Question

The value of $ \left| \begin{matrix} {{a}^{2}} & 2ab & {{b}^{2}} \\ {{b}^{2}} & {{a}^{2}} & 2ab \\ 2ab & {{b}^{2}} & {{a}^{2}} \\ \end{matrix} \right| $ is

• A. $ {{({{a}^{2}}+{{b}^{2}})}^{3}} $
• B. $ {{({{a}^{3}}+{{b}^{3}})}^{2}} $
• C. $ {{({{a}^{4}}+{{b}^{4}})}^{2}} $
• D. $ {{({{a}^{2}}+{{b}^{2}})}^{4}} $

Answer 1

Answer: (B)

Let $ \Delta =\left| \begin{matrix} {{a}^{2}} & 2ab & {{b}^{2}} \\ {{b}^{2}} & {{a}^{2}} & 2ab \\ 2ab & {{b}^{2}} & {{a}^{2}} \\ \end{matrix} \right| $
Applying, $ {{C}_{1}}\to {{C}_{2}}+{{C}_{3}}, $ we get $ \Rightarrow $
$ \Delta \left| \begin{matrix} {{(a+b)}^{2}} & 2ab & {{b}^{2}} \\ {{(a+b)}^{2}} & {{a}^{2}} & 2ab \\ {{(a+b)}^{2}} & {{b}^{2}} & {{a}^{2}} \\ \end{matrix} \right| $
$ ={{(a+b)}^{2}}\left| \begin{matrix} 1 & 2ab & {{b}^{2}} \\ 1 & {{a}^{2}} & 2ab \\ 1 & {{b}^{2}} & {{a}^{2}} \\ \end{matrix} \right| $
Apply $ {{R}_{2}}\to {{R}_{1}}-{{R}_{1}},\,{{R}_{3}}\Rightarrow {{R}_{3}}-{{R}_{1}}, $
we get $ \Delta ={{(a+b)}^{2}}\left| \begin{matrix} 1 & 2ab & {{b}^{2}} \\ 0 & {{a}^{2}}-2ab & 2ab-{{b}^{2}} \\ 0 & {{b}^{2}}-2ab & {{a}^{2}}-{{b}^{2}} \\ \end{matrix} \right| $
$ ={{(a+b)}^{2}}\left| \begin{matrix} a(a-2b) & b(2a-b) \\ b(b-2a) & (a-b)\,(a+b) \\ \end{matrix} \right| $
$ ={{(a+b)}^{2}}\{a(a-b)(a+b)(a-2b) $
$ -{{b}^{2}}(b-2a)\,(2a-b)\} $
$ ={{(a+b)}^{2}}\{a({{a}^{2}}-{{b}^{2}})(a-2b)+{{b}^{2}}{{(2a-b)}^{2}}\} $
$ ={{(a+b)}^{2}}\{({{a}^{2}}-{{b}^{2}})\,({{a}^{2}}-2ab) $
$ +{{b}^{2}}(4{{a}^{2}}+{{b}^{2}}-4ab)\} $
$ ={{(a+b)}^{2}}\{{{a}^{4}}-{{a}^{2}}{{b}^{2}}-2{{a}^{3}}b+2a{{b}^{3}} $
$ +4{{a}^{2}}{{b}^{2}}+{{b}^{4}}-4a{{b}^{3}}\} $
$ ={{(a+b)}^{2}}\{{{a}^{4}}+{{b}^{4}}+3{{a}^{2}}{{b}^{2}}-2{{a}^{3}}b-2a{{b}^{3}}\} $
$ ={{(a+b)}^{2}}\{{{({{a}^{2}}+{{b}^{2}})}^{2}}-2ab({{a}^{2}}+{{b}^{2}})+{{a}^{2}}{{b}^{2}}\} $
$ ={{(a+b)}^{2}}\{({{a}^{2}}+{{b}^{2}})({{a}^{2}}+{{b}^{2}}-2ab)+{{a}^{2}}{{b}^{2}}\} $
$ ={{(a+b)}^{2}}\{({{a}^{2}}+{{b}^{2}}){{(a-b)}^{2}}+{{a}^{2}}{{b}^{2}}\} $
$ =({{a}^{2}}+{{b}^{2}})({{a}^{2}}-{{b}^{2}})+{{a}^{2}}{{b}^{2}}{{(a+b)}^{2}} $
$ =({{a}^{4}}-{{b}^{4}})({{a}^{2}}-{{b}^{2}})+{{a}^{2}}{{b}^{2}}{{(a+b)}^{2}} $
$ ={{a}^{6}}-{{a}^{2}}{{b}^{4}}-{{a}^{4}}{{b}^{2}}+{{b}^{6}}+{{a}^{4}}{{b}^{2}}+{{a}^{2}}{{b}^{4}}+2{{a}^{3}}{{b}^{3}} $
$ ={{a}^{6}}+2{{a}^{3}}{{b}^{3}}+{{b}^{6}}={{({{a}^{3}}+{{b}^{3}})}^{2}} $