Question

If $\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ (a+\lambda {)}^2& (b+\lambda {)}^2& (c+\lambda {)}^2\\ (a-\lambda {)}^2& (b-\lambda {)}^2& (c-\lambda {)}^2\end{array}\right|=k\lambda \left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|,\lambda \ne 0$
then k is equal to :

• A. $4\lambda abc$
• B. $-4\lambda abc$
• C. $4{\lambda }^2$
• D. $-4{\lambda }^2$

Answer 1

Answer: (C)

$4{\lambda }^2$

$\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ (a+\lambda {)}^2& (b+\lambda {)}^2& (c+\lambda {)}^2\\ (a-\lambda {)}^2& (b-\lambda {)}^2& (c-\lambda {)}^2\end{array}\right|$ .
$R_2\to R_2-R_3$
$=4\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a\lambda & b\lambda & c\lambda \\ {(a-\lambda )}^2& {(b-\lambda )}^2& {(c-\lambda )}^2\end{array}\right|$
$R_3\to R_3{+2R}_2-R_1$

$=4{\lambda }^3\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|$
$4{\lambda }^3\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|=k\lambda \left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|$

$\Rightarrow (4{\lambda }^3-k\lambda )\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|=0$

Since $\lambda \ne 0$

$\Rightarrow k=4{\lambda }^2$