Question

Let $\left[a\right]$ denotes the integral part of $a$ and $x=a_{3}y+a_{2}z,y=a_{1}z+a_{3}x$ and $z=a_{2}x+a_{1}y$, where $x,y,z$ are not all zero. If $a_1=m-\left[m\right],m$ being a non-integral constant, then the least integral value of $|a_1{a}_2{a}_3|$ is

Answer 1

Given , $x=a_{3}y+a_{2}z$
$y=a_{1}z+a_{3}x$
$z=a_{2}x+a_{1}y$
Since, $x,y,z$ are not all zero, therefore given system of equations has a non-trivial solution.
$\therefore \left|\begin{array}{ccc}1& -a_3& -a_2\\ a_3& -1& a_1\\ a_2& a_1& -1\end{array}\right|=0$

$\Rightarrow a_1^2+a_2^2+a_3^2+2a_1{a}_2{a}_3=1\cdots \left(1\right)$
Since, $a_1=m-\left[m\right]$ and m is not an integer,
$\therefore 0<a_1<1\Rightarrow 0<1-a_1^2<1$
From Eq. $\left(1\right)$,
$1-a_2^2-a_3^2=a_1^2+2a_1{a}_2{a}_3$
$\Rightarrow 1-a_2^2-a_3^2+a_2^2{a}_3^2=a_1^2+2a_1{a}_2{a}_3+a_2^2{a}_3^2$
$\Rightarrow (1-a_2^2)(1-a_3^2)=(a_1+a_2{a}_3{)}^2$
Similarly, $(1-a_1^2)(1-a_3^2)=(a_2+a_1{a}_3{)}^2\cdots (2)$
and $(1-a_1^2)(1-a_2^2)=(a_3+a_1{a}_2{)}^2\cdots (3$

From Eq. $\left(3\right)$,
$1-a_2^2=\frac{(a_3+a_1{a}_2{)}^2}{1-a_1^2}>0$
From Eq. $\left(2\right)$, $1-a_3^2>0$

So, $1-a_1^2+1-a_2^2+1-a_3^2>0$
$\Rightarrow a_1^2+a_2^2+a_3^2<3$
$\Rightarrow 1-2a_1{a}_2{a}_3<3$ [From Eq. $\left(1\right)$]
$\Rightarrow a_1{a}_2{a}_3>-1\Rightarrow |a_1{a}_2{a}_3|>1$
So the least integral value of $|a_1{a}_2{a}_3|$ is $2$