Question

Using properties of determinants it can be proved
$\left|\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ c& c& a+b\end{array}\right|=4abc$

• A. True
• B. False

Answer 1

The correct option is A True

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

$=\left|\begin{array}{ccc}0& -2c& -2b\\ b& c+a& b\\ c& c& a+b\end{array}\right|[R_1^{{}^{\prime }}=R_1-(R_2+R_3\left)\right]$

expanding w.r.t. row 1, we get,

$=0+2c\left(b\right(a+b)-bc))-2b(bc-c(c+a))=2abc+2b^{2}c-2b{c}^2-2b^{2}c+2b{c}^2+2abc=4abc$