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  • Need Solution for R.D.Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 36 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 36 Maths Textbook Solution.

Answers (1)

Answer:(a b+b c+c a)^{3}

Hint Use determinant formula

Given:\left|\begin{array}{ccc} -b c & b^{2}+b c & c^{2}+a b \\ a^{2}+a c & -a c & c^{2}+a c \\ a^{2}+a b & b^{2}+a b & -a b \end{array}\right|=(a b+b c+c a)^{3}

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} -b c & b^{2}+b c & c^{2}+a b \\ a^{2}+a c & -a c & c^{2}+a c \\ a^{2}+a b & b^{2}+a b & -a b \end{array}\right| \\ &R_{1} \rightarrow a R_{1}, R_{2} \rightarrow b R_{2}, R_{3} \rightarrow c R_{3} \\ &=\frac{1}{a b c}\left|\begin{array}{ccc} -a b c & a b^{2}+a b c & a c^{2}+a b c \\ a^{2}+a b c & -a b c & c^{2}+a b c \\ a^{2}+a b c & b^{2} c+a b c & -a b c \end{array}\right| \end{aligned}

\begin{aligned} &\text { Common a from } \mathrm{C}_{1}, b \text { from } \mathrm{C}_{2} \text { and } \mathrm{c} \text { from } \mathrm{C}_{3}\\ &=\frac{a b c}{a b c}\left|\begin{array}{ccc} b c & a b+a c & a c+a b \\ a b+b c & -a c & b c+a b \\ a c+b c & b c+a c & -a b \end{array}\right|\\ &\text { Use } R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &=\left|\begin{array}{ccc} a b+b c+a c & a b+b c+a c & a b+b c+a c \\ a b+b c & -a c & b c+a b \\ a c+b c & b c+a c & -a b \end{array}\right| \end{aligned}

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

\begin{aligned} &(a b+b c+a c) \text { common from } \mathrm{C}_{2} \& C_{3}\\ &=(a b+b c+a c)^{3}\left|\begin{array}{ccc} 1 & 0 & 0 \\ a b+b c & -1 & 1 \\ a c+b c & 0 & -1 \end{array}\right|\\ &\text { Expanding w.r.t } \mathrm{R}_{1}\\ &=(a b+b c+a c)^{3}\{1(1-0)+0\}\\ &=(a b+b c+a c)^{3}\\ &=R \cdot H \cdot S \end{aligned}

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