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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

$q u e s t i o n i n t e l i g e n t e$ $(C l e v e r q u e s t i o n)$
Commented by Shrinava last updated on 16/Jan/23

$but a, b \in Z for please$
Commented by Shrinava last updated on 16/Jan/23

$cool dear professor thank you$
	Answered by a.lgnaoui last updated on 16/Jan/23

	$A = (a^{2} - {2 b}^{2}) 9 A = (9 a^{2} - 18 b^{2})$ ${6 A}^{2} = 6 {(a^{2} - 2 b^{2})}^{2}; A^{3} = {(a^{2} - 2 b^{2})}^{3}$ ${27 I}_{2} = 27$ $A^{3} - {6 A}^{2} + 9 A - 27 = 0$ $A = 5, 2646329$ $a^{2} - 2 b^{2} = 5, 2646329$ $a^{2} = 2 b^{2} + 5, 264633$ $a^{6} - {8 b}^{6} - 3 ({2 a}^{2} b^{2} (a^{2} - {2 b}^{2}) - 6 (a^{4} + {4 b}^{4} - {4 a}^{2} b^{2}) + 9 (a^{2} - {2 b}^{2}) - 27 = 0$ $a^{6} - {(a^{2} - x_{0})}^{3} - {6 a}^{2} \times (\frac{a^{2} - x_{0}}{2}) x_{0} - 6 [a^{4} + 4 (\frac{a^{2} - x_{0})}{2}) (\frac{a^{2} - x_{0})}{2} - a^{2})] + {9 x}_{0} - 27 = 0$ $a^{6} - {(a^{2} - x_{0})}^{3} - a^{2} x_{0} (a^{2} - x_{0}) - 6 (a^{2} - x_{0}) (\frac{a^{2} - x_{0}}{2} - a^{2}) + {9 x}_{0} - 27 = 0$ $= {3 a}^{4} x_{0} - {3 a}^{2} x_{0}^{2} + x_{0}^{3} - (a^{2} - x_{0}) [a^{2} x_{0} + 3 (a^{2} - x_{0}) - {6 a}^{2})] + {9 x}_{0} - 27 = 0$ $= {3 a}^{2} x_{0} (a^{2} - x_{0}) + x_{0}^{3} - (a^{2} - x_{0}) [a^{2} (x_{0} - 3) - {3 x}_{0}] + {9 x}_{0} - 27 = 0$ $(a^{2} - x_{0}) [{3 a}^{2} x_{0} - a^{2} (x_{0} - 3) - {3 x}_{0}] + x_{0}^{3} + {9 x}_{0} - 27 = 0$ $(a^{2} - x_{0}) [a^{2} ({2 x}_{0} - 3) - {3 x}_{0}] + x_{0}^{3} + 9 (x_{0} - 3) = 0$ $(a^{2} - x_{0}) [{2 x}_{0} (a^{2} - x_{0}) - 3 (a^{2} - x_{0}) + {2 x}_{0}^{2} - {6 x}_{0}] + x_{0}^{3} + 9 (x_{0} - 3)$ $a^{2} - x_{0} = z x_{0} - 3 = y_{0}$ $z ({2 x}_{0} z - 3 + {2 x}_{0} y_{0})] + {(y_{0} + 3)}^{3} + {9 y}_{0}$ ${2 x}_{0} z^{2} + ({2 x}_{0} y_{0} - 3) z + {(y_{0} + 3)}^{2} + {9 y}_{0} = 0$ $z^{2} + (y_{0} - \frac{3}{2 x_{0}} - \frac{3}{2 x_{0}}) z + {(y_{0} + 3)}^{2} + 9 y_{0} = 0$ ${(z + \frac{x_{0} y_{0} - 3}{2})}^{2} - \frac{{(x_{0} y_{0} - 3)}^{2}}{4} + x_{0}^{3} + 9 y_{0} = 0$ $x_{0} y_{0} = 5, 264633 \times 2, 264633 = 11, 922461$ ${2 x}_{0} = y_{0} = 2, 264633$ ${(z + \frac{8, 922461}{2})}^{2} - \frac{{(8, 922461)}^{2}}{4} + 253, 218$ ${(z + 4, 46123)}^{2} - 273, 12057$ $(z + 4, 46123 - \sqrt{273, 12057}) = 0$ $z = 16, 52636 - 4, 46123 = 12, 065129$ $a^{2} - x_{0} = 12, 065129$ $a^{2} = 12, 065129 + 5, 264633$ $= 17, 329762$ $a = 4, 163 . .; b = 2, 277$
	Commented by aba last updated on 16/Jan/23

	$why A = a^{2} - {2 b}^{2} ?$
	Commented by aba last updated on 16/Jan/23

	$A is a matrice$
	Commented by aba last updated on 16/Jan/23

	$A^{3} + 9 A = {6 A}^{2} + {27 I}_{2} \Rightarrow A^{3} - {9 A}^{2} + 27 A - {27 I}_{2} + {3 A}^{2} - 18 A = 0$ $\Rightarrow {(A - {3 I}_{2})}^{3} + 3 A (A - 6 I 2) = 0$
	Commented by a.lgnaoui last updated on 16/Jan/23

	$A = \| \begin{matrix} a + 2 b 2 b \\ - b a - 2 b \end{matrix} \| = (a + 2 b) (a - 2 b) + 2 b^{2} = a^{2} - 4 b^{2} + 2 b^{2}$ $= a^{2} - 2 b^{2} A = 5, 264633$
	Commented by Frix last updated on 16/Jan/23

	$A matrix is a matrix and a determinant is$ $a determinant . They are n o t interchangeable .$
	Answered by Frix last updated on 16/Jan/23

	$I think {there}^{'} s no solution for a, b \in Z .$ ${There}^{'} s not even one for a, b \in R .$
	Answered by aleks041103 last updated on 17/Jan/23

	$c h a r a c t e r i s t i c p o l y n o m i a l o f A :$ $p (x) = \| \begin{matrix} a + 2 b - x & 2 b \\ - b & a - 2 b - x \end{matrix} \| =$ $= (a + 2 b - x) (a - 2 b - x) + 2 b^{2} =$ $= x^{2} - 2 a x + (a^{2} - 2 b^{2})$ $H a m i l t o n - C a y l e y t h e o r e m :$ $p (A) = 0$ $\Rightarrow A^{2} - 2 a A + (a^{2} - 2 b^{2}) I = 0$ $\Rightarrow A^{2} = 2 a A + (2 b^{2} - a^{2}) I$ $\Rightarrow A^{3} = {A A}^{2} = A (2 a A + (2 b^{2} - a^{2}) I) =$ $= 2 {a A}^{2} + (2 b^{2} - a^{2}) A =$ $= 2 a (2 a A + (2 b^{2} - a^{2}) I) + (2 b^{2} - a^{2}) A =$ $= (3 a^{2} + 2 b^{2}) A + 2 a (2 b^{2} - a^{2}) I$ $W e w a n t$ $A^{3} - 6 A^{2} + 9 A - 27 I = 0$ $\Rightarrow (3 a^{2} + 2 b^{2}) A + 2 a (2 b^{2} - a^{2}) I - 6 (2 a A + (2 b^{2} - a^{2}) I) + 9 A - 27 I = 0$ $(3 a^{2} + 2 b^{2} - 12 a + 9) A = (2 a^{3} - 4 {a b}^{2} - 6 a^{2} + 12 b^{2} + 27) I$ $T h i s m e a n s t h a t e i t h e r$ $A \sim I, i . e . A i s d i a g o n a l i f f$ $3 a^{2} + 2 b^{2} - 12 a + 9 \neq 0 a n d 2 a^{3} - 4 {a b}^{2} - 6 a^{2} + 12 b^{2} + 27 \neq 0$ $o r w e g e t n o i n f o i f b o t h a r e 0 .$ $b u t f o r a, b \in Z,$ $2 a^{3} - 4 {a b}^{2} - 6 a^{2} + 12 b^{2} + 27 i s o d d a n d c a n n o t$ $b e 0$ $\Rightarrow A i s d i a g o n a l, i . e .$ $A = (\begin{matrix} a + 2 b & 2 b \\ - b & a - 2 b \end{matrix}) = (\begin{matrix} c & 0 \\ 0 & c \end{matrix})$ $\Rightarrow b = 0 a n d A = (\begin{matrix} a & 0 \\ 0 & a \end{matrix})$ $n o w i f A^{3} - 6 A^{2} + 9 A - 27 I = 0, t h e n$ $a^{3} - 6 a^{2} + 9 a - 27 = 0$ $t h e o n l y s o l u t i o n i n R i s n o t i n Z .$ $\Rightarrow ∄ a, b \in Z, s u c h t h a t f o r$ $A = (\begin{matrix} a + 2 b & 2 b \\ - b & a - 2 b \end{matrix})$ $i s s a t i s f i e d$ $A^{3} + 9 A = 6 A^{2} + 27 I$
	Commented by Frix last updated on 17/Jan/23

	$Great!$
	Commented by Shrinava last updated on 20/Jan/23

	$perfect dear professor Aleks thank you$
	Answered by Frix last updated on 17/Jan/23

	$Just to complete it :$ $A^{3} - 6 A^{2} + 9 A - 27 I_{2} = [\begin{matrix} {t e r m}_{1} & {t e r m}_{2} \\ {t e r m}_{3} & {t e r m}_{4} \end{matrix}]$ $All terms = 0$ $(1) a^{3} + 6 a^{2} b + 6 {a b}^{2} + 4 b^{3} - 6 a^{2} - 24 a b - 12 b^{2} + 9 a + 18 b - 27 = 0$ $(2) 6 a^{2} b + 4 b^{3} - 24 a b + 18 b = 0$ $(3) - 3 a^{2} b - 2 b^{3} + 12 a b - 9 b = 0$ $(4) a^{3} - 6 a^{2} b + 6 {a b}^{2} - 4 b^{3} - 6 a^{2} + 24 a b - 12 b^{2} + 9 a - 18 b - 27 = 0$ $From (2) or (3) we get$ $a = 2 + \frac{\sqrt{3 - 2 b^{2}}}{\sqrt{3}} s with s = \pm 1$ $From (1) or (4) we get$ $\frac{2 (8 b^{2} - 3) \sqrt{9 - 6 b^{2}}}{9} s - 25 = 0$ $We get$ $s = - 1 \land b \approx \pm 1 . 58009 i \land a \approx . 367684$ $s = + 1 \land b \approx \pm (1.73133 \pm . 790045 i) \land a \approx 2.81616 \pm 1 . 11729 i$
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