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  • Explain Solution R.D. Sharma Class 12 Chapter deteminants Exercise 5.2 Question 13 maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 2 sub question 13 maths Textbook Solution.

Answers (1)

Answer:\left|\begin{array}{lll} a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{array}\right|=2\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|

Hint: We will make R_{1}  of L.H.S according to R.H.S

Given:\left|\begin{array}{lll} a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{array}\right|=2\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|

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

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

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

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

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

Hence it is proved that

 

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

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