#### Provide Solution For  R. D. Sharma Maths Class 12 Chapter  determinants Exercise 5.2 Question 25 Maths Textbook Solution.

Answer:$\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)$

Hint First we will try to convert some elements into zero

Given:$\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)$

Solution:

$\text { L.H.S }\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &=\left|\begin{array}{ccc} 3 x+4 & 3 x+4 & 3 x+4 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|\\ &\text { On taking } 3 x+4 \text { common from } \mathrm{R}_{1}\\ &=(3 x+4)\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}, \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-C_{3} \\ &=(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ x-x-4 & x+4-x & x \\ x-x & x-x-4 & x+4 \end{array}\right| \\ &=(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -4 & 4 & x \\ 0 & -4 & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking }(-4) \text { common from } \mathrm{C}_{1} \text { and }(-4) \text { from } \mathrm{C}_{2}\\ &=(3 x+4)(-4)(-4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & -1 & x \\ 0 & 1 & x+4 \end{array}\right|\\ &=16(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & -1 & x \\ 0 & 1 & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=16(3 x+4)\left[0\left|\begin{array}{cc} -1 & x \\ 1 & x+4 \end{array}\right|-0\left|\begin{array}{cc} 1 & x \\ 0 & x+4 \end{array}\right|+1\left|\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right|\right]\\ &=16(3 x+4)[0-0+1(1-0)]\\ &=16(3 x+4) \times 1\\ &=16(3 x+4)\\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

$\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)$