#### Please Solve R.D. Sharma class 12 Chapter determinants Exercise  5.2 Question 32 Maths textbook Solution.

Answer: $4\: abc$

Hint: Use determinant formula

Given: $\left|\begin{array}{ccc} b+c & a & a \\ b & a+c & b \\ c & c & a+b \end{array}\right|=4 a b c$

Solution:

$\text { L.H.S }\left|\begin{array}{ccc} b+c & a & a \\ b & a+c & b \\ c & c & a+b \end{array}\right|$

\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &|\Delta|=\left|\begin{array}{ccc} 2(b+c) & 2(c+a) & 2(a+b) \\ b & a+c & b \\ c & c & a+b \end{array}\right|\\ &\text { Now } R_{2} \rightarrow R_{2}-R_{1} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-R_{1} \text { and } 2 \text { common }\\ &|\Delta|=2\left|\begin{array}{ccc} b+c & c+a & a+b \\ -c & 0 & -a \\ -b & -a & 0 \end{array}\right| \end{aligned}

\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &|\Delta|=2\left|\begin{array}{ccc} 0 & c & b \\ -c & 0 & -a \\ -b & -a & 0 \end{array}\right|\\ &\text { Expanding w.r.t } \mathrm{C}_{1}\\ &|\Delta|=2[-c(0-a b)+b(a c-0)]\\ &|\Delta|=2(a b c+a b c)\\ &|\Delta|=2(2 a b c)=4 a b c\\ &=R \cdot H . S \end{aligned}