Question

The value of deteminant $\left|\begin{array}{ccc}a+b& a& b\\ a& a+c& c\\ b& c& b+c\end{array}\right|$ is equal to

• A. $4abc$
• B. $abc$
• C. $a^2{b}^2{c}^2$
• D. $4a^{2}bc$

Answer 1

The correct option is A $4abc$

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

$=2ab\left|\begin{array}{ccc}a+b& a+c& b+c\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|$
$C_2\to C_2-C_1$
$=2ab\left|\begin{array}{ccc}a+b& c-b& b+c\\ 1& -1& 1\\ 1& 0& 0\end{array}\right|$

$\begin{array}{c}=2ab[c-b+b+c]\end{array}$
$=4abc$