Using properties of determinants, prove the following: a^ 2 & bc … - Vidyadip

Question

Using properties of determinants, prove the following:
$ \begin{vmatrix} \text{a}^{2} & \text{bc} & \text{ac + c}^{2} \\ \text{a}^{2} + \text{ab} & \text{b}^{2} & \text{ac} \\ \text{ab} & \text{b}^{2} + \text{bc} & \text{c}^{2} \end{vmatrix} = \text{4 a}^{2}\text{b}^{2}\text{c}^{2}. $

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Answer

$ \begin{vmatrix} \text{a}^{2} & \text{bc} & \text{ac + c}^{2} \\ \text{a}^{2} + \text{ab} & \text{b}^{2} & \text{ac} \\ \text{ab} & \text{b}^{2} + \text{bc} & \text{c}^{2} \end{vmatrix} = \text{abc} \begin{vmatrix} \text{a} & \text{c} & \text{a + c} \\ \text{a+b} & \text{b} & \text{a} \\ \text{b} & \text{b} + {c} & \text{c} \end{vmatrix}$
$\text{Talking b & c common from C}_{1}, \text{C}_{\text{2 and}} \text{C}_{3} $
$= \text{2abc} \begin{vmatrix} \text{a + c} & \text{c} & \text{a + c} \\ \text{a+b} & \text{b} & \text{a} \\ \text{b + c} & \text{b + c} & \text{c} \end{vmatrix}$
$\text{C}_{1}\rightarrow\text{C}_{1} + \text{C}_{2} + \text{C}_{3} \text{ and taking 2 common from C}_{1} $
$=2 \text{abc} \begin{vmatrix} \text{a + c} & \text{c} & \text{0} \\ \text{a+b} & \text{b} & \text{-b} \\ \text{b + c} & \text{b + c} & \text{-b} \end{vmatrix} \text{C}_{3}\rightarrow\text{C}_{3} - \text{C}_{1}$
$=\text{2 abc} \begin{vmatrix} \text{a + c} & \text{c} & \text{0} \\ \text{a - c} & \text{-c} & \text{0} \\ \text{b + c} & \text{b + c} & \text{-b} \end{vmatrix} \text{R}_{2}\rightarrow\text{R}_{2} - \text{R}_{3}$
Expand by $\text{C}_{3}, = \text{2abc( -b) (-ac - c}^{2} - \text{ac + c}^{2}) = \text{4a}^{2}\text{b}^{2}\text{c}^{2}$

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