Question

The determinant $ \left| \begin{matrix} 4+{{x}^{2}} & -6 & -2 \\ -6 & 9+{{x}^{2}} & 3 \\ -2 & 3 & 1+{{x}^{2}} \\ \end{matrix} \right| $ is not divisible by

• A. $ x $
• B. $ {{x}^{3}} $
• C. $ 14+{{x}^{2}} $
• D. $ {{x}^{5}} $

Answer 1

Answer: (D)

$ \left| \begin{matrix} 4+{{x}^{2}} & -6 & -2 \\ -6 & 9+{{x}^{2}} & 3 \\ -2 & 3 & 1+{{x}^{2}} \\ \end{matrix} \right| $
$ =(4+{{x}^{2}})[(1+{{x}^{2}})(9+{{x}^{2}})-9] $
$ +6[-6(1+{{x}^{2}})+6]-2[-18+2(9+{{x}^{2}})] $
$ =(4+{{x}^{2}})(9+9{{x}^{2}}+{{x}^{2}}{{x}^{4}}-9) $
$ +6(-6-6{{x}^{2}}+6)-2(-18+18+2{{x}^{2}}) $
$ =(4+{{x}^{2}})(10{{x}^{2}}+{{x}^{4}})-36{{x}^{2}}-4{{x}^{2}} $
$ =40{{x}^{2}}+4{{x}^{4}}+10{{x}^{4}}+{{x}^{6}}-40{{x}^{2}} $
$ =14{{x}^{4}}+{{x}^{6}} $
$ ={{x}^{4}}({{x}^{2}}+14) $
Which is not divisible by $ {{x}^{5}} $ .