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)

Using $R_3\to R_3-R_2$ and $R_2\to R_2-R_1$, we get
$\Delta =\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ 2a\lambda +{\lambda }^2& 2b\lambda +{\lambda }^2& 2c\lambda +{\lambda }^2\\ -4a\lambda & -4b\lambda & -4c\lambda \end{array}\right|$
Take $-4\lambda$ common from $R_3$, and applying $R_2\to R_2-2\lambda R_3$, we get
$\Delta =-4{\lambda }^3\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ 1& 1& 1\\ a& b& c\end{array}\right|$
$=\lambda \left(4{\lambda }^2\right)\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|$
$\therefore k=4{\lambda }^2$