Question

$\left|\begin{array}{ccc}{(b+c)}^2& a^2& a^2\\ b^2& {(c+a)}^2& b^2\\ c^2& c^2& {(a+b)}^2\end{array}\right|$ $=\lambda$ $abc{(a+b+c)}^3$ then$\lambda$ =......

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

$\left|\begin{array}{ccc}{(b+c)}^2& a^2& a^2\\ b^2& {(c+a)}^2& b^2\\ c^2& c^2& {(a+b)}^2\end{array}\right|=\lambda abc{(a+b+c)}^3$

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

$C_2\to C_2-C_1,C_3\to C_3-C_1$

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

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

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

$C_2\to C_2+\frac{1}{b}{C}_1$

$C_3\to C_3+\frac{1}{c}{C}_1$

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

Expanding

${(a+b+c)}^2\left[2bc\right(a^2+ab+ac\left)\right]$

$\Rightarrow 2abc(a+b+c)=$RHS

$\Rightarrow \lambda =2$

Option C is correct