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  • Please Solve R. D. Sharma class 12 Chapter determinants Exercise 5.2 Question 15 Maths textbook Solution.

Please Solve R. D. Sharma class 12 Chapter determinants Exercise 5.2 Question 2 sub question 15 Maths textbook Solution.

Answers (1)

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

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

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

Solution:

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

=\left|\begin{array}{ccc} a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|

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

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

\begin{aligned} &=(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 2 b-b+c+a & -b-c-a & 2 b \\ 2 c-2 c & 2 c-c+a+b & c-a-b \end{array}\right| \\ &=(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ a+b+c & -(a+b+c) & 2 b \\ 0 & a+b+c & c-a-b \end{array}\right| \end{aligned}

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

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

Hence it is proved that

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

 

 

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