Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Using the property of determinants and without expanding, prove that determinant 1 bc a(b+c) 1 ca b(c+a) 1 ab c(a+b) = 0 Ex 4.2 Q : 2

Q : 4           Using the property of determinants and without expanding, prove that 

                    \begin{vmatrix}1 &bc &a(b+c) \\1 &ca &b(c+a) \\1 &ab & c(a+b) \end{vmatrix}=0

Answers (1)

best_answer

We have determinant: 

\triangle = \begin{vmatrix} 1 &bc &a(b+c) \\ 1& ca &b(c+a) \\ 1& ab &c(a+b) \end{vmatrix} 

Applying C_{3} \rightarrow C_{2} + C_{3} we have then;

\triangle = \begin{vmatrix} 1 &bc & ab+bc+ca \\ 1& ca &ab+bc+ca \\ 1& ab &ab+bc+ca \end{vmatrix}

So, here column 3 and column 1 are proportional.

Therefore, \triangle = 0.

Posted by

Divya Prakash Singh

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 Central Sector Scheme of Scholarship 2025 applications open on scholarships.gov.in
anam-khan

CBSE Board 2025: Class 10th, 12th supplementary date sheet out; exams from July 15
anam-khan

CBSE issues final reminder to schools for supplementary exam LOC submission; submit by today without late fee

Latest Question