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  • Explain Solution R.D.Sharma Class 12 Chapter deteminants Exercise 5.2 Question 38 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 38 maths Textbook Solution.

Answers (1)

Answer:(5 x+4)(4-x)^{2}

Hint Use determinant formula

Given:\left|\begin{array}{ccc} x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right|=(5 x+4)(4-x)^{2}

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right| \\ &\text { Use C }_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} 5 x+4 & 2 x & 2 x \\ 5 x+4 & x+4 & 2 x \\ 5 x+4 & 2 x & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &(5 x+4) \text { common from } \mathrm{C}_{1} \\ &=(5 x+4)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 1 & x+4 & 2 x \\ 1 & 2 x & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &\text { Now } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-R_{1} \\ &=(5 x+4)\left|\begin{array}{ccc} 0 & x-4 & 0 \\ 0 & 4+x-2 x & 2 x-x-4 \\ 1 & 2 x & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &(x-4) \text { from } \mathrm{R}_{1} \text { common }\\ &(x-4)(5 x+4)\left|\begin{array}{ccc} 0 & x-4 & 0 \\ 0 & 4+x-2 x & 2 x-x-4 \\ 1 & 2 x & x+4 \end{array}\right| \end{aligned}

Expanding w.r.t C_{1}

\begin{aligned} &=(x-4)(5 x+4)[(x-4)(x-4)-0] \\ &=(5 x+4)(x-4)^{2} \\ &=R . H . S \end{aligned}

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