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  • Need Solution for R.D.Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 37 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 37  Maths Textbook Solution.

Answers (1)

Answer:(5 x+\lambda)(\lambda-x)^{2}

Hint: Use determinant formula

Given: \left|\begin{array}{ccc} x+\lambda & 2 x & 2 x \\ 2 x & x+\lambda & 2 x \\ 2 x & 2 x & x+\lambda \end{array}\right|=(5 x+\lambda)(\lambda-x)^{2}

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} x+\lambda & 2 x & 2 x \\ 2 x & x+\lambda & 2 x \\ 2 x & 2 x & x+\lambda \end{array}\right| \\ &\text { Use } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} 5 x+\lambda & 2 x & 2 x \\ 5 x+\lambda & x+\lambda & 2 x \\ 5 x+\lambda & 2 x & x+\lambda \end{array}\right| \end{aligned}

\begin{aligned} &(5 x+\lambda) \text { common from } \mathrm{C}_{1}\\ &=(5 x+\lambda)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 1 & x+\lambda & 2 x \\ 1 & 2 x & x+\lambda \end{array}\right|\\ &\text { Now } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-R_{1}\\ &=(5 x+\lambda)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 0 & x-\lambda & 0 \\ 0 & 0 & x-\lambda \end{array}\right|\\ &=(5 x+\lambda)(-1)(-1)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 0 & -x+\lambda & 0 \\ 0 & 0 & -x+\lambda \end{array}\right| \end{aligned}

\text { Expanding w.r.t } \mathrm{R}_{1}

\begin{aligned} &=(5 x+\lambda)\left[1(\lambda-x)^{2}-0+0\right] \\ &=(5 x+\lambda)(\lambda-x)^{2} \\ &=R \cdot H . S \end{aligned}

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