.790045i)∧a≈2.81616±1.11729i

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