Using properties of determinants, show that = b+c & a & a b & c+a… - Vidyadip

Question

Using properties of determinants, show that
$\Delta=\begin{vmatrix} \text{b+c} & \text{a} & \text{a} \\ \text{b} & \text{c+a} & \text{b} \\ \text{c} & \text{c} & \text{a+b} \end{vmatrix}=\text{4 abc}.$

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Answer

$\text{LHS }=\Delta=\begin{vmatrix} \text{b+c} & \text{a} & \text{a} \\ \text{b} & \text{c+a} & \text{b} \\ \text{c} & \text{c} & \text{a+b} \end{vmatrix};\DeclareMathOperator*{\median}{\text{ performing}} \median_{\text{R}_{1}\rightarrow\text{R}_{1}-\text{R}_{2}-\text{R}_{3}}\Delta=\begin{vmatrix} \text{0} & \text{-2c} & \text{-2b} \\ \text{b} & \text{c+a} & \text{b} \\ \text{c} & \text{c} & \text{a+b} \end{vmatrix}$
$=\frac{1}{\text{c}}\begin{vmatrix} 0 & \text{-2c} & \text{-2b} \\ \text{bc} & \text{(c+a)c} & \text{bc} \\ \text{c} & \text{c} & \text{a+b} \end{vmatrix}$
using $R_2\rightarrow R_2-bR_3$ gives $\Delta=\frac{1}{\text{c}}$ $\begin{vmatrix} 0 & \text{-2c} & \text{-2b} \\ \text{0} & \text{c(a+c-b)} & \text{b(c-a-b)} \\ \text{c} & \text{c} & \text{a+b} \end{vmatrix}$
$=\frac{\text{1}}{\text{c}}\cdot\text{c}\begin{vmatrix} \text{-2c} & \text{-2b} \\ \text{(c+a-b)c} & \text{(c-a-b)b} \\ \end{vmatrix}$ = 2bc [(-c+a+b) + (c+a-b)] = 4 abc
= RHS.

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