Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • By using properties of determinants, show that determinant 1 1 1 a b c a^3 b^3 c^3 = (a-b)(b-c)(c-a)(a+b+c) Ex 4.2 Q : 8 (ii)

Q : 8        By using properties of determinants, show that:

              (ii) \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c)

Answers (1)

best_answer

Given determinant :

                                          \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix},

Applying column transformation C_{1} \rightarrow C_{1}-C_{3} and then C_{2} \rightarrow C_{2}-C_{3}

We get,

\triangle =\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ a^3-c^3 &b^3-c^3 & c^3 \end{vmatrix}

=\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ (a-c)(a^2+ac+c^2) &(b-c)(b^2+bc+c^2) & c^3 \end{vmatrix}

=(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 1& 1 & c \\ (a^2+ac+c^2) &(b^2+bc+c^2) & c^3 \end{vmatrix}

Now, applying column transformation C_{1} \rightarrow C_{1} - C_{2}, we have:

=(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a^2-b^2+ac-bc) &(b^2+bc+c^2) & c^3 \end{vmatrix}

=(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a-b)(a+b+c) &(b^2+bc+c^2) & c^3 \end{vmatrix}

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

=-(a-c)(b-c)(a-b)(a+b+c) = (a-b)(b-c)(c-a)(a+b+c)

Hence proved.

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

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