Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
46 views
in Mathematics by (60.1k points)
closed by

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

1 Answer

+1 vote
by (59.3k points)
selected by
 
Best answer

\(\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right| C_{1} \rightarrow C_{1}+C_{2}+C_{3}\)

\(=\left|\begin{array}{ccc} 1+x+x^{2} & x & x^{2} \\ 1+x+x^{2} & 1 & x \\ 1+x+x^{2} & x^{2} & 1 \end{array}\right|=\left(1+x+x^{2}\right)\left|\begin{array}{ccc} 1 & x & x^{2} \\ 1 & 1 & x \\ 1 & x^{2} & 1 \end{array}\right| \)

\(R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3} \)

\(=\left(1+x+x^{2}\right)\left|\begin{array}{ccc} 0 & x-1 & x^{2}-x \\ 0 & 1-x^{2} & x-1 \\ 1 & x^{2} & 1 \end{array}\right|\)

\(=\left(1+x+x^{2}\right)\left|\begin{array}{ccc} 0 & -(1-x) & -x(1-x) \\ 0 & (1+x)(1-x) & -(1-x) \\ 1 & {x}^{2} & 1 \end{array}\right|\)

\(=(1-x)^2\left(1+x+x^2\right)\left|\begin{array}{ccc} 0 & -1 & -x \\ 0 & 1+x & -1 \\ 1 & x^2 & 1 \end{array}\right| \)

\(=(1-x)^2\left(1+x+x^2\right) \times 1\left|\begin{array}{cc} -1 & -x \\ 1+x & -1 \end{array}\right| \)

\( =(1-x)^2\left(1+x+x^2\right)(1+x(1+x))\)

\(=(1-x)^2\left(1+x+x^2\right)\left(1+x+x^2\right)\)

\(=(1-x)^2\left(1+x+x^2\right)^2=\left\{\left(1-x+x^2\right)\left(1+x+x^2\right)\right\}^2=\left(1-x^3\right)^2=\text { R.H.S. } \)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...