Question

Let $x=a_{2}z+a_{3}y,y=a_{1}z+a_{3}x$ and $z=a_{2}x+a_{1}y$, where x, y, z are not all zero. If a1 = k - [k], k being a non-integral constant, where [.] denotes greatest integer function, then $a_1{a}_2{a}_3$ is

• A. > 1
• B. > -1
• C. < 1
• D. < -1

Answer 1

The correct option is B > -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 zero, therefore given system of equation have non-trivial solution.
$\left|\begin{array}{ccc}1& -a_3& -a_2\\ a_3& -1& a_1\\ a_2& a_1& 1\end{array}\right|=0$
$\Rightarrow a_1^2+a_2^2+a_3^2+1a_1{a}_2{a}_3=1$ $\cdots \left(i\right)$
Since $a_1=m-\left[m\right],m\notin I$
$\therefore 0<a_1<1\Rightarrow 0<1-a_1^2<1$ $\cdots \left(ii\right)$
Now, from (i)
$1-a_2^2-a_3^2=a_1^2+2a_1{a}_2{a}_3$
$1-a_2^2-a_3^2+a_2^2{a}_3^2=a_1^2+2a_1{a}_2{a}_3+a_2^2{a}_3^2$
$\Rightarrow (1-a_2^2)(1-a_3^2)=(a_1+a_2{a}_3{)}^2$ $\cdots \left(iii\right)$
Similarly,
$(1-a_1^2)(1-a_3^2)=(a_2+a_1{a}_3{)}^2$ $\cdots \left(iv\right)$
$(1-a_1^2)(1-a_2^2)=(a_3+a_1{a}_2{)}^2$ $\cdots \left(v\right)$
Using equation (v)
$(1-a_2^2)=\frac{(a_3+a_1{a}_2{)}^2}{1-a_1^2}>0$
From (iv),
$1-a_3^2>0\Rightarrow 3-(a_1^2+a_2^2+a_3^2)>0$
$a_1^2+a_2^2+a_3^2<3\Rightarrow 1-2a_1{a}_2{a}_3<3$
$\Rightarrow a_1{a}_2{a}_3>-1$