Determinants Matrices Question 17
Question: If $ [a] $ denotes the integral part of a and $ x=a_3y+a_2z, $
$ y=a_1z+a_3z $ and $ z=a_2x+a_1y, $ where x, y, z are not all zero. If $ a_1=m-[m], $ m being a non-integral constant, then $ a_1a_2a_3 $ is
Options:
A) $ >1 $
B) $ >-1 $
C) $ <1 $
D) $ <-1 $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Given, $ x=a_3y+a_2z $
… (i) $ y=a_1z+a_3x $
… (ii) $ z=a_2x+a_1y $
… (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}+2a_1a_2a_3=1 $ ……. (iv) Since, $ a_1=m-[m] $ and m is not an integer.
$ \therefore 0<a_1<1\Rightarrow 1-a_1^{2}<1 $
… (v) From Eq. (iv), $ 1-a_2^{2}-a_3^{2}=a_1^{2}+2a_1a_2a_3 $
$ \Rightarrow 1-a_2^{2}-a_3^{2}+a_2^{2}a_3^{2}=a_1^{2}+2a_1a_2a_3+a_2^{2}a_3^{2} $
$ \Rightarrow (1-a_2^{2})(1-a_3^{2})={{(a_1+a_2a_3)}^{2}}. $
……..(vi) Similarly, $ (1-a_1^{2})(1-a_3^{2})={{(a_2+a_1a_3)}^{2}} $ …(vii) $ (1-a_1^{2})(1-a_2^{2})={{(a_3+a_1a_2)}^{2}} $
…(viii)
From Eq. (viii), $ 1-a_2^{2}\Rightarrow 0.\frac{{{(a_3+a_1a_2)}^{2}}}{1-a_1^{2}} $
From Eq. (viii),
$ 1-a_3^{2}>0\Rightarrow 3-(a_1^{2}+a_2^{2}+a_3^{2})>0 $
$ \Rightarrow a_1^{2}+a_2^{2}+a_3^{2}<3\Rightarrow 1-2a_1a_2a_3<3 $
[From Eq. (iv)]
$ \Rightarrow a_1a_2a_3>-1 $
BETA
