Using the properties of determinants prove that: 1 +a^ 2 -b^ 2 & … - Vidyadip

Question

Using the properties of determinants prove that:
$\begin{vmatrix} 1 +\text{a}^{2}-\text{b}^{2} & 2\text{ab} & -2{b} \\ \text{2ab} & 1 - \text{a}^{2} + \text{b}^{2} & \text{2a} \\ \text{2b} & \text{-2a} & \text{1 - a}^{2} - \text{b}^{2} \end{vmatrix} = (1 +\text{ a}^{2} + \text{b}^{2})^{3}$

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Answer

L.H.S. = $\begin{vmatrix} 1 +\text{a}^{2}-\text{b}^{2} & 2\text{ab} & -2{b} \\ \text{2ab} & 1 - \text{a}^{2} + \text{b}^{2} & \text{2a} \\ \text{2b} & \text{-2a} & \text{1 - a}^{2} - \text{b}^{2} \end{vmatrix} $
$\text{R}_{1}\rightarrow\text{R}_{1}+\text{b}.\text{R}_{3}, \text{R}_{2}\rightarrow\text{R}_{2} - \text{a}\text{ R}_{3}$
$\begin{vmatrix} \text{1 + a}^{2} + \text{b}^{2} & 0 & \text{-b(1 + a}^{2} + \text{b}^{2} \\ 0 &\text{1 + a}^{2} + \text{b}^{2}& \text{a (1 + a}^{2} + \text{b}^{2}) \\ \text{2b} & \text{-2a} & \text{1 - a}^{2} - \text{b}^{2} \end{vmatrix} $
Expanding and getting
$\triangle = {(1 + \text{a}}^{2} + \text{b}^{2})^{3} = \text{R.H.S}$

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