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 Exam Dates 2026 LIVE: CBSE Class 10th, 12th revised date sheet at cbse.gov.in; how to download, updates
anam-khan

CBSE board exam 2026 scheduled on March 3 postponed for Class 10, 12; revised date sheet out

Latest Question