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Mathematics Question on Determinants

The value of ∣a22abb2 b2a22ab 2abb2a2 ∣\left| \begin{matrix} {{a}^{2}} & 2ab & {{b}^{2}} \\\ {{b}^{2}} & {{a}^{2}} & 2ab \\\ 2ab & {{b}^{2}} & {{a}^{2}} \\\ \end{matrix} \right| is

A

(a2+b2)3{{({{a}^{2}}+{{b}^{2}})}^{3}}

B

(a3+b3)2{{({{a}^{3}}+{{b}^{3}})}^{2}}

C

(a4+b4)2{{({{a}^{4}}+{{b}^{4}})}^{2}}

D

(a2+b2)4{{({{a}^{2}}+{{b}^{2}})}^{4}}

Answer

(a3+b3)2{{({{a}^{3}}+{{b}^{3}})}^{2}}

Explanation

Solution

Let Δ=∣a22abb2 b2a22ab 2abb2a2 ∣\Delta =\left| \begin{matrix} {{a}^{2}} & 2ab & {{b}^{2}} \\\ {{b}^{2}} & {{a}^{2}} & 2ab \\\ 2ab & {{b}^{2}} & {{a}^{2}} \\\ \end{matrix} \right|
Applying, C1→C2+C3,{{C}_{1}}\to {{C}_{2}}+{{C}_{3}}, we get ⇒\Rightarrow
Δ∣(a+b)22abb2 (a+b)2a22ab (a+b)2b2a2 ∣\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∣12abb2 1a22ab 1b2a2 ∣={{(a+b)}^{2}}\left| \begin{matrix} 1 & 2ab & {{b}^{2}} \\\ 1 & {{a}^{2}} & 2ab \\\ 1 & {{b}^{2}} & {{a}^{2}} \\\ \end{matrix} \right|
Apply R2→R1−R1, R3⇒R3−R1,{{R}_{2}}\to {{R}_{1}}-{{R}_{1}},\,{{R}_{3}}\Rightarrow {{R}_{3}}-{{R}_{1}},
we get Δ=(a+b)2∣12abb2 0a2−2ab2ab−b2 0b2−2aba2−b2 ∣\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∣a(a−2b)b(2a−b) b(b−2a)(a−b) (a+b) ∣={{(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)2a(a2−b2)(a−2b)+b2(2a−b)2={{(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)2a4+b4+3a2b2−2a3b−2ab3={{(a+b)}^{2}}\\{{{a}^{4}}+{{b}^{4}}+3{{a}^{2}}{{b}^{2}}-2{{a}^{3}}b-2a{{b}^{3}}\\}
=(a+b)2(a2+b2)2−2ab(a2+b2)+a2b2={{(a+b)}^{2}}\\{{{({{a}^{2}}+{{b}^{2}})}^{2}}-2ab({{a}^{2}}+{{b}^{2}})+{{a}^{2}}{{b}^{2}}\\}
=(a+b)2(a2+b2)(a2+b2−2ab)+a2b2={{(a+b)}^{2}}\\{({{a}^{2}}+{{b}^{2}})({{a}^{2}}+{{b}^{2}}-2ab)+{{a}^{2}}{{b}^{2}}\\}
=(a+b)2(a2+b2)(a−b)2+a2b2={{(a+b)}^{2}}\\{({{a}^{2}}+{{b}^{2}}){{(a-b)}^{2}}+{{a}^{2}}{{b}^{2}}\\}
=(a2+b2)(a2−b2)+a2b2(a+b)2=({{a}^{2}}+{{b}^{2}})({{a}^{2}}-{{b}^{2}})+{{a}^{2}}{{b}^{2}}{{(a+b)}^{2}}
=(a4−b4)(a2−b2)+a2b2(a+b)2=({{a}^{4}}-{{b}^{4}})({{a}^{2}}-{{b}^{2}})+{{a}^{2}}{{b}^{2}}{{(a+b)}^{2}}
=a6−a2b4−a4b2+b6+a4b2+a2b4+2a3b3={{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}}
=a6+2a3b3+b6=(a3+b3)2={{a}^{6}}+2{{a}^{3}}{{b}^{3}}+{{b}^{6}}={{({{a}^{3}}+{{b}^{3}})}^{2}}