Question

If$\begin{vmatrix}a^2 & b^2 & c^2 \\(a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\(a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2 \end{vmatrix}=k \lambda\begin{vmatrix}a^2 & b^2 & c^2 \\a & b & c \\1 & 1 & 1\end{vmatrix}$
$\lambda \neq 0$, then $k$ is equal to:

• A. $4 \lambda a b c$
• B. $-4 \lambda a b c$
• C. $4 \lambda^2$
• D. $-4 \lambda^2$

Answer 1

Answer: (C)

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