#### Need Solution for R.D.Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 27  Maths Textbook Solution.

Answer:$\left|\begin{array}{ccc} a & b-c & c-b \\ a-c & b & c-a \\ a-b & b-a & c \end{array}\right|=(a+b-c)(b+c-a)(c+a-b)$

Hint First we will make elements of $C_{1}(a+b-c) \text { or } 0$

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

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a & b-c & c-b \\ a-c & b & c-a \\ a-b & b-a & c \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}+C_{3}\\ &=\left|\begin{array}{ccc} a+b-c & 0 & c-b \\ a+b-c & b+c-a & c-a \\ 0 & b+c-a & c \end{array}\right| \end{aligned}

$\text { On taking }(a+b-c) \text { common from } \mathrm{C}_{1} \text { and }(b+c-a) \text { common from } \mathrm{C}_{2}$

$=(a+b-c)(b+c-a)\left|\begin{array}{ccc} 1 & 0 & c-b \\ 1 & 1 & c-a \\ 0 & 1 & c \end{array}\right|$

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+b-c)(b+c-a)\left[1\left|\begin{array}{cc} 1 & c-a \\ 1 & c \end{array}\right|-0\left|\begin{array}{cc} 1 & c-a \\ 0 & c \end{array}\right|+(c-b)\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right|\right]\\ &=(a+b-c)(b+c-a)[1\{c-(c-a)\}-0+(c-b)(1-0)]\\ &=(a+b-c)(b+c-a)(c-c+a+c-b)\\ &=(a+b-c)(b+c-a)(c+a-b)\\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

$\left|\begin{array}{ccc} a & b-c & c-b \\ a-c & b & c-a \\ a-b & b-a & c \end{array}\right|=(a+b-c)(b+c-a)(c+a-b)$