Math Question and Answers


Question and Answers Forum

All Questions Topic List
Matrices and Determinants Questions
Previous in All Question Next in All Question
Previous in Matrices and Determinants Next in Matrices and Determinants
Question Number 56339 by Tinkutara last updated on 14/Mar/19

	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19
	$x=a_3 y+a_2 (a_2 x+a_1 y) x−a_2 ^2 x=y(a_3 +a_1 a_2 ) x(1−a_2 ^2 )=y(a_3 +a_1 a_2 ) (x/y)=((a_3 +a_1 a_2 )/(1−a_2 ^2 )) y=a_1 (a_2 x+a_1 y)+a_3 x y(1−a_1 ^2 )=x(a_3 +a_1 a_2 ) (x/y)=((1−a_1 ^2 )/(a_3 +a_1 a_2 )) hence ((a_3 +a_1 a_2 )/(1−a_2 ^2 ))=((1−a_1 ^2 )/(a_3 +a_1 a_2 )) a_3 ^2 +2a_1 a_2 a_3 +a_1 ^2 a_2 ^2 =1−a_1 ^2 −a_2 ^2 +a_1 ^2 a_2 ^2 a_1 a_2 a_3 =(1/2){1−(a_1 ^2 +a_2 ^2 +a_3 ^2 )} 1)a_1 =m−[m]={m}←fractional part so 1 >a_1 ^2 >0 but no mention of a_2 and a_3 in question so conclusion can not be drawn... let other check...$
	$x = a_{3} y + a_{2} (a_{2} x + a_{1} y)$ $x - a_{2}^{2} x = y (a_{3} + a_{1} a_{2})$ $x (1 - a_{2}^{2}) = y (a_{3} + a_{1} a_{2})$ $\frac{x}{y} = \frac{a_{3} + a_{1} a_{2}}{1 - a_{2}^{2}}$ $y = a_{1} (a_{2} x + a_{1} y) + a_{3} x$ $y (1 - a_{1}^{2}) = x (a_{3} + a_{1} a_{2})$ $\frac{x}{y} = \frac{1 - a_{1}^{2}}{a_{3} + a_{1} a_{2}}$ $h e n c e$ $\frac{a_{3} + a_{1} a_{2}}{1 - a_{2}^{2}} = \frac{1 - a_{1}^{2}}{a_{3} + a_{1} a_{2}}$ $a_{3}^{2} + 2 a_{1} a_{2} a_{3} + a_{1}^{2} a_{2}^{2} = 1 - a_{1}^{2} - a_{2}^{2} + a_{1}^{2} a_{2}^{2}$ $a_{1} a_{2} a_{3} = \frac{1}{2} {1 - (a_{1}^{2} + a_{2}^{2} + a_{3}^{2})}$ $1) a_{1} = m - [m] = {m} \leftarrow f r a c t i o n a l p a r t$ $s o 1 > a_{1}^{2} > 0$ $b u t n o m e n t i o n o f a_{2} a n d a_{3} i n q u e s t i o n$ $s o c o n c l u s i o n c a n n o t b e d r a w n . . .$ $l e t o t h e r c h e c k . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum