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, $x=a_{3}y+a_{2}z\dots (i)$
$y=a_1{z}^1{a}_{3}x\dots (ii)$
$z=a_{2}x+a_{1}y\dots (iii)$
Since, $x,y,z$ are not all zero, therefore given system of equations has non-trivial solution.
$\therefore \left|\begin{array}{ccc}1& -a_3& -a_2\\ a_3& -1& a_1\\ a_2& a_1& -1\end{array}\right|=0$
$\Rightarrow a_1^2+a_2^2+a_3^2+2a_1{a}_2{a}_3=1\dots (iv)$
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), $1-a_2{}_2^2-a_3^{1_2}=a_1{}^2+2a_1{a}_2{a}_3$
$\Rightarrow 1-a_2^2-a_3^2+a_2^2{a}_3^2=a_1^2+2a_1{a}_2{a}_3+a_2{}^2{a}_3^2$
$\Rightarrow \left(1-a_2^2\right)\left(1-a_3^2\right)={\left(a_1+a_2{a}_3\right)}^2\dots (vi)$
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), $1-a_3^2>0$
$\Rightarrow 3-\left(a_1^2+a_2^2+a_3^2\right)>0$
$\Rightarrow a_1^2+a_2^2+a_3^2<3$
$\Rightarrow 1-2a_1{a}_2{a}_3<3$ [ From Eq. (iv)]
$\Rightarrow a_1{a}_2{a}_3>-1$