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 30 Maths Textbook Solution.

Need Solution for R. D. Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 30 Maths Textbook Solution.

Answers (1)

Answer: \left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{2}

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

Given:\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{2}

Solution:

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

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

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

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

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

Hence it is proved that

\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question