Question

If $2s=a+b+c$ then $\left|\begin{array}{ccc}{a}^2& (s-a{)}^2& (s-a{)}^2\\ (s-b{)}^2& b^2& (s-b{)}^2\\ (s-c{)}^2& (s-c{)}^2& c^2\end{array}\right|$ is equal to

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

Answer 1

The correct option is B $2s^3(s-a)(s-b)(s-c)$

$\left|\begin{array}{ccc}{a}^2& (s-a{)}^2& (s-a{)}^2\\ (s-b{)}^2& b^2& (s-b{)}^2\\ (s-c{)}^2& (s-c{)}^2& c^2\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$=\left|\begin{array}{ccc}{a}^2-(s-a{)}^2& 0& (s-a{)}^2\\ (s-b{)}^2-b^2& b^2-(s-b{)}^2& (s-b{)}^2\\ 0& (s-c{)}^2-c^2& c^2\end{array}\right|$
$=\left|\begin{array}{ccc}s(2a-s)& 0& (s-a{)}^2\\ s(s-2b)& s(2b-s)& (s-b{)}^2\\ 0& s(s-2c)& c^2\end{array}\right|$
$=s^2\left|\begin{array}{ccc}(2a-s)& 0& (s-a{)}^2\\ (s-2b)& (2b-s)& (s-b{)}^2\\ 0& (s-2c)& c^2\end{array}\right|$

On expanding the above determinant we get,
$=2s^3(s-a)(s-b)(s-c)$

Hence, option 'B' is correct.