Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain Solution R D Sharma Class 12 Chapter deteminants Exercise 5.2 Question 28 maths Textbook Solution.

Explain Solution R D Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 28 maths Textbook Solution.

Answers (1)

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

Hint First we will try to make some elements of R_{1} \text { or } C_{1} \quad(a+b)^{2}=a^{2}+2 a b+b^{2}

Because from R.H.S \begin{aligned} &\left(a^{3}+b^{3}\right)^{2}=\left[(a+b)\left(a^{2}-a b+b^{2}\right)\right]^{2} \\ &=(a+b)^{2}\left(a^{2}-a b+b^{2}\right)^{2} \end{aligned}

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

Solution:

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

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

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

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+b)^{2}\left[0\left|\begin{array}{cc} a^{2}-2 a b & 2 a b \\ b^{2}-a^{2} & a^{2} \end{array}\right|-0\left|\begin{array}{cc} b^{2}-a^{2} & 2 a b \\ 2 a b-b^{2} & a^{2} \end{array}\right|+1\left|\begin{array}{cc} b^{2}-a^{2} & a^{2}-2 a b \\ 2 a b-b^{2} & b^{2}-a^{2} \end{array}\right|\right]\\ &=(a+b)^{2}\left[0-0+1\left\{\left(b^{2}-a^{2}\right)\left(b^{2}-a^{2}\right)\right\}-\left(a^{2}-2 a b\right)\left(2 a b-b^{2}\right)\right]\\ &=(a+b)^{2}\left(b^{4}+a^{4}-2 a^{2} b^{2}-2 a^{3} b+a^{2} b^{2}+4 a^{2} b^{2}-2 a b^{3}\right) \end{aligned}

\begin{aligned} &=(a+b)^{2}\left[\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}+2 a^{2} b^{2}-2 a^{3} b-2 a b^{3}+a^{2} b^{2}\right] \\ &=(a+b)^{2}\left[\left(a^{2}\right)^{2}+(a b)^{2}+\left(b^{2}\right)^{2}-2 a^{3} b-2 a b^{3}+2 a^{2} b^{2}\right] \\ &=(a+b)^{2}\left(a^{2}-a b+b^{2}\right)^{2} \\ &=\left(a^{3}+b^{3}\right)^{2} \\ &=R \cdot H . S \end{aligned}

Hence it is proved that

\left|\begin{array}{ccc} a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right|=\left(a^{3}+b^{3}\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

'CBSE failed us, again!': Board faces backlash as Class 12 re-evaluation portal remains inaccessible
anam-khan

CBSE Class 12 re-evaluation portal faces technical glitch, services to resume by 2 pm
anam-khan

CBSE extends deadline to download Class 12 answer sheets till May 23; re-evaluation from May 26

Latest Question