જો left| begin{array}{ccc} a^2 & b^2 & c^2 ( | vedclass

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(C) ધારો કે $\Delta = \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ (a + \lambda)^2 & (b + \lambda)^2 & (c + \lambda)^2 \ (a - \lambda)^2 & (b - \lamb

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$4\lambda^2$

Vedclass · Answer

(C) ધારો કે $\Delta = \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ (a + \lambda)^2 & (b + \lambda)^2 & (c + \lambda)^2 \ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{array} ight|$.$R_2 	o R_2 - R_3$ પ્રક્રિયા કરતા:$\Delta = \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ (a + \lambda)^2 - (a - \lambda)^2 & (b + \lambda)^2 - (b - \lambda)^2 & (c + \lambda)^2 - (c - \lambda)^2 \ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{array} ight|$.$(x + y)^2 - (x - y)^2 = 4xy$ સૂત્રનો ઉપયોગ કરતા:$\Delta = \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ 4a\lambda & 4b\lambda & 4c\lambda \ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{array} ight|$.$R_2$ માંથી $4\lambda$ સામાન્ય લેતા:$\Delta = 4\lambda \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ a & b & c \ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{array} ight|$.$R_3 	o R_3 - R_1 + 2\lambda R_2$ પ્રક્રિયા કરતા:$(a - \lambda)^2 = a^2 + \lambda^2 - 2a\lambda$ હોવાથી,$R_3 - R_1 + 2\lambda R_2 = \lambda^2$ મળે છે:$\Delta = 4\lambda \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ a & b & c \ \lambda^2 & \lambda^2 & \lambda^2 \end{array} ight|$.$R_3$ માંથી $\lambda^2$ સામાન્ય લેતા:$\Delta = 4\lambda^3 \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ a & b & c \ 1 & 1 & 1 \end{array} ight|$.આપેલ સમીકરણ $k\lambda \left| \begin{array}{ccc} a^2 & b^2 & c^2 \ a & b & c \ 1 & 1 & 1 \end{array} ight|$ સાથે સરખાવતા,$k\lambda = 4\lambda^3$ મળે,તેથી $k = 4\lambda^2$.

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