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  • Explain Solution R.D. Sharma Class 12 Chapter deteminants Exercise 5.2 Question 14 maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 2 sub question 14 maths Textbook Solution.

Answers (1)

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

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

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

Solution:\text { L.H.S }\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right|

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} 2 a+2 b+2 c & a & b \\ 2 a+2 b+2 c & b+c+2 a & b \\ 2 a+2 b+2 c & a & c+a+2 b \end{array}\right| \\ &=\left|\begin{array}{ccc} 2(a+b+c) & a & b \\ 2(a+b+c) & b+c+2 a & b \\ 2(a+b+c) & a & c+a+2 b \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking common } 2(a+b+c) \text { from } \mathrm{C}_{1}\\ &=2(a+b+c)\left|\begin{array}{ccc} 1 & a & b \\ 1 & b+c+2 a & b \\ 1 & a & c+a+2 b \end{array}\right| \end{aligned}

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

\begin{aligned} &\text { On taking common }(a+b+c) \text { from } \mathrm{R}_{2}\\ &=2(a+b+c)(a+b+c)\left|\begin{array}{ccc} 0 & -(a+b+c) & 0 \\ 0 & 1 & -1 \\ 1 & a & c+a+2 b \end{array}\right|\\ &=2(a+b+c)^{2}\left|\begin{array}{ccc} 0 & -(a+b+c) & 0 \\ 0 & 1 & -1 \\ 1 & a & c+a+2 b \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=2(a+b+c)^{2}\left[0\left|\begin{array}{cc} 1 & -1 \\ a & c+a+2 b \end{array}\right|-\{-(a+b+c)\}\left|\begin{array}{cc} 0 & -1 \\ 1 & c+a+2 b \end{array}\right|+0 \mid \begin{array}{ll} 0 & 1 \\ 1 & a \end{array}\right]\\ &=2(a+b+c)^{2}[0+(a+b+c)\{0-(-1) 1\}+0]\\ &=2(a+b+c)^{2}[(a+b+c)(1)]\\ &=2(a+b+c)^{2}(a+b+c)\\ &=2(a+b+c)^{3}\\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

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

 

 

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