Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve R. D. Sharma class 12 Chapter determinants Exercise 5.2 Question 24 Maths textbook Solution.

Please Solve R. D. Sharma class 12 Chapter determinants Exercise  5.2  Question 24 Maths textbook Solution.

Answers (1)

Answer:\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}

Hint We will convert some elements of the determinant into zero

Given: \left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right| \\ &=\left|\begin{array}{ccc} a^{2} & b c & c(a+c) \\ a(a+b) & b^{2} & a c \\ a b & b(b+c) & c^{2} \end{array}\right| \end{aligned}

\begin{aligned} &\text { Taking out 'a' common from } \mathrm{C}_{1}, \text { and } \mathrm{b} \text { from } \mathrm{C}_{2} \text { and } \mathrm{c} \text { from } \mathrm{C}_{3}\\ &=a b c\left|\begin{array}{ccc} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{array}\right| \end{aligned}

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

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

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

\text { On taking }(-a) \text { common from } \mathrm{C}_{2} \text { and }(-b) \text { from } \mathrm{C}_{3}

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

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=2 a^{2} b^{2} c\left[(a+c)\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right|-1\left|\begin{array}{ll} a+b & 1 \\ b+c & 1 \end{array}\right|+0\left|\begin{array}{ll} a+b & 1 \\ b+c & 0 \end{array}\right|\right]\\ &=2 a^{2} b^{2} c[(a+c)(1-0)-1(a+b-b-c)+0]\\ &=2 a^{2} b^{2} c[a+c-1(a-c)]\\ &=2 a^{2} b^{2} c(a+c-a+c)\\ &=2 a^{2} b^{2} c \times 2 c\\ &=4 a^{2} b^{2} c^{2}\\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}

 

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question