If $\begin{aligned} &\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ (a+\lambda)^2 & (b+\lambda)^2 &…

If $\begin{aligned} &\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ (a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda^2) \\ (a-\lambda)^2 & (b-\lambda^2) & (-\lambda^2) \end{array}\right| \\ &=k \lambda\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{array}\right|, \lambda \neq 0 \end{aligned}$ then $k$ is equal to:
  1. $4 \lambda \mathrm{abc}$
  2. $-4 \lambda \mathrm{abc}$
  3. $4 \lambda^2$
  4. $-4 \lambda^2$

Solution

Let $\Delta=\left|\begin{array}{ccc}a^2 & b^2 & c^2 \\ (a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2\end{array}\right|$ $ \begin{aligned} &\text { Apply } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3 \\ &\Delta=\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ (a+\lambda)^2-(a-\lambda)^2 & (b+\lambda)^2-(b-\lambda)^2 & (c+\lambda)^2-(c-\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2 \end{array}\right| \\ &=\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ 4 a \lambda & 4 b \lambda & 4 c \lambda \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2 \end{array}\right| \\ &\left(\because(x+y)^2-(x-y)^2=4 x y\right) \\ & \end{aligned} $ Taking out 4 common from $\mathrm{R}_2$ $=4\left|\begin{array}{ccc}a^2 & b^2 & c^2 \\ a \lambda & b \lambda & c \lambda \\ a^2+\lambda^2-2 a \lambda & b^2+\lambda^2-2 b \lambda & c^2+\lambda^2-2 c \lambda\end{array}\right|$ $ \begin{aligned} &\text { Apply } \mathrm{R}_3 \rightarrow\left[\mathrm{R}_3-\left(\mathrm{R}_1-2 \mathrm{R}_2\right)\right] \\ &=4\left|\begin{array}{lll} a^2 & b^2 & c^2 \\ a \lambda & b \lambda & c \lambda \\ \lambda^2 & \lambda^2 & \lambda^2 \end{array}\right| \end{aligned} $ Taking out $\lambda$ common from $\mathrm{R}_2$ and $\lambda^2$ from $\mathrm{R}_3$. $ =4 \lambda\left(\lambda^2\right)\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{array}\right| $ $ \begin{aligned} &=k \lambda\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{array}\right| \\ &\Rightarrow k=4 \lambda^2 \end{aligned} $

Asked in: JEE Main 2014 (12 Apr Online)

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