Question

$\left|\begin{array}{ccc}b+c& c& b\\ c& c+a& a\\ b& a& a+b\end{array}\right|=kabc$, where $k$ is a constant, then $k$ is equal to

Answer 1

Let $\Delta =\left|\begin{array}{ccc}b+c& c& b\\ c& c+a& a\\ b& a& a+b\end{array}\right|$
Applying $R_1\to R_1-(R_2+R_3)$
$\therefore \Delta =\left|\begin{array}{ccc}0& -2a& -2a\\ c& c+a& a\\ b& a& a+b\end{array}\right|$
Taking $-2a$ common from $R_1$, then
$\Delta =(-2a)\left|\begin{array}{ccc}0& 1& 1\\ c& c+a& a\\ b& a& a+b\end{array}\right|$
Applying $C_2\to C_2-C_3$
$\therefore \Delta =-2a\left|\begin{array}{ccc}0& 0& 1\\ c& c& a\\ b& -b& a+b\end{array}\right|$
Expanding along $R_1$, we get
$\Delta =(-2a).1.\left|\begin{array}{cc}c& c\\ b& -b\end{array}\right|$
$=(-2a)(-2bc)$
$\therefore \Delta =4abc$

Hence, $k=4$.