Question

Using properties of determinant, prove that:

$\left|\begin{array}{ccc}3a& -a+b& -a+c\\ -b+a& 3b& -b+c\\ -c+a& -c+b& 3c\end{array}\right|$ $=3(a+b+c)(ab+bc+ca)$

Answer 1

On applying $C_1\to C_1+C_2+C_3$ to the given matrix, we get

$\left|\begin{array}{ccc}a+b+c& -a+b& -a+c\\ a+b+c& 3b& -b+c\\ a+b+c& -c+b& 3c\end{array}\right|$

$=(a+b+c)\left|\begin{array}{ccc}1& -a+b& -a+c\\ 1& 3b& -b+c\\ 1& -c+b& 3c\end{array}\right|$

$R_1\to R_1-R_2$ and $R_3\to R_3-R_2$

$=(a+b+c)\left|\begin{array}{ccc}1-1& -a+b-3b& -a+c+b-c\\ 1& 3b& -b+c\\ 1-1& -c+b-3b& 3c+b-c\end{array}\right|$

$=(a+b+c)\left|\begin{array}{ccc}0& -a-2b& b-a\\ 1& 3b& -b+c\\ 0& -c-2b& 2c+b\end{array}\right|$

On expanding

$=(a+b+c)(-1)\left[\right(2c+b\left)\right(-a-2b)+(c+2b\left)\right(b-a\left)\right]$

$=(a+b+c)(-1)(-3ab-3bc-3ac)$

$=3(a+b+c)(ab+bc+ac)$