#### Need Solution for R.D.  Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 2 sub question 9 Maths Textbook Solution.

Answer:$\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|=a^{2}(a+x+y+z)$

Hint:We will try to make some elements of the determinant into zero

Given :$\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|$

Solution:$\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking common }(a+x+y+z) \text { from } \mathrm{C}_{1}\\ &=(a+x+y+z)\left|\begin{array}{ccc} 1 & y & z \\ 1 & a+y & z \\ 1 & y & a+z \end{array}\right| \end{aligned}

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \\ &=(a+x+y+z)\left|\begin{array}{ccc} 0 & y-(a+y) & z-z \\ 0 & a+y-y & z-(a+z) \\ 1 & y & a+z \end{array}\right| \\ &=(a+x+y+z)\left|\begin{array}{ccc} 0 & y-a-y & 0 \\ 0 & a & z-a-z \\ 1 & y & a+z \end{array}\right| \\ &=(a+x+y+z)\left|\begin{array}{ccc} 0 & -a & 0 \\ 0 & a & -a \\ 1 & y & a+z \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+x+y+z)\left[0\left|\begin{array}{cc} a & -a \\ y & a+z \end{array}\right|-(-a)\left|\begin{array}{cc} 0 & -a \\ 1 & a+z \end{array}\right|+0\left|\begin{array}{ll} 0 & a \\ 1 & y \end{array}\right|\right]\\ &=(a+x+y+z)[0+a\{0 \times(a+z)-(-a) 1\}+0]\\ &=(a+x+y+z)\{a(0+a)\}\\ &=(a+x+y+z) a^{2}\\ &=a^{2}(a+x+y+z) \end{aligned}

Hence $\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|=a^{2}(a+x+y+z)$