Question

Show that:

$\left|\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ c& c& a+b\end{array}\right|$ $=4abc$

Answer 1

To prove:

$\left|\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ c& c& a+b\end{array}\right|$ $=4abc$

Proof:

Lets take LHS and then equate it to RHS.

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

$R_1=R_1-(R_2+R_3)$

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

$=\left|\begin{array}{ccc}0& -2c& -2b\\ b& c+a& b\\ c& c& a+b\end{array}\right|$

$=0\left\{\right(c+a\left)\right(a+b)-bc)\}-(-2c\left)\right\{b(a+b)-bc\}+(-2b\left)\right\{bc-\left(c\right(c+a\left)\right\}$

$=0+2c(ab+b^2-bc)-2b(bc-c^2-ac)$

$=2abc+2b^{2}c-2b{c}^2-2b^{2}c+2b{c}^2+2abc$

$=4abc$

$=$ RHS

$\therefore \left|\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ c& c& a+b\end{array}\right|=4abc$

Hence, proved.