Question

The value of $\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$ will be

• A. $(a+b+c{)}^2$
• B. $(a+b+c{)}^3$
• C. $(a-b-c{)}^2$
• D. $(a+b-c{)}^2$

Answer 1

The correct option is B $(a+b+c{)}^3$

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

$R_1\to R_1+(R_2+R_3)$

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

Taking common term (a + b + c) from the first row

$=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$

Subtracting first column from second and third column

$=(a+b+c)\left|\begin{array}{ccc}1& 0& 0\\ 2b& -b-c-a& 0\\ 2c& 0& -c-a-b\end{array}\right|$

On expansion in terms of

$R_1=(a+b+c)\left|\begin{array}{cc}-b-c-a& 0\\ 0& -c-a-b\end{array}\right|$

.
$=(a+b+c)\left\{\right(-b-c-a\left)\right(-c-a-b)-0\}$
$=(a+b+c{)}^3$