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

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

Answers (1)

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)

 

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