Question

Using properties of determinants prove the following questions.

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

Answer 1

Let $A=\left|\begin{array}{ccc}3a& -a+b& -a+c\\ -b+a& 3b& -b+c\\ -c+a& -c+b& 3c\end{array}\right|=\left|\begin{array}{ccc}a+b+c& -a+b& -a+c\\ a+b+c& 3b& -b+c\\ a+b+c& -c+b& 3c\end{array}\right|$
(using $C_1\to C_1+C_2+C_3$)
$=(a+b+c)\left|\begin{array}{ccc}1& -a+b& -a+c\\ 1& 3b& -b+c\\ 1& -c+b& 3c\end{array}\right|$
(Taking out $(a+b+c)$ common from $C_1$)
$=(a+b+c)\left|\begin{array}{ccc}1& -a+b& -a+c\\ 0& 2b+a& a-b\\ 0& a-c& 2c+a\end{array}\right|$
(using $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$)
Expanding along $C_1$, we get
$=(a+b+c)\left[\right(2b+a\left)\right(2c+a)-(a-b\left)\right(a-c\left)\right]=(a+b+c)[4bc+2ab+2ac+a^2-a^2+ac+ba-bc]=(a+b+c)(3ab+3bc+3ac)=3(a+b+c)(ab+bc+ca)=RHS$