Prove that:

Solution:

Let $\Delta=\mid \begin{array}{lll}a+b & b+c & c+a\end{array}$

$b+c \quad c+a \quad a+b$

$c+a \quad a+b \quad b+c \mid$

Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we get'

$\Delta=\mid \begin{array}{lll}a & b & c\end{array}$

$\begin{array}{lll}b & c & a\end{array}$

$c \quad a \quad b|+| \begin{array}{lll}b & c & a\end{array}$

$\begin{array}{lll}c & a & b\end{array}$

$a \quad b \quad c$

$=\mid \begin{array}{lll}a & b & c\end{array}$

$\begin{array}{lll}b & c & a\end{array}$

$\begin{array}{lll}c & a & b\end{array}|+(-1)| \begin{array}{lll}a & c & b\end{array}$

$\begin{array}{lll}b & a & c\end{array}$

$\begin{array}{ccc}c & b & a\end{array}$            [Applying $\mathrm{C}_{1} \leftrightarrow \mathrm{C}_{3}$ in second determinant to get negative value of the deteminant]

$=\mid \begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}$

$\begin{array}{lll}b & c & a\end{array}$

$\begin{array}{lll}\mathrm{c} & \mathrm{a} & \mathrm{b}\end{array}|+(-1)(-1)| \begin{array}{lll}a & b & c\end{array}$

$\begin{array}{lll}b & c & a\end{array}$

$\begin{array}{ccc}c & a & b\end{array}$         [Applying $\mathrm{C}_{2} \leftrightarrow \mathrm{C}_{3}$ ]

$=2 \mid a b c$

$\begin{array}{lll}b & c & a\end{array}$

$\begin{array}{lll}c & a & b\end{array}=$ RHS