Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution For R. D. Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 25 Maths Textbook Solution.

Provide Solution For  R. D. Sharma Maths Class 12 Chapter  determinants Exercise 5.2 Question 25 Maths Textbook Solution.

Answers (1)

Answer:\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)

Hint First we will try to convert some elements into zero

Given:\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)

Solution:

\text { L.H.S }\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &=\left|\begin{array}{ccc} 3 x+4 & 3 x+4 & 3 x+4 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|\\ &\text { On taking } 3 x+4 \text { common from } \mathrm{R}_{1}\\ &=(3 x+4)\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}, \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-C_{3} \\ &=(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ x-x-4 & x+4-x & x \\ x-x & x-x-4 & x+4 \end{array}\right| \\ &=(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -4 & 4 & x \\ 0 & -4 & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking }(-4) \text { common from } \mathrm{C}_{1} \text { and }(-4) \text { from } \mathrm{C}_{2}\\ &=(3 x+4)(-4)(-4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & -1 & x \\ 0 & 1 & x+4 \end{array}\right|\\ &=16(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & -1 & x \\ 0 & 1 & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=16(3 x+4)\left[0\left|\begin{array}{cc} -1 & x \\ 1 & x+4 \end{array}\right|-0\left|\begin{array}{cc} 1 & x \\ 0 & x+4 \end{array}\right|+1\left|\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right|\right]\\ &=16(3 x+4)[0-0+1(1-0)]\\ &=16(3 x+4) \times 1\\ &=16(3 x+4)\\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)

 

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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question