Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need Solution for R.D.Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 29 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 29 Maths Textbook Solution.

Answers (1)

Answer:\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|=1+a^{2}+b^{2}+c^{2}

Hint We will try to convert some elements of determinant into zero

Given:\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|=1+a^{2}+b^{2}+c^{2}

Solution:

\text { L.H.S }\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|

\begin{aligned} &=\left|\begin{array}{ccc} 1+a^{2} & a b & a c \\ a b & 1+b^{2} & b c \\ c a & c b & 1+c^{2} \end{array}\right|\\ &\text { Multiply } C_{1} \text { by } a, C_{2} \text { by } b \text { and } C_{3} \text { by } c\\ &=\frac{1}{a b c}\left|\begin{array}{ccc} a\left(1+a^{2}\right) & a b^{2} & a c^{2} \\ a^{2} b & b\left(1+b^{2}\right) & b c^{2} \\ c a^{2} & c b^{2} & c\left(1+c^{2}\right) \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking } a \text { common from } \mathrm{R}_{1}, \mathrm{~b} \text { from } \mathrm{R}_{2} \text { and } \mathrm{c} \text { from } \mathrm{R}_{3}\\ &=\frac{a b c}{a b c}\left|\begin{array}{ccc} 1+a^{2} & b^{2} & c^{2} \\ a^{2} & 1+b^{2} & c^{2} \\ a^{2} & b^{2} & 1+c^{2} \end{array}\right|\\ &=\left|\begin{array}{ccc} 1+a^{2} & b^{2} & c^{2} \\ a^{2} & 1+b^{2} & c^{2} \\ a^{2} & b^{2} & 1+c^{2} \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3}\\ &=\left|\begin{array}{ccc} 1+a^{2}+b^{2}+c^{2} & b^{2} & c^{2} \\ 1+a^{2}+b^{2}+c^{2} & 1+b^{2} & c^{2} \\ 1+a^{2}+b^{2}+c^{2} & b^{2} & 1+c^{2} \end{array}\right| \end{aligned}

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

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

Hence it is proved that

\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|=1+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 issues final reminder to schools for supplementary exam LOC submission; submit by today without late fee
anam-khan

CBSE Supplementary Exam 2025: Registrations open for Class 10, 12 private students today
anam-khan

CBSE opens application to access 10th answer scripts; verification, revaluation for Class 12 starts on May 31

Latest Question