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

Question

Using properties of determinants, prove that
$\begin{vmatrix} \text{b + c } & \text{c + a} & \text{a + b} 0.3em] \text{q } + \text{r} & \text{r + p} & \text{p + q} 0.3em] \text{y + z} & \text{z + x} &\text{x + y} \end{vmatrix}= \text{2}$
$\begin{vmatrix} \text{a } & \text{b} & \text{c} 0.3em] \text{p} & \text{q} & \text{r} 0.3em] \text{x} & \text{y} &\text{z} \end{vmatrix}$

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Answer

Operating $C → C_{1 —}(C_2 + C_3)$, we get
$\text{LHS}=\begin{vmatrix} \text{-2a } & \text{c + a} & \text{a + b} 0.3em] \text{-2p } & \text{r + p} & \text{p + q} 0.3em] \text{-2x} & \text{z + x} &\text{x + y} \end{vmatrix}=-2$
$\begin{vmatrix} \text{a } & \text{c + a} & \text{a + b} 0.3em] \text{p } & \text{r + p} & \text{p + q} \$0.3em] \text{x} & \text{z + x} &\text{x + y} \end{vmatrix}$
$ \begin{matrix} \text{C}{_2} → C_{2} — C_{1} \ \ \ \\ \text{C}{_3} → C_{3} — C_{1}\ \ \ \end{matrix} \Rightarrow\text{LHS}=-2\begin{vmatrix} \text{a } & \text{c} & \text{b} \$0.3em] \text{p } & \text{r} & \text{q} \$0.3em] \text{x} & \text{z} &\text{ y} \end{vmatrix}$
$\text{C}_2\leftrightarrow\text{C}_3$
$=+2\begin{vmatrix} \text{a } & \text{c} & \text{b} \$0.3em] \text{p } & \text{r} & \text{q} \$0.3em] \text{x} & \text{z} &\text{ y} \end{vmatrix}=\text{RHS}$

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