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(C) 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)^

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$4\lambda^2$

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(C) Let $\Delta = \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ (a + \lambda)^2 & (b + \lambda)^2 & (c + \lambda)^2 \ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{array} ight|$.Apply $R_2 	o R_2 - R_3$:$\Delta = \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ (a + \lambda)^2 - (a - \lambda)^2 & (b + \lambda)^2 - (b - \lambda)^2 & (c + \lambda)^2 - (c - \lambda)^2 \ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{array} ight|$.Using $(x + y)^2 - (x - y)^2 = 4xy$,we get:$\Delta = \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ 4a\lambda & 4b\lambda & 4c\lambda \ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{array} ight|$.Taking $4\lambda$ common from $R_2$:$\Delta = 4\lambda \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ a & b & c \ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{array} ight|$.Apply $R_3 	o R_3 - R_1 + 2R_2$ (where $R_2$ is the row $a, b, c$):Since $(a - \lambda)^2 = a^2 + \lambda^2 - 2a\lambda$,then $R_3 - R_1 + 2\lambda R_2 = \lambda^2$:$\Delta = 4\lambda \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ a & b & c \ \lambda^2 & \lambda^2 & \lambda^2 \end{array} ight|$.Taking $\lambda^2$ common from $R_3$:$\Delta = 4\lambda^3 \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ a & b & c \ 1 & 1 & 1 \end{array} ight|$.Comparing with $k\lambda \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ a & b & c \ 1 & 1 & 1 \end{array} ight|$,we get $k\lambda = 4\lambda^3$,so $k = 4\lambda^2$.

Similar Questions

Using properties of determinants,prove that:
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