Question

$\left|\begin{array}{ccc}a+b& b+c& c+a\\ b+c& c+a& a+b\\ c+a& a+b& b+c\end{array}\right|=K\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$, then $K=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

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

[$C_1^{\prime }=C_1+C_2+C_3$]

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

$=2\left|\begin{array}{ccc}a+b+c& b+c& c+a\\ a+b+c& c+a& a+b\\ a+b+c& a+b& b+c\end{array}\right|$

[$C_1^{\prime }=C_1-C_2$]

$=2\left|\begin{array}{ccc}a& b+c& c+a\\ b& c+a& a+b\\ c& a+b& b+c\end{array}\right|$

[$C_3^{\prime }=C_3-C_1$]

$=2\left|\begin{array}{ccc}a& b+c& c\\ b& c+a& a\\ c& a+b& b\end{array}\right|$

[$C_2^{\prime }=C_2-C_3$]

$=2\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$

Comparing it with the given problem we have $K=2$