A) \[>1\]
B) \[>-1\]
C) \[<1\]
D) \[<-1\]
Correct Answer: B
Solution :
| [b] Given, \[x={{a}_{3}}y+{{a}_{2}}z\] ... (i) |
| \[y={{a}_{1}}z+{{a}_{3}}x\] ... (ii) |
| \[z={{a}_{2}}x+{{a}_{1}}y\] ... (iii) |
| Since, x, y, z are not all zero, therefore given system of equations has non-trivial solution. |
| \[\therefore \,\,\,\left| \begin{matrix} 1 & -{{a}_{3}} & -{{a}_{2}} \\ {{a}_{3}} & -1 & {{a}_{1}} \\ {{a}_{2}} & {{a}_{1}} & -1 \\ \end{matrix} \right|=0\] |
| \[\Rightarrow \,\,\,{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}+2{{a}_{1}}{{a}_{2}}{{a}_{3}}=1\] ?. (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), \[1-{{a}_{2}}^{2}-{{a}_{3}}^{2}={{a}_{1}}^{2}+2{{a}_{1}}{{a}_{2}}{{a}_{3}}\] |
| \[\Rightarrow \,\,1-{{a}_{2}}^{2}-{{a}_{3}}^{2}+{{a}_{2}}^{2}{{a}_{3}}^{2}={{a}_{1}}^{2}+2{{a}_{1}}{{a}_{2}}{{a}_{3}}+{{a}_{2}}^{2}{{a}_{3}}^{2}\] |
| \[\Rightarrow \,\,(1-{{a}_{2}}^{2})(1-{{a}_{3}}^{2})={{({{a}_{1}}+{{a}_{2}}{{a}_{3}})}^{2}}.\] ?..(vi) |
| Similarly, \[(1-{{a}_{1}}^{2})(1-{{a}_{3}}^{2})={{({{a}_{2}}+{{a}_{1}}{{a}_{3}})}^{2}}\] ...(vii) |
| \[(1-{{a}_{1}}^{2})(1-{{a}_{2}}^{2})={{({{a}_{3}}+{{a}_{1}}{{a}_{2}})}^{2}}\] ...(viii) |
| From Eq. (viii), \[1-{{a}_{2}}^{2}\,\,\,\,\,\Rightarrow \,\,\,0.\frac{{{({{a}_{3}}+{{a}_{1}}{{a}_{2}})}^{2}}}{1-{{a}_{1}}^{2}}\] |
| From Eq. (viii), |
| \[1-{{a}_{3}}^{2}>0\Rightarrow 3-({{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2})>0\] |
| \[\Rightarrow \,\,\,{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}<3\Rightarrow 1-2{{a}_{1}}{{a}_{2}}{{a}_{3}}<3\] |
| [From Eq. (iv)] |
| \[\Rightarrow \,\,\,\,{{a}_{1}}{{a}_{2}}{{a}_{3}}>-1\] |
You need to login to perform this action.
You will be redirected in
3 sec
