Question

If [a] denotes the integral part of $a$ and $x=a_{3} y+a_{2} z$, $y = a _{1} z + a _{3} z$ and $z = a _{2} x + a _{1} y$, where $x , y , z$ are not all zero. If $a _{1}= m -[ m ], m$ being a non-integral constant, then $a _{1} a _{2} a _{3}$ is

• A. > 1
• B. >-1
• C. <1
• D. < -1

Answer 1

Answer: (B)

Given, $x=a_{3} y+a_{2} z\,\,\,\,\, \dots(i)$
$y =a_1z^{1}a_3 x \,\,\,\,\, \dots(ii)$
$z = a _{2} x + a _{1} y\,\,\,\,\,\,\dots(iii)$
Since, $x , y , z $ are not all zero, therefore given system of equations has non-trivial solution.
$\therefore \begin{vmatrix} 1 & - a _{3} & - a _{2} \\ a _{3} & -1 & a _{1} \\ a _{2} & a _{1} & -1 \end{vmatrix}=0 $
$\Rightarrow a _{1}^{2}+ a _{2}^{2}+ a _{3}^{2}+2 a _{1} a _{2} a _{3}=1\,\,\,\, \dots(iv)$
Since, $a _{1}= m -[ m ]$ and $m$ is not an integer.
$\therefore 0< \,a _{1}<\,1 $
$\Rightarrow 0<\, 1- a _{1_{2}}{ }^{2}<\,1 \,\,\,\, \dots(v)$
From Eq. (iv), $1-a_{2}{ }_{2}^{2}-a_{3}^{1_{2}}=a_{1}{ }^{2}+2 a_{1} a_{2} a_{3}$
$\Rightarrow 1- a _{2}^{2}- a _{3}^{2}+ a _{2}^{2} a _{3}^{2}= a _{1}^{2}+2 a _{1} a _{2} a _{3}+ a _{2}{ }^{2} a _{3}^{2}$
$\Rightarrow \left(1-a_{2}^{2}\right)\left(1-a_{3}^{2}\right)=\left(a_{1}+a_{2} a_{3}\right)^{2}\,\,\,\,\, \dots(vi)$
Similarly, $\left(1- a _{1}^{2}\right)\left(1- a _{3}^{2}\right)=\left( a _{2}+ a _{1} a _{3}\right)^{2}\,\,\,\, \dots(vii)$
$\left(1-a_{1}^{2}\right)\left(1-a_{2}^{2}\right)=\left(a_{3}+a_{1} a_{2}\right)^{2}\,\,\,\, \dots(viii)$
From Eq. (viii), $1- a _{2}{ }^{2}=>\,0 \cdot \frac{\left(a_{3}+a_{1} a_{2}\right)^{2}}{1-a_{1}^{2}}$
From Eq. (viii), $1- a _{3}^{2}>\,0$
$ \Rightarrow 3-\left( a _{1}^{2}+ a _{2}^{2}+ a _{3}^{2}\right)>\,0$
$\Rightarrow a _{1}^{2}+ a _{2}^{2}+ a _{3}^{2}<\,3$
$ \Rightarrow 1-2 a _{1} a _{2} a _{3}<\,3$ [ From Eq. (iv)]
$\Rightarrow a _{1} a _{2} a _{3}>\,-1$