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

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}