Question

If $\left|\begin{array}{ccc}b+c& c+a& a+b\\ a+b& b+c& c+a\\ c+a& a+b& b+c\end{array}\right|=k\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$, then the value of $k$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

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