Question

If $\left|\begin{array}{ccc}a+b& b+c& c+a\\ b+c& c+a& a+b\\ c+a& a+b& b+c\end{array}\right|=\lambda \left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ then $\lambda$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Given $\left|\begin{array}{ccc}a+b& b+c& c+a\\ b+c& c+a& a+b\\ c+a& a+b& b+c\end{array}\right|=\lambda \left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ ....(1)
Taking $LHS=\left|\begin{array}{ccc}a+b& b+c& c+a\\ b+c& c+a& a+b\\ c+a& a+b& b+c\end{array}\right|$
$C_1\to C_1+C_2+C_3$
$\left|\begin{array}{ccc}2(a+b+c)& b+c& c+a\\ 2(a+b+c)& c+a& a+b\\ 2(a+b+c)& a+b& b+c\end{array}\right|$
$=2(a+b+c)\left|\begin{array}{ccc}1& b+c& c+a\\ 1& c+a& a+b\\ 1& a+b& b+c\end{array}\right|$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$=2(a+b+c)\left|\begin{array}{ccc}0& b-a& c-b\\ 0& c-b& a-c\\ 1& a+b& b+c\end{array}\right|$
$=2(a+b+c)\left[\right(b-a\left)\right(a-c)-{(c-b)}^2]$
$LHS=2(a+b+c)[-a^2-b^2-c^2+ab+bc+ca]$
Now consider, $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$
$C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}a+b+c& b& c\\ a+b+c& c& a\\ a+b+c& a& b\end{array}\right|$
$=(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& c& a\\ 1& a& b\end{array}\right|$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$=(a+b+c)\left|\begin{array}{ccc}0& b-c& c-a\\ 0& c-a& a-b\\ 1& a& b\end{array}\right|$
$=(a+b+c)\left[\right(b-c\left)\right(a-b)-{(c-a)}^2]$
$=(a+b+c)[-a^2-b^2-c^2+ab+bc+ca]$
So by (1)

$2(a+b+c)[-a^2-b^2-c^2+ab+bc+ca]=\lambda (a+b+c)[-a^2-b^2-c^2+ab+bc+ca]$
$\Rightarrow \lambda =2$