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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants4 Marks
Question
Prove that:
$\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}=(\text{a}^3+\text{b}^3)^2$
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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}$
$=\text{a}^2\begin{vmatrix}\text{a}^2&2\text{ab}\\\text{b}^2&\text{a}^2\end{vmatrix}-(2\text{ab})\begin{vmatrix}\text{b}^2&2\text{ab}\\2\text{ab}&\text{a}^2\end{vmatrix}+\text{b}^2\begin{vmatrix}\text{b}^2&\text{a}^2\\2\text{ab}&\text{b}^2\end{vmatrix}$ [Expanding]
$=\text{a}^2(\text{a}^4-2\text{ab}^3)-(2\text{ab})(\text{b}^2\text{a}^2-4\text{a}^2\text{b}^2)+\text{b}^2(\text{b}^4-2\text{a}^3\text{b})$
$=\text{a}^6-2\text{a}^3\text{b}^3-2\text{a}^3\text{b}^3+8\text{a}^3\text{b}^3+\text{b}^6-2\text{a}^3\text{b}^3$
$=\text{a}^6+2\text{a}^3\text{b}^3+(\text{b}^3)^2$
$=(\text{a}^3+\text{b}^3)^2$
$=\text{R.H.S}$

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